NCERT Solutions for Class 12 Maths Chapter 3 Matrices

NCERT Maths Class 12 Chapter 3 Solutions, "Metrices," is based on the concept of Matrix and their properties. The chapter consists of the following exercises:

Exercise 3.1: This exercise introduces types of matrices and covers basic operations like addition, subtraction, and multiplication.

Exercise 3.2: This exercise explores matrix properties like commutative, associative, and identity matrix use.

Exercise 3.3: This exercise focuses on matrix multiplication, its properties, and practicing inverse calculations.

Exercise 3.4: This exercise teaches and applies the concept of matrix transpose and its properties.

Miscellaneous Exercise: Mix of questions testing all matrix concepts and problem-solving skills.

Overall, this chapter is an important topic in linear algebra and covers the fundamental concepts of matrices, including matrix notation, matrix operations, matrix properties, and matrix multiplication.

Mastering Class 12 Maths Chapter 3: Matrices - MCQs, Question and Answers, and Tips for Success

Exercise 3.1

1. In the matrix \[\mathbf{A=\left[ \begin{matrix} 2 & 5 & 19 & -7  \\ 35 & -2 & \dfrac{5}{2} & 12  \\ \sqrt{3} & 1 & -5 & 17  \\ \end{matrix} \right]}\], write

i. The order of the matrix.

Ans: The order of a matrix is \[m\times n\] where \[m\] is the number of rows and \[n\] is the number of columns. Therefore, here the order is \[3\times 4\].

ii. The number of elements.

Ans: Since the order of the given matrix is \[3\times 4\] therefore, the number of elements in it is \[3\times 4=12\].

iii. Write the elements \[\mathbf{{{a}_{13}},{{a}_{21}},{{a}_{33}},{{a}_{24}},{{a}_{23}}}\]

Ans: The elements are given as \[{{a}_{mn}}\] . Therefore, here \[{{a}_{13}}=19\] , \[{{a}_{21}}=35\] , \[{{a}_{33}}=-5\] , \[{{a}_{24}}=12\] , \[{{a}_{23}}=\dfrac{5}{2}\].

2. If a matrix has \[24\] elements, what are the possible order it can have? What if it has \[13\] elements?

Ans: The order of a matrix is \[m\times n\] where \[m\] is the number of rows and \[n\] is the number of columns. To find the possible orders of a matrix, we have to find all the ordered pairs of natural numbers whose product is \[24\] .

\[\therefore \left( 1\times 24 \right),\left( 24\times 1 \right),\left( 2\times 12 \right),\left( 12\times 2 \right),\left( 3\times 8 \right),\left( 8\times 3 \right),\left( 4\times 6 \right),\left( 6\times 4 \right)\] are all the possible ordered pairs here.

If the matrix had \[13\] elements, then the ordered pairs would be \[\left( 1\times 13 \right)\] and \[\left( 13\times 1 \right)\].

3. If a matrix has \[18\] elements, what are the possible orders it can have? What if it has \[5\] elements?

Ans: The order of a matrix is \[m\times n\] where \[m\] is the number of rows and \[n\] is the number of columns. To find the possible orders of a matrix, we have to find all the ordered pairs of natural numbers whose product is \[18\] .

\[\therefore \left( 1\times 18 \right),\left( 18\times 1 \right),\left( 2\times 9 \right),\left( 9\times 2 \right),\left( 3\times 6 \right),\left( 6\times 3 \right)\] are all the possible ordered pairs here.

If the matrix had \[5\] elements, then the ordered pairs would be \[\left( 1\times 5 \right)\] and \[\left( 5\times 1 \right)\].

4. Construct a \[3\times 4\] matrix, whose elements are given by 

i. \[\mathbf{{{a}_{ij}}=\dfrac{1}{2}\left| -3i+j \right|}\]

Ans: Given that \[{{a}_{ij}}=\dfrac{1}{2}\left| -3i+j \right|\] ,

\[\therefore {{a}_{11}}=\dfrac{1}{2}\left| -3\times 1+1 \right|=1\]

\[{{a}_{21}}=\dfrac{1}{2}\left| -3\times 2+1 \right|=\dfrac{5}{2}\]

\[{{a}_{31}}=\dfrac{1}{2}\left| -3\times 3+1 \right|=4\]

\[{{a}_{12}}=\dfrac{1}{2}\left| -3\times 1+2 \right|=\dfrac{1}{2}\]

\[{{a}_{22}}=\dfrac{1}{2}\left| -3\times 2+2 \right|=2\]

\[{{a}_{32}}=\dfrac{1}{2}\left| -3\times 3+2 \right|=\dfrac{7}{2}\]

\[{{a}_{13}}=\dfrac{1}{2}\left| -3\times 1+3 \right|=0\]

\[{{a}_{23}}=\dfrac{1}{2}\left| -3\times 2+3 \right|=\dfrac{3}{2}\]

\[{{a}_{33}}=\dfrac{1}{2}\left| -3\times 3+3 \right|=3\]

\[{{a}_{14}}=\dfrac{1}{2}\left| -3\times 1+4 \right|=\dfrac{1}{2}\]

\[{{a}_{24}}=\dfrac{1}{2}\left| -3\times 2+4 \right|=1\]

\[{{a}_{34}}=\dfrac{1}{2}\left| -3\times 3+4 \right|=\dfrac{5}{2}\]

Thus, the required matrix is \[A=\left[ \begin{matrix} 1 & \dfrac{1}{2} & 0 & \dfrac{1}{2}  \\ \dfrac{5}{2} & 2 & \dfrac{3}{2} & 1  \\ 4 & \dfrac{7}{2} & 3 & \dfrac{5}{2}  \\ \end{matrix} \right]\].

ii. \[\mathbf{{{a}_{ij}}=2i-j}\]

Ans: A \[3\times 4\] matrix is given by \[A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} & {{a}_{14}}  \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} & {{a}_{24}}  \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} & {{a}_{34}}  \\ \end{matrix} \right]\]

Given that \[{{a}_{ij}}=2i-j\] ,

\[\therefore {{a}_{11}}=2\times 1-1=1\]

\[{{a}_{21}}=2\times 2-1=3\]

\[{{a}_{31}}=2\times 3-1=5\]

\[{{a}_{12}}=2\times 1-2=0\]

\[{{a}_{22}}=2\times 2-2=4\]

\[{{a}_{32}}=2\times 3-2=4\]

\[{{a}_{13}}=2\times 1-3=-1\]

\[{{a}_{23}}=2\times 2-3=1\]

\[{{a}_{33}}=2\times 3-3=3\]

\[{{a}_{14}}=2\times 1-4=-2\]

\[{{a}_{24}}=2\times 2-4=0\]

\[{{a}_{34}}=2\times 3-4=2\]

Thus, the required matrix is \[A=\left[ \begin{matrix} 1 & 0 & -1 & -2  \\ 3 & 2 & 1 & 0  \\ 5 & 4 & 3 & 2  \\ \end{matrix} \right]\].

5. Find the value of \[x,y,z\] from the following equation:

i. \[\mathbf{\left[ \begin{matrix} 4 & 3  \\ x & 5  \\ \end{matrix} \right]=\left[ \begin{matrix} y & z  \\  1 & 5  \\ \end{matrix} \right]}\]

Ans: Given \[\left[ \begin{matrix} 4 & 3  \\  x & 5  \\ \end{matrix} \right]=\left[ \begin{matrix} y & z  \\ 1 & 5  \\ \end{matrix} \right]\] 

Comparing the corresponding elements we get,

\[x=1,y=4,z=3\]

ii. \[\mathbf{\left[ \begin{matrix} x+y & 2  \\ 5+z & xy  \\ \end{matrix} \right]=\left[ \begin{matrix} 6 & 2  \\ 5 & 8  \\ \end{matrix} \right]}\]

Ans: Given \[\left[ \begin{matrix} x+y & 2  \\ 5+z & xy  \\ \end{matrix} \right]=\left[ \begin{matrix} 6 & 2  \\ 5 & 8  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[x+y=6,xy=8,5+z=5\]

Now, \[\because 5+z=5\]

\[\Rightarrow z=0\]

We know that, \[{{\left( x-y \right)}^{2}}={{\left( x+y \right)}^{2}}-4xy\]

\[\Rightarrow {{\left( x-y \right)}^{2}}=36-32\]

\[\Rightarrow \left( x-y \right)=\pm 2\]

When \[\left( x-y \right)=2\] and \[\left( x+y \right)=6\],

We get \[x=4,y=2\]

When \[\left( x-y \right)=-2\] and \[\left( x+y \right)=6\],

We get \[x=2,y=4\]

\[\therefore x=4,y=2,z=0\] or \[\therefore x=2,y=4,z=0\]

iii. \[\mathbf{\left[ \begin{matrix} x+y+z  \\ x+z  \\ y+z  \\ \end{matrix} \right]=\left[ \begin{matrix} 9  \\ 5  \\ 7  \\ \end{matrix} \right]}\]

Ans: Given \[\left[ \begin{matrix} x+y+z  \\ x+z  \\ y+z  \\ \end{matrix} \right]=\left[ \begin{matrix} 9  \\ 5  \\ 7  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[x+y+z=9\]                      …(1)

\[x+z=5\]                              …(2)

\[y+z=7\]                              …(3)

From equation (1) and (2),

\[y+5=9\]

\[\Rightarrow y=4\]

From equation (3) we have,

\[4+z=7\]

\[\Rightarrow z=3\]

\[x+z=5\]

\[\Rightarrow x=2\]

\[\therefore x=2,y=4,z=3\]

6. Find the value of \[\mathbf{a,b,c,d}\] from the equation:

\[\mathbf{\left[ \begin{matrix} a-b & 2a+c  \\ 2a-b & 3c+d  \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 5  \\ 0 & 13  \\ \end{matrix} \right]}\]

Ans: Given \[\left[ \begin{matrix} a-b & 2a+c  \\ 2a-b & 3c+d  \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 5  \\ 0 & 13  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[a-b=-1\]                              …(1)

\[2a-b=0\]                              …(2)

\[2a+c=5\]                              …(3)

\[3c+d=13\]                                        …(4)

From equation (2),

\[b=2a\]

From equation (1),

\[a-2a=-1\]

\[\Rightarrow a=1\]

\[\Rightarrow b=2\]

From equation (3),

\[2\times 1+c=5\]

\[\Rightarrow c=3\]

From equation (4),

\[3\times 3+d=13\]

\[\Rightarrow d=4\]

\[\therefore a=1,b=2,c=3,d=4\]

7. \[\mathbf{A={{\left[ {{a}_{y}} \right]}_{m\times n}}}\] is a square matrix, if

1. m<n

2. m>n

3. m=n

4. None of these

Ans: A given matrix is said to be a square matrix if the number of rows is equal to the number of columns.

\[\therefore A={{\left[ {{a}_{y}} \right]}_{m\times n}}\] is a square matrix if, \[m=n\].

Thus, option (C) is correct.

8. Which of the given values of \[x\] and \[y\] make the following pair of matrices equal \[\mathbf{\left[ \begin{matrix} 3x+7 & 5  \\ y+1 & 2-3x  \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & y-2  \\  8 & 4  \\ \end{matrix} \right]}\]

1. \[\mathbf{x=\dfrac{-1}{3},y=7}\]

2. Not possible to find

3. \[\mathbf{y=7,x=\dfrac{-2}{3}}\]

4. \[\mathbf{x=\dfrac{-1}{3},y=\dfrac{-2}{3}}\]

Ans: Given \[\left[ \begin{matrix} 3x+7 & 5  \\ y+1 & 2-3x  \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & y-2  \\  8 & 4  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[3x+7=0\]

\[\Rightarrow x=-\dfrac{7}{3}\]

\[y-2=5\]

\[\Rightarrow y=7\]

\[y+1=8\]

\[\Rightarrow y=7\]

\[2-3x=4\]

\[\Rightarrow x=-\dfrac{2}{3}\]

Since we get two different values of \[x\] ,which is not possible. It is not possible to find the values of \[x\] and \[y\] for which the given matrices are equal.

Thus, the correct option is (B).

9. The number of all possible matrices of order \[\mathbf{3\times 3}\] with each entry \[0\] or \[1\] is:

1. \[\mathbf{27}\]

2. \[\mathbf{18}\]

3. \[\mathbf{81}\]

4. \[\mathbf{512}\]

Ans: Given a matrix of the order \[3\times 3\] has nine elements and each of these elements can be either \[0\] or \[1\] .

Now, each of the nine elements can be filled in two possible ways.

Therefore, the required number of possible matrices is \[{{2}^{9}}=512\].

Exercise 3.2

1. Let \[\mathbf{A=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right],C=\left[ \begin{matrix} -2 & 5  \\ 3 & 4  \\ \end{matrix} \right]}\]

Find each of the following

i. \[\mathbf{A+B}\]

Ans: Given \[A=\left[ \begin{matrix}  2 & 4  \\  3 & 2  \\ \end{matrix} \right],B=\left[ \begin{matrix}  1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\therefore A+B=\left[ \begin{matrix}  2 & 4  \\  3 & 2  \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow A+B=\left[ \begin{matrix} 2+1 & 4+3  \\ 3-2 & 2+5  \\ \end{matrix} \right]\]

\[\Rightarrow A+B=\left[ \begin{matrix}  3 & 7  \\ 1 & 7  \\ \end{matrix} \right]\]

ii. \[\mathbf{A-B}\]

Ans: Given \[A=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\therefore A-B=\left[ \begin{matrix}  2 & 4  \\ 3 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow A-B=\left[ \begin{matrix} 2-1 & 4-3  \\ 3+2 & 2-5  \\ \end{matrix} \right]\]

\[\Rightarrow A-B=\left[ \begin{matrix} 1 & 1  \\ 5 & -3  \\ \end{matrix} \right]\]

iii. \[\mathbf{3A-C}\]

Ans: Given \[A=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right],C=\left[ \begin{matrix} -2 & 5  \\ 3 & 4  \\ \end{matrix} \right]\]

\[\therefore 3A-C=3\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix} -2 & 5  \\ 3 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow 3A-C=\left[ \begin{matrix} 6 & 12  \\ 9 & 6  \\ \end{matrix} \right]-\left[ \begin{matrix} -2 & 5  \\ 3 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow 3A-C=\left[ \begin{matrix} 6+2 & 12-5  \\ 9-3 & 6-4  \\ \end{matrix} \right]\]

\[\Rightarrow 3A-C=\left[ \begin{matrix} 8 & 7  \\ 6 & 2  \\ \end{matrix} \right]\]

iv. \[\mathbf{AB}\]

Ans: Given \[A=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\therefore AB=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow AB=\left[ \begin{matrix} 2\left( 1 \right)+4\left( -2 \right) & 2\left( 3 \right)+4\left( 5 \right)  \\ 3\left( 1 \right)+2\left( -2 \right) & 3\left( 3 \right)+2\left( 5 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow AB=\left[ \begin{matrix}  -6 & 26  \\ -1 & 19  \\ \end{matrix} \right]\]

v. \[\mathbf{BA}\]

Ans: Given \[A=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\therefore BA=\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow BA=\left[ \begin{matrix} 1\left( 2 \right)+3\left( 3 \right) & 1\left( 4 \right)+3\left( 2 \right)  \\ -2\left( 2 \right)+5\left( 3 \right) & -2\left( 4 \right)+5\left( 2 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow BA=\left[ \begin{matrix} 11 & 10  \\ 11 & 2  \\ \end{matrix} \right]\]

2. Compute the following:

i. \[\mathbf{\left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]+\left[ \begin{matrix} a & b  \\ b & a  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]+\left[ \begin{matrix} a & b  \\ b & a  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]+\left[ \begin{matrix} a & b  \\ b & a  \\ \end{matrix} \right]=\left[ \begin{matrix} a+a & b+b  \\ -b+b & a+a  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]+\left[ \begin{matrix} a & b  \\ b & a  \\ \end{matrix} \right]=\left[ \begin{matrix} 2a & 2b  \\ 0 & 2a  \\ \end{matrix} \right]\]

ii. \[\mathbf{\left[ \begin{matrix}  {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}}  \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}}  \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc  \\ -2ac & -2ab  \\ \end{matrix} \right]}\]

Ans: We have to find 

\[\left[ \begin{matrix}  {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}}  \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}}  \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc  \\ -2ac & -2ab  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix} {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}}  \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}}  \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc  \\ -2ac & -2ab  \\ \end{matrix} \right]=\left[ \begin{matrix} {{a}^{2}}+{{b}^{2}}+2ab & {{b}^{2}}+{{c}^{2}}+2bc  \\ {{a}^{2}}+{{c}^{2}}-2ac & {{a}^{2}}+{{b}^{2}}-2ab  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}}  \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}}  \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc  \\ -2ac & -2ab  \\ \end{matrix} \right]=\left[ \begin{matrix} {{\left( a+b \right)}^{2}} & {{\left( b+c \right)}^{2}}  \\ {{\left( a-c \right)}^{2}} & {{\left( a-b \right)}^{2}}  \\ \end{matrix} \right]\]

iii. \[\left[ \begin{matrix} -1 & 4 & -6  \\ 8 & 5 & 16  \\ 2 & 8 & 5  \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6  \\ 8 & 0 & 5  \\ 3 & 2 & 4  \\ \end{matrix} \right]\]

Ans: We have to find \[\left[ \begin{matrix} -1 & 4 & -6  \\ 8 & 5 & 16  \\ 2 & 8 & 5  \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6  \\ 8 & 0 & 5  \\ 3 & 2 & 4  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix} -1 & 4 & -6  \\ 8 & 5 & 16  \\ 2 & 8 & 5  \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6  \\ 8 & 0 & 5  \\ 3 & 2 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix} -1+12 & 4+7 & -6+6  \\ 8+8 & 5+0 & 16+5  \\ 2+3 & 8+2 & 5+4  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} -1 & 4 & -6  \\ 8 & 5 & 16  \\ 2 & 8 & 5  \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6  \\ 8 & 0 & 5  \\ 3 & 2 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix} 11 & 11 & 0  \\ 16 & 5 & 21  \\ 5 & 10 & 9  \\ \end{matrix} \right]\]

iv. \[\left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x  \\ {{\sin }^{2}}x & {{\cos }^{2}}x  \\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}x & {{\cos }^{2}}x  \\ {{\cos }^{2}}x & {{\sin }^{2}}x  \\ \end{matrix} \right]\]

Ans: We have to find \[\left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x  \\   {{\sin }^{2}}x & {{\cos }^{2}}x  \\ \end{matrix} \right]+\left[ \begin{matrix}   {{\sin }^{2}}x & {{\cos }^{2}}x  \\ {{\cos }^{2}}x & {{\sin }^{2}}x  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x  \\ {{\sin }^{2}}x & {{\cos }^{2}}x  \\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}x & {{\cos }^{2}}x  \\ {{\cos }^{2}}x & {{\sin }^{2}}x  \\ \end{matrix} \right]=\left[ \begin{matrix} {{\cos }^{2}}x+{{\sin }^{2}}x & {{\sin }^{2}}x+{{\cos }^{2}}x  \\ {{\sin }^{2}}x+{{\cos }^{2}}x & {{\cos }^{2}}x+{{\sin }^{2}}x  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x  \\ {{\sin }^{2}}x & {{\cos }^{2}}x  \\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}x & {{\cos }^{2}}x  \\ {{\cos }^{2}}x & {{\sin }^{2}}x  \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 1  \\ 1 & 1  \\ \end{matrix} \right]\]

3. Compute the indicated products

i. \[\mathbf{\left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]\left[ \begin{matrix} a & -b  \\ b & a  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]\left[ \begin{matrix} a & -b  \\ b & a  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]\left[ \begin{matrix} a & -b  \\ b & a  \\ \end{matrix} \right]=\left[ \begin{matrix} a\left( a \right)+b\left( b \right) & a\left( -b \right)+b\left( a \right)  \\ -b\left( a \right)+a\left( b \right) & -b\left( -b \right)+a\left( a \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]\left[ \begin{matrix} a & -b  \\ b & a  \\ \end{matrix} \right]=\left[ \begin{matrix}   {{a}^{2}}+{{b}^{2}} & 0  \\  0 & {{a}^{2}}+{{b}^{2}}  \\ \end{matrix} \right]\]

ii. \[\mathbf{\left[ \begin{matrix} 1  \\   2  \\   3  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 3 & 4  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix}    1  \\   2  \\   3  \\ \end{matrix} \right]\left[ \begin{matrix}  2 & 3 & 4  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix}    1  \\   2  \\   3  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   1\left( 2 \right) & 1\left( 3 \right) & 1\left( 4 \right)  \\   2\left( 2 \right) & 2\left( 3 \right) & 2\left( 4 \right)  \\   3\left( 2 \right) & 3\left( 3 \right) & 3\left( 4 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   1  \\   2  \\   3  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   2 & 3 & 4  \\   4 & 6 & 8  \\   6 & 9 & 12  \\ \end{matrix} \right]\]

iii. \[\mathbf{\left[ \begin{matrix}   1 & -2  \\  2 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   2 & 3 & 1  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix}   1 & -2  \\   2 & 3  \\\end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   2 & 3 & 1  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix}   1 & -2  \\   2 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   2 & 3 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1\left( 1 \right)-2\left( 2 \right) & 1\left( 2 \right)-2\left( 3 \right) & 1\left( 3 \right)-2\left( 1 \right)  \\   2\left( 1 \right)+3\left( 2 \right) & 2\left( 2 \right)+3\left( 3 \right) & 2\left( 3 \right)+3\left( 1 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   1 & -2  \\   2 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   2 & 3 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   -3 & -4 & 1  \\   8 & 13 & 9  \\ \end{matrix} \right]\]

iv. \[\mathbf{\left[ \begin{matrix}  2 & 3 & 4  \\   3 & 4 & 5  \\   4 & 5 & 6  \\ \end{matrix} \right]\left[ \begin{matrix}    1 & -3 & 5  \\   0 & 2 & 4  \\   3 & 0 & 5  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix}  2 & 3 & 4  \\   3 & 4 & 5  \\   4 & 5 & 6  \\ \end{matrix} \right]\left[ \begin{matrix}    1 & -3 & 5  \\   0 & 2 & 4  \\   3 & 0 & 5  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix}   2 & 3 & 4  \\   3 & 4 & 5  \\   4 & 5 & 6  \\ \end{matrix} \right]\left[ \begin{matrix}  1 & -3 & 5  \\   0 & 2 & 4  \\   3 & 0 & 5  \\ \end{matrix} \right]=\left[ \begin{matrix}   2\left( 1 \right)+3\left( 0 \right)+4\left( 3 \right) & 2\left( -3 \right)+3\left( 2 \right)+4\left( 0 \right) & 2\left( 5 \right)+3\left( 4 \right)+4\left( 5 \right)  \\   3\left( 1 \right)+4\left( 0 \right)+5\left( 3 \right) & 3\left( -3 \right)+4\left( 2 \right)+5\left( 0 \right) & 3\left( 5 \right)+4\left( 4 \right)+5\left( 5 \right)  \\   4\left( 1 \right)+5\left( 0 \right)+5\left( 3 \right) & 4\left( -3 \right)+5\left( 2 \right)+6\left( 0 \right) & 4\left( 5 \right)+5\left( 4 \right)+6\left( 5 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2 & 3 & 4  \\   3 & 4 & 5  \\   4 & 5 & 6  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & -3 & 5  \\   0 & 2 & 4  \\   3 & 0 & 5  \\ \end{matrix} \right]=\left[ \begin{matrix}   14 & 0 & 42  \\  18 & -1 & 56  \\   22 & -2 & 70  \\ \end{matrix} \right]\]

v. \[\mathbf{\left[ \begin{matrix}   2 & 1  \\   3 & 2  \\   -1 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 1  \\   -1 & 2 & 1  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix}   2 & 1  \\   3 & 2  \\   -1 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 1  \\   -1 & 2 & 1  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix}   2 & 1  \\   3 & 2  \\   -1 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 1  \\   -1 & 2 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}  2\left( 1 \right)+1\left( -1 \right) & 2\left( 0 \right)+1\left( 2 \right) & 2\left( 1 \right)+1\left( 1 \right)  \\   3\left( 1 \right)+2\left( -1 \right) & 3\left( 0 \right)+2\left( 2 \right) & 3\left( 1 \right)+2\left( 1 \right)  \\   -1\left( 1 \right)+1\left( -1 \right) & -1\left( 0 \right)+1\left( 2 \right) & -1\left( 1 \right)+1\left( 1 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2 & 1  \\   3 & 2  \\   -1 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 1  \\   -1 & 2 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 2 & 3  \\   1 & 4 & 5  \\   -2 & 2 & 0  \\ \end{matrix} \right]\]

vi. \[\mathbf{\left[ \begin{matrix}   3 & -1 & 3  \\   -1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & -3  \\   1 & 0  \\   3 & 1  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix}  3 & -1 & 3  \\   -1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & -3  \\   1 & 0  \\   3 & 1  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix}   3 & -1 & 3  \\   -1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & -3  \\  1 & 0  \\   3 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}  3\left( 2 \right)-1\left( 1 \right)+3\left( 3 \right) & 3\left( -3 \right)-1\left( 0 \right)+3\left( 1 \right)  \\  -1\left( 2 \right)+0\left( 1 \right)+2\left( 3 \right) & -1\left( -3 \right)+0\left( 0 \right)+2\left( 1 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   3 & -1 & 3  \\   -1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & -3  \\   1 & 0  \\   3 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix} 14 & -6  \\   4 & 5  \\ \end{matrix} \right]\]

4.If,

\[\mathbf{A=\left[\begin{matrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1  \\ \end{matrix}\right],B=\left[ \begin{matrix}  3 & -1 & 2  \\  4 & 2 & 5  \\  2 & 0 & 3  \\ \end{matrix}\right],C=\left[ \begin{matrix}  4 & 1 & 2  \\  0 & 3 & 2  \\  1 & -2 & 3  \\ \end{matrix}\right]}\],

then Compute ( A+B ) and ( B-C ).Also, verify that A+( B-C)=( A+B )-C}.

Ans: Given  \[A=\left[ \begin{matrix}   1 & 2 & -3  \\   5 & 0 & 2  \\   1 & -1 & 1  \\ \end{matrix} \right],B=\left[ \begin{matrix}   3 & -1 & 2  \\   4 & 2 & 5  \\   2 & 0 & 3  \\ \end{matrix} \right],C=\left[ \begin{matrix}   4 & 1 & 2  \\   0 & 3 & 2  \\   1 & -2 & 3  \\ \end{matrix} \right]\]

\[\therefore A+B=\left[ \begin{matrix}   1 & 2 & -3  \\   5 & 0 & 2  \\   1 & -1 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   3 & -1 & 2  \\   4 & 2 & 5  \\   2 & 0 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow A+B=\left[ \begin{matrix}   1+3 & 2-1 & -3+2  \\   5+4 & 0+2 & 2+5 \\  1+2 & -1+0 & 1+3  \\ \end{matrix} \right]\]

\[\Rightarrow A+B=\left[ \begin{matrix}   4 & 1 & -1  \\   9 & 2 & 7  \\   3 & -1 & 4  \\ \end{matrix} \right]\]

And \[B-C=\left[ \begin{matrix}   3 & -1 & 2  \\   4 & 2 & 5  \\   2 & 0 & 3  \\ \end{matrix} \right]-\left[ \begin{matrix}   4 & 1 & 2  \\   0 & 3 & 2  \\   1 & -2 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow B-C=\left[ \begin{matrix}   3-4 & -1-1 & 2-2  \\   4-0 & 2-3 & 5-2  \\   2-1 & 0+2 & 3-3  \\ \end{matrix} \right]\]

\[\therefore B-C=\left[ \begin{matrix}   -1 & -2 & 0  \\   4 & -1 & 3  \\   1 & 2 & 0  \\ \end{matrix} \right]\]

Now,

\[A+\left( B-C \right)=\left[ \begin{matrix}  1 & 2 & -3  \\   5 & 0 & 2  \\   1 & -1 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   -1 & -2 & 0  \\   4 & -1 & 3  \\   1 & 2 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow A+\left( B-C \right)=\left[ \begin{matrix}   1-1 & 2-2 & -3+0  \\   5+4 & 0-1 & 2+3  \\   1+1 & -1+2 & 1+0  \\ \end{matrix} \right]\]

\[\therefore A+\left( B-C \right)=\left[ \begin{matrix}   0 & 0 & -3  \\   9 & -1 & 5  \\   2 & 1 & 1  \\ \end{matrix} \right]\]                                                  …(1)

And \[\left( A+B \right)-C=\left[ \begin{matrix}   4 & 1 & -1  \\   9 & 2 & 7  \\   3 & -1 & 4  \\ \end{matrix} \right]-\left[ \begin{matrix}  4 & 1 & 2  \\   0 & 3 & 2  \\   1 & -2 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow \left( A+B \right)-C=\left[ \begin{matrix}   4-4 & 1-1 & -1-2  \\   9-0 & 2-3 & 7-2  \\   3-1 & -1+2 & 4-3  \\ \end{matrix} \right]\]

\[\therefore \left( A+B \right)-C=\left[ \begin{matrix}   0 & 0 & -3  \\   9 & -1 & 5  \\   2 & 1 & 1  \\ \end{matrix} \right]\]                                                  …(2)

Thus, from equation (1) and (2),

\[A+\left( B-C \right)=\left( A+B \right)-C\]

Hence proved.

5. If \[A=\left[ \begin{matrix}   \dfrac{2}{3} & 1 & \dfrac{5}{3}  \\   \dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3}  \\   \dfrac{7}{3} & 2 & \dfrac{2}{3}  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{3}{5} & 1  \\   \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5}  \\    \dfrac{7}{5} & \dfrac{6}{5} & \dfrac{2}{5}  \\ \end{matrix} \right]\] then compute \[3A-5B\] .

Ans: Given that \[A=\left[ \begin{matrix}   \dfrac{2}{3} & 1 & \dfrac{5}{3}  \\   \dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3}  \\   \dfrac{7}{3} & 2 & \dfrac{2}{3}  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{3}{5} & 1  \\   \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5}  \\    \dfrac{7}{5} & \dfrac{6}{5} & \dfrac{2}{5}  \\ \end{matrix} \right]\] 

\[\therefore 3A-5B=3\left[ \begin{matrix}   \dfrac{2}{3} & 1 & \dfrac{5}{3}  \\   \dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3}  \\   \dfrac{7}{3} & 2 & \dfrac{2}{3}  \\ \end{matrix} \right]-5\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{3}{5} & 1  \\   \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5}  \\   \dfrac{7}{5} & \dfrac{6}{5} &\dfrac{2}{5}  \\ \end{matrix} \right]\]

\[\Rightarrow 3A-5B=\left[ \begin{matrix}   2 & 3 & 5  \\   1 & 2 & 4  \\   7 & 6 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix}   2 & 3 & 5  \\   1 & 2 & 4  \\   7 & 6 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow 3A-5B=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

6. Simplify \[\mathbf{\cos \theta \left[ \begin{matrix}   \cos \theta  & \sin \theta   \\   -\sin \theta  & \cos \theta   \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix}  \sin \theta  & -\cos \theta   \\  \cos \theta  & \sin \theta   \\ \end{matrix} \right]}\] .

Ans: We have to simplify \[\cos \theta \left[ \begin{matrix}   \cos \theta  & \sin \theta   \\   -\sin \theta  & \cos \theta   \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix}  \sin \theta  & -\cos \theta   \\  \cos \theta  & \sin \theta   \\ \end{matrix} \right]\] 

\[\therefore \cos \theta \left[ \begin{matrix} \cos \theta  & \sin \theta   \\   -\sin \theta  & \cos \theta   \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix}   \sin \theta  & -\cos \theta   \\   \cos \theta  & \sin \theta   \\ \end{matrix} \right]=\left[ \begin{matrix}   {{\cos }^{2}}\theta  & \cos \theta \sin \theta   \\   -\cos \theta \sin \theta  & {{\cos }^{2}}\theta   \\ \end{matrix} \right]+\left[ \begin{matrix}   {{\sin }^{2}}\theta  & -\sin \theta \cos \theta   \\   \sin \theta \cos \theta  & {{\sin }^{2}}\theta   \\ \end{matrix} \right]\]

\[\Rightarrow \cos \theta \left[ \begin{matrix}   \cos \theta  & \sin \theta   \\   -\sin \theta  & \cos \theta   \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix}   \sin \theta  & -\cos \theta   \\  \cos \theta  & \sin \theta   \\ \end{matrix} \right]=\left[ \begin{matrix}  {{\cos }^{2}}\theta +{{\sin }^{2}}\theta  & \cos \theta \sin \theta -\cos \theta \sin \theta   \\  -\cos \theta \sin \theta +\cos \theta \sin \theta  & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta   \\ \end{matrix} \right]\]

\[\Rightarrow \cos \theta \left[ \begin{matrix}  \cos \theta  & \sin \theta   \\   -\sin \theta  & \cos \theta   \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix}  \sin \theta  & -\cos \theta   \\  \cos \theta  & \sin \theta   \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

7. Find \[X\] and \[Y\] ,if

i. \[{X+Y=}\left [ \begin{matrix}   7 & 0  \\   2 & 5  \\ \end{matrix} \right]\] and \[X-Y=\left[ \begin{matrix}   3 & 0  \\   0 & 3  \\ \end{matrix} \right]\]

Ans: Given:

\[X+Y=\left[ \begin{matrix}   7 & 0  \\   2 & 5  \\ \end{matrix} \right]\]        …(1)

\[X-Y=\left[ \begin{matrix}   3 & 0  \\   0 & 3  \\ \end{matrix} \right]\]   …(2)

Adding equations (1) and (2), we get,

\[2X=\left[ \begin{matrix}   7 & 0  \\   2 & 5  \\ \end{matrix} \right]+\left[ \begin{matrix}   3 & 0  \\   0 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow 2X=\left[ \begin{matrix}   10 & 0  \\   2 & 8  \\ \end{matrix} \right]\]

\[\Rightarrow X=\dfrac{1}{2}\left[ \begin{matrix}   10 & 0  \\   2 & 8  \\ \end{matrix} \right]\]

\[\therefore X=\left[ \begin{matrix}   5 & 0  \\   1 & 4  \\ \end{matrix} \right]\]

Now, since \[X+Y=\left[ \begin{matrix}    7 & 0  \\   2 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}  5 & 0  \\   1 & 4  \\ \end{matrix} \right]+Y=\left[ \begin{matrix}   7 & 0  \\   2 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow Y=\left[ \begin{matrix}   7 & 0  \\   2 & 5  \\ \end{matrix} \right]-\left[ \begin{matrix}   5 & 0  \\   1 & 4  \\ \end{matrix} \right]\]

\[\therefore Y=\left[ \begin{matrix}   2 & 0  \\   1 & 1  \\ \end{matrix} \right]\]

\[\therefore X=\left[ \begin{matrix}   5 & 0  \\   1 & 4  \\ \end{matrix} \right],Y=\left[ \begin{matrix}   2 & 0  \\   1 & 1  \\ \end{matrix} \right]\]

ii. \[\mathbf{2X+3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]}\] and \[\mathbf{3X+2Y=\left[ \begin{matrix}   2 & -2  \\   -1 & 5  \\ \end{matrix} \right]}\]

Ans: Given:

\[2X+3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]\]                               …(1)

\[3X+2Y=\left[ \begin{matrix}   2 & -2  \\   -1 & 5  \\ \end{matrix} \right]\]                            …(2)

Multiplying equation (1) with two we get,

\[2\left( 2X+3Y \right)=2\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow \left( 4X+6Y \right)=\left[ \begin{matrix}   4 & 6  \\   8 & 0  \\ \end{matrix} \right]\]                      …(3)

Multiplying equation (2) with three we get,

\[3\left( 3X+2Y \right)=3\left[ \begin{matrix}   2 & -2  \\   -1 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow \left( 9X+6Y \right)=\left[ \begin{matrix}   6 & -6  \\   -3 & 15  \\ \end{matrix} \right]\]                 …(4)

From equation (3) and (4),

\[\left( 4X+6Y \right)-\left( 9X+6Y \right)=\left[ \begin{matrix}   4 & 6  \\   8 & 0  \\ \end{matrix} \right]-\left[ \begin{matrix} 6 & -6  \\   -3 & 15  \\ \end{matrix} \right]\]

\[\Rightarrow -5X=\left[ \begin{matrix}   -2 & 12  \\   11 & -15  \\ \end{matrix} \right]\]

\[\Rightarrow X=\dfrac{-1}{5}\left[ \begin{matrix}   -2 & 12  \\   11 & -15  \\ \end{matrix} \right]\]

\[\therefore X=\left[ \begin{matrix}  \dfrac{2}{5} & \dfrac{-12}{5}  \\   \dfrac{-11}{5} & 3  \\ \end{matrix} \right]\]

Now, since \[2X+3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]\]

\[\therefore 2\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{-12}{5}  \\   \dfrac{-11}{5} & 3  \\ \end{matrix} \right]+3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \dfrac{4}{5} & \dfrac{-24}{5}  \\  \dfrac{-22}{5} & 6  \\ \end{matrix} \right]+3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow 3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]-\left[ \begin{matrix}   \dfrac{4}{5} & \dfrac{-24}{5}  \\   \dfrac{-22}{5} & 6  \\ \end{matrix} \right]\]

\[\Rightarrow 3Y=\left[ \begin{matrix}   \dfrac{6}{5} & \dfrac{39}{5}  \\    \dfrac{42}{5} & -6  \\ \end{matrix} \right]\]

\[\therefore Y=\left[ \begin{matrix}  \dfrac{2}{5} & \dfrac{13}{5}  \\   \dfrac{14}{5} & -2  \\ \end{matrix} \right]\]

\[\therefore X=\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{-12}{5}  \\   \dfrac{-11}{5} & 3  \\ \end{matrix} \right],Y=\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{13}{5}  \\   \dfrac{14}{5} & -2  \\ \end{matrix} \right]\]

8. Find \[\mathbf{X}\] ,if \[\mathbf{Y=\left[ \begin{matrix}   3 & 2  \\   1 & 4  \\ \end{matrix} \right]}\] and \[\mathbf{2X+Y=\left[ \begin{matrix}   1 & 0  \\   -3 & 2  \\ \end{matrix} \right]}\] .

Ans: Given \[Y=\left[ \begin{matrix}   3 & 2  \\   1 & 4  \\ \end{matrix} \right]\]

And \[2X+Y=\left[ \begin{matrix}   1 & 0  \\   -3 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow 2X+\left[ \begin{matrix}   3 & 2  \\   1 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   -3 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow 2X=\left[ \begin{matrix}   1 & 0  \\   -3 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix}   3 & 2  \\   1 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow 2X=\left[ \begin{matrix}   -2 & -2  \\   -4 & -2  \\ \end{matrix} \right]\]

\[\Rightarrow X=\dfrac{1}{2}\left[ \begin{matrix}   -2 & -2  \\   -4 & -2  \\ \end{matrix} \right]\]

\[\therefore X=\left[ \begin{matrix}   -1 & -1  \\   -2 & -1  \\ \end{matrix} \right]\]

9. Find \[X\] and \[Y\] ,if \[\mathbf{2\left[ \begin{matrix}   1 & 3  \\   0 & x  \\ \end{matrix} \right]+\left[ \begin{matrix}   y & 0  \\   1 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 6  \\   1 & 8  \\ \end{matrix} \right]}\]

Ans: Given: \[2\left[ \begin{matrix}   1 & 3  \\   0 & x  \\ \end{matrix} \right]+\left[ \begin{matrix}   y & 0  \\    1 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 6  \\   1 & 8  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2 & 6  \\   0 & 2x  \\ \end{matrix} \right]+\left[ \begin{matrix}   y & 0  \\   1 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 6  \\   1 & 8  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2+y & 6+0  \\   0+1 & 2x+2  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 6  \\   1 & 8  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2+y & 6  \\   1 & 2x+2  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 6  \\   1 & 8  \\ \end{matrix} \right]\]

Comparing the corresponding elements of these two matrices, we get:

\[2+y=5\]

\[\Rightarrow y=3\]

\[2x+2=8\]

\[\Rightarrow x=3\]

\[\therefore x=3,y=3\]

10. Solve the equation for \[X,Y,Z\] and \[t\] if \[\mathbf{2\left[ \begin{matrix}   x & z  \\   y & t  \\ \end{matrix} \right]+3\left[ \begin{matrix}   1 & -1  \\   0 & 2  \\ \end{matrix} \right]=3\left[ \begin{matrix}   3 & 5  \\   4 & 6  \\ \end{matrix} \right]}\]

Ans: Given: \[2\left[ \begin{matrix}   x & z  \\   y & t  \\ \end{matrix} \right]+3\left[ \begin{matrix}   1 & -1  \\   0 & 2  \\ \end{matrix} \right]=3\left[ \begin{matrix}   3 & 5  \\   4 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2x & 2z  \\   2y & 2t  \\ \end{matrix} \right]+\left[ \begin{matrix}   3 & -3  \\   0 & 6  \\ \end{matrix} \right]=\left[ \begin{matrix}   9 & 15  \\   12 & 18  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2x+3 & 2z-3  \\   2y & 2t+6  \\ \end{matrix} \right]=\left[ \begin{matrix}   9 & 15  \\   12 & 18  \\ \end{matrix} \right]\]

Equating the corresponding elements of these two matrices, we get:

\[2x+3=9\]

\[\Rightarrow x=3\]

\[2y=12\]

\[\Rightarrow y=6\]

\[2z-3=15\]

\[\Rightarrow z=9\]

\[2t+6=18\]

\[\Rightarrow t=6\]

\[\therefore x=3,y=6,z=9,t=6\]

11. If \[\mathbf{x\left[ \begin{matrix}   2  \\   3  \\ \end{matrix} \right]+y\left[ \begin{matrix}   -1  \\   1  \\ \end{matrix} \right]=\left[ \begin{matrix}  10  \\   5  \\ \end{matrix} \right]}\] ,find values of \[x\] and \[y\] .

Ans: Given: \[x\left[ \begin{matrix}   2  \\   3  \\ \end{matrix} \right]+y\left[ \begin{matrix}   -1  \\   1  \\ \end{matrix} \right]=\left[ \begin{matrix}  10  \\   5  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2x  \\   3x  \\ \end{matrix} \right]+\left[ \begin{matrix}   -y  \\   y  \\ \end{matrix} \right]=\left[ \begin{matrix}   10  \\   5  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2x-y  \\   3x+y  \\ \end{matrix} \right]=\left[ \begin{matrix}   10  \\   5  \\ \end{matrix} \right]\]

Equating the corresponding elements of these two matrices, we get:

\[2x-y=10\] and

\[3x+y=5\]

Adding these two equations, we get:

\[5x=15\]

\[\Rightarrow x=3\]

Now, since \[3x+y=5\]

\[\Rightarrow y=5-3x\]

\[\Rightarrow y=5-9\]

\[\Rightarrow y=-4\]

\[\therefore x=3,y=-4\]

12. Given \[\mathbf{3\left[ \begin{matrix}   x & y  \\   z & w  \\ \end{matrix} \right]=\left[ \begin{matrix}   x & 6  \\   -1 & 2w  \\ \end{matrix} \right]+\left[ \begin{matrix}   4 & x+y  \\   z+w & 2w+3  \\ \end{matrix} \right]}\] ,find the values of \[x,y,z\] and \[w\] .

Ans: Given: \[3\left[ \begin{matrix}   x & y  \\   z & w  \\ \end{matrix} \right]=\left[ \begin{matrix}   x & 6  \\   -1 & 2w  \\ \end{matrix} \right]+\left[ \begin{matrix}   4 & x+y  \\   z+w & 2w+3  \\ \end{matrix} \right]\] 

\[\Rightarrow \left[ \begin{matrix}   3x & 3y  \\   3z & 3w  \\ \end{matrix} \right]=\left[ \begin{matrix}   x+4 & 6+x+y  \\   -1+z+w & 2w+3  \\ \end{matrix} \right]\]

Equating the corresponding elements of these two matrices, we get:

\[3x=x+4\]

\[\Rightarrow x=2\]

\[3y=6+x+y\]

\[\Rightarrow y=4\]

\[3w=2w+3\]

\[\Rightarrow w=3\]

\[3z=-1+z+w\]

\[\Rightarrow z=1\]

\[\therefore x=2,y=4,z=1,w=3\]

13. If \[\mathbf{F\left( x \right)=\left[ \begin{matrix}   \cos x & -\sin x & 0  \\  \sin x & \cos x & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]}\] , show that \[\mathbf{F\left( x \right)F\left( y \right)=F\left( x+y \right)}\] .

Ans: Here, \[F\left( x \right)=\left[ \begin{matrix}   \cos x & -\sin x & 0  \\  \sin x & \cos x & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[F\left( y \right)=\left[ \begin{matrix}   \cos y & -\sin y & 0  \\   \sin y & \cos y & 0 \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[\therefore F\left( x \right)F\left( y \right)=\left[ \begin{matrix}   \cos x & -\sin x & 0  \\   \sin x & \cos x & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   \cos y & -\sin y & 0  \\   \sin y & \cos y & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow F\left( x \right)F\left( y \right)=\left[ \begin{matrix}   \cos x\cos y-\sin x\sin y+0 & -\cos x\sin y-\sin x\cos y+0 & 0  \\   \sin x\cos y-\cos x\sin y+0 & -\sin x\sin y+\cos x\cos y+0 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow F\left( x \right)F\left( y \right)=\left[ \begin{matrix}   \cos \left( x+y \right) & -\sin \left( x+y \right) & 0  \\   \sin \left( x+y \right) & \cos \left( x+y \right) & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]                            …(1)

And \[F\left( x+y \right)=\left[ \begin{matrix}   \cos \left( x+y \right) & -\sin \left( x+y \right) & 0  \\   \sin \left( x+y \right) & \cos \left( x+y \right) & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]                               …(2)

From equation (1) and (2),

\[F\left( x \right)F\left( y \right)=F\left( x+y \right)\]

Hence proved.

14. Show that

i. \[\mathbf{\left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]\ne \left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]}\]

Ans: LHS: \[\left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   5\left( 2 \right)-1\left( 3 \right) & 5\left( 1 \right)-1\left( 4 \right)  \\   6\left( 2 \right)+7\left( 3 \right) & 6\left( 1 \right)+7\left( 4 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   7 & 1  \\   33 & 34  \\ \end{matrix} \right]\]                …(1)

RHS: \[\left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]=\left[ \begin{matrix}   2\left( 5 \right)+1\left( 6 \right) & 2\left( -1 \right)+1\left( 7 \right)  \\   3\left( 5 \right)+4\left( 6 \right) & 3\left( -1 \right)+4\left( 7 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]=\left[ \begin{matrix}   16 & 5  \\   39 & 25  \\ \end{matrix} \right]\]        …(2)

From equation (1) and (2),

\[LHS\ne RHS\]

Hence proved.

ii. \[\mathbf{\left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]\left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]\ne \left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]}\]

Ans: LHS: \[\left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]\left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   1\left( -1 \right)+2\left( 0 \right)+3\left( 2 \right) & 1\left( 1 \right)+2\left( -1 \right)+3\left( 3 \right) & 1\left( 0 \right)+2\left( 1 \right)+3\left( 4 \right)  \\   0\left( -1 \right)+1\left( 0 \right)+0\left( 2 \right) & 0\left( 1 \right)+1\left( -1 \right)+0\left( 3 \right) & 0\left( 0 \right)+1\left( 1 \right)+0\left( 4 \right)  \\   1\left( -1 \right)+1\left( 0 \right)+0\left( 2 \right) & 1\left( 1 \right)+1\left( -1 \right)+0\left( 3 \right) & 1\left( 0 \right)+1\left( 1 \right)+0\left( 4 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]\left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 8 & 14  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]\]                            …(1)

RHS: \[\left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]=\left[ \begin{matrix}  -1\left( 1 \right)+1\left( 0 \right)+0\left( 1 \right) & -1\left( 2 \right)+1\left( 1 \right)+0\left( 1 \right) & -1\left( 3 \right)+1\left( 0 \right)+0\left( 0 \right)  \\   0\left( 1 \right)-1\left( 0 \right)+1\left( 1 \right) & 0\left( 2 \right)-1\left( 1 \right)+1\left( 1 \right) & 0\left( 3 \right)-1\left( 0 \right)+1\left( 0 \right)  \\  2\left( 1 \right)+3\left( 0 \right)+4\left( 1 \right) & 2\left( 2 \right)+3\left( 1 \right)+4\left( 1 \right) & 2\left( 3 \right)+3\left( 0 \right)+4\left( 0 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]=\left[ \begin{matrix}   -1 & -1 & 3  \\   1 & 0 & 0  \\   6 & 11 & 6  \\ \end{matrix} \right]\]                          …(2)

From equation (1) and (2),

\[LHS\ne RHS\]

Hence proved.

15. Find \[\mathbf{{{A}^{2}}-5A+6I}\] if \[\mathbf{A=\left[ \begin{matrix}   2 & 0 & 1  \\   2 & 1 & 3  \\   1 & -1 & 0  \\ \end{matrix} \right]}\] .

Ans: Given: \[A=\left[ \begin{matrix}   2 & 0 & 1  \\   2 & 1 & 3  \\   1 & -1 & 0  \\ \end{matrix} \right]\]

\[{{A}^{2}}=AA\]

\[\therefore {{A}^{2}}=\left[ \begin{matrix}   2 & 0 & 1  \\   2 & 1 & 3  \\   1 & -1 & 0  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 0 & 1  \\   2 & 1 & 3  \\   1 & -1 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   2\left( 2 \right)+0\left( 2 \right)+1\left( 1 \right) & 2\left( 0 \right)+0\left( 1 \right)+1\left( -1 \right) & 2\left( 1 \right)+0\left( 3 \right)+1\left( 0 \right)  \\   2\left( 2 \right)+1\left( 2 \right)+3\left( 1 \right) & 2\left( 0 \right)+1\left( 1 \right)+3\left( -1 \right) & 2\left( 1 \right)+1\left( 3 \right)+3\left( 0 \right)  \\   1\left( 2 \right)-1\left( 2 \right)+0\left( 1 \right) & 1\left( 0 \right)-1\left( 1 \right)+0\left( -1 \right) & 1\left( 1 \right)-1\left( 3 \right)+0\left( 0 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   5 & -1 & 2  \\   9 & -2 & 5  \\   0 & -1 & -2  \\ \end{matrix} \right]\]

\[\therefore {{A}^{2}}-5A+6I=\left[ \begin{matrix}   5 & -1 & 2  \\   9 & -2 & 5  \\   0 & -1 & -2  \\ \end{matrix} \right]-5\left[ \begin{matrix}   2 & 0 & 1  \\   2 & 1 & 3  \\   1 & -1 & 0  \\ \end{matrix} \right]+6\left[ \begin{matrix}   1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix}   5 & -1 & 2  \\   9 & -2 & 5  \\   0 & -1 & -2  \\ \end{matrix} \right]-\left[ \begin{matrix}   10 & 0 & 5  \\   10 & 5 & 15  \\   5 & -5 & 0  \\ \end{matrix} \right]+\left[ \begin{matrix}   6 & 0 & 0  \\   0 & 6 & 0  \\   0 & 0 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix}   5-10 & -1-0 & 2-5  \\   9-10 & -2-5 & 5-15  \\   0-5 & -1+5 & -2-0  \\ \end{matrix} \right]+\left[ \begin{matrix}   6 & 0 & 0  \\   0 & 6 & 0  \\   0 & 0 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix}   -5 & -1 & -3  \\   -1 & -7 & -10 \\   -5 & 4 & -2  \\ \end{matrix} \right]+\left[ \begin{matrix}   6 & 0 & 0  \\   0 & 6 & 0  \\   0 & 0 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix}   1 & -1 & -3  \\   -1 & -1 & -10 \\   -5 & 4 & 4  \\ \end{matrix} \right]\]

16. If \[\mathbf{A=\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]}\] , prove that \[\mathbf{{{A}^{3}}-6{{A}^{2}}+7A+21=0}\] .

Ans: Given: \[A=\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\]

\[{{A}^{2}}=AA\]

\[\therefore {{A}^{2}}=\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   1+0+4 & 0+0+0 & 2+0+6  \\   0+0+2 & 0+4+0 & 0+2+3  \\   2+0+6 & 0+0+0 & 4+0+9  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   5 & 0 & 8  \\   2 & 4 & 5  \\   8 & 0 & 13  \\ \end{matrix} \right]\]

Now, \[{{A}^{3}}={{A}^{2}}\cdot A\]

\[\therefore {{A}^{3}}=\left[ \begin{matrix}   5 & 0 & 8  \\   2 & 4 & 5  \\   8 & 0 & 13  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{3}}=\left[ \begin{matrix}   5+0+16 & 0+0+0 & 10+0+24  \\   2+0+10 & 0+8+0 & 4+4+15  \\   8+0+26 & 0+0+0 & 16+0+39  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{3}}=\left[ \begin{matrix}   21 & 0 & 34  \\   12 & 8 & 23  \\   34 & 0 & 55  \\ \end{matrix} \right]\]

\[\therefore {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix}   21 & 0 & 34  \\   12 & 8 & 23  \\   34 & 0 & 55  \\ \end{matrix} \right]-6\left[ \begin{matrix}   5 & 0 & 8  \\   2 & 4 & 5  \\   8 & 0 & 13  \\ \end{matrix} \right]+7\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]+2\left[ \begin{matrix}   1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix}   21 & 0 & 34  \\   12 & 8 & 23  \\   34 & 0 & 55  \\ \end{matrix} \right]-\left[ \begin{matrix}   30 & 0 & 48  \\   12 & 24 & 30  \\   48 & 0 & 78  \\ \end{matrix} \right]+\left[ \begin{matrix}   7 & 0 & 14  \\   0 & 14 & 7  \\   14 & 0 & 21  \\ \end{matrix} \right]+\left[ \begin{matrix}   2 & 0 & 0  \\   0 & 2 & 0  \\   0 & 0 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix}   21-30+7+2 & 0-0+0+0 & 34-48+14+0  \\   12-12+0+0 & 8-24+14+2 & 23-30+7+0  \\   34-48+14+0 & 0-0+0+0 & 55-78+21+2  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

\[\therefore {{A}^{3}}-6{{A}^{2}}+7A+21=0\]

Hence proved.

17.  If \[\mathbf{A=\left[ \begin{matrix}   3 & -2  \\   4 & -2  \\ \end{matrix} \right]}\] and \[\mathbf{I=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]}\] , find \[\mathbf{k}\] so that \[\mathbf{{{A}^{2}}=kA-21}\] .

Ans: Given:\[A=\left[ \begin{matrix}   3 & -2  \\   4 & -2  \\ \end{matrix} \right]\]

\[{{A}^{2}}=AA\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   3 & -2  \\   4 & -2  \\ \end{matrix} \right]\left[ \begin{matrix}   3 & -2  \\   4 & -2  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   9-8 & -6+4  \\   12-8 & -8+4  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   1 & -2  \\   4 & -4  \\ \end{matrix} \right]\]

Now, \[{{A}^{2}}=kA-21\]

\[\Rightarrow \left[ \begin{matrix}   1 & -2  \\   4 & -4  \\ \end{matrix} \right]=k\left[ \begin{matrix}   3 & -2  \\   4 & -2  \\ \end{matrix} \right]-2\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   1 & -2  \\   4 & -4  \\ \end{matrix} \right]=\left[ \begin{matrix}   3k & -2k  \\   4k & -2k  \\ \end{matrix} \right]-\left[ \begin{matrix}   2 & 0  \\   0 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   1 & -2  \\   4 & -4  \\ \end{matrix} \right]=\left[ \begin{matrix}   3k-2 & -2k  \\   4k & -2k-2  \\ \end{matrix} \right]\]

Equating the corresponding elements, we have:

\[3k-2=1\]

\[\Rightarrow k=1\]

Thus, the value of \[k\] is \[1\].

18. If \[\mathbf{A=\left[ \begin{matrix}   0 & -\tan \dfrac{\alpha }{2}  \\   \tan \dfrac{\alpha }{2} & 0  \\ \end{matrix} \right]}\] and \[I\] is the identity matrix of order \[2\] , show that \[\mathbf{I+A=\left( I-A \right)\left[ \begin{matrix}    \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]}\]

Ans: Given: \[A=\left[ \begin{matrix}   0 & -\tan \dfrac{\alpha }{2}  \\   \tan \dfrac{\alpha }{2} & 0  \\ \end{matrix} \right]\]

LHS: \[I+A=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   0 & -\tan \dfrac{\alpha }{2}  \\   \tan \dfrac{\alpha }{2} & 0  \\ \end{matrix} \right]\]

\[\Rightarrow I+A=\left[ \begin{matrix}   1 & -\tan \dfrac{\alpha }{2}  \\   \tan \dfrac{\alpha }{2} & 1  \\ \end{matrix} \right]\]                                           …(1)

RHS: \[\left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left( \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]-\left[ \begin{matrix}   0 & -\tan \dfrac{\alpha }{2}  \\   \tan \dfrac{\alpha }{2} & 0  \\ \end{matrix} \right] \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow \left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}  1 & \tan \dfrac{\alpha }{2}  \\   -\tan \dfrac{\alpha }{2} & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow \left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\    \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   \cos \alpha +\sin \alpha \tan \dfrac{\alpha }{2} & -\sin \alpha +\cos \alpha \tan \dfrac{\alpha }{2}  \\   -\cos \alpha \tan \dfrac{\alpha }{2}+\sin \alpha  & \sin \alpha \tan \dfrac{\alpha }{2}+\cos \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow \left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   1-2{{\sin }^{2}}\dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}-\cos \dfrac{\alpha }{2}\tan \dfrac{\alpha }{2} & -2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}+\left( 2{{\cos }^{2}}\dfrac{\alpha }{2}-1 \right)\tan \dfrac{\alpha }{2}  \\   -\left( 2{{\cos }^{2}}\dfrac{\alpha }{2}-1 \right)\tan \dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2} & 2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}\tan \dfrac{\alpha }{2}+1-2{{\sin }^{2}}\dfrac{\alpha }{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   1-2{{\sin }^{2}}\dfrac{\alpha }{2}+2{{\sin }^{2}}\dfrac{\alpha }{2} & -2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}-\tan \dfrac{\alpha }{2}  \\   -\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}+\tan \dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2} & 2{{\sin }^{2}}\dfrac{\alpha }{2}+1-2{{\sin }^{2}}\dfrac{\alpha }{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\    \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & -\tan \dfrac{\alpha }{2}  \\    \tan \dfrac{\alpha }{2} & 1  \\ \end{matrix} \right]\]                                        …(2)

Thus, from equation (1) and (2).

\[LHS=RHS\]

Hence proved.

19. A trust fund has Rs \[30,000\] that must be invested in two different types of bonds. The first bond pays \[5%\] interest per year, and the second bond pays \[7%\] interest per year. Using matrix multiplication, determine how to divide Rs \[30,000\] among the two types of bonds. If the trust fund must obtain an annual total interest of:

i. Rs \[\mathbf{1,800}\]

Ans: Let Rs \[x\] be invested in the first round.

Then, the sum of money invested in the second bond pays Rs \[\left( 30000-x \right)\]

It is given that the first bond pays \[5%\] interest per year, and the second bond pays \[7%\] interest per year.

We know that,

Simple interest for one year is \[\dfrac{principle\times rate}{100}\] .

Therefore, in order to obtain an annual total interest of Rs \[1,800\] ,

\[\left[ \begin{matrix}   x & \left( 30000-x \right)  \\ \end{matrix} \right]\left[ \begin{matrix}   \dfrac{5}{100}  \\   \dfrac{7}{100}  \\ \end{matrix} \right]=1800\]

\[\Rightarrow \dfrac{5x}{100}+\dfrac{7\left( 30000-x \right)}{100}=1800\]

\[\Rightarrow 210000-2x=180000\]

\[\Rightarrow x=15000\]

Thus, in order to obtain an annual total interest of Rs \[1,800\] , the trust fund should invest Rs \[15000\] in the first bond and the remaining Rs \[15000\] in the second bond.

ii. Rs \[\mathbf{2,000}\]

Ans: Let Rs \[x\] be invested in the first round.

Then, the sum of money invested in the second bond pays Rs \[\left( 30000-x \right)\]

It is given that the first bond pays \[5%\] interest per year, and the second bond pays \[7%\] interest per year.

We know that,

Simple interest for one year is \[\dfrac{principle\times rate}{100}\] .

Therefore, in order to obtain an annual total interest of Rs \[2,000\] ,

\[\left[ \begin{matrix}   x & \left( 30000-x \right)  \\ \end{matrix} \right]\left[ \begin{matrix}   \dfrac{5}{100}  \\   \dfrac{7}{100}  \\ \end{matrix} \right]=2000\]

\[\Rightarrow \dfrac{5x}{100}+\dfrac{7\left( 30000-x \right)}{100}=2000\]

\[\Rightarrow 210000-2x=200000\]

\[\Rightarrow x=5000\]

Thus, in order to obtain an annual total interest of Rs \[2,000\] , the trust fund should invest Rs \[5000\] in the first bond and the remaining Rs \[25000\] in the second bond.

20. The bookshop of a particular school has \[10\] dozen chemistry books, \[8\] dozen physics books, \[10\] dozen economics books. Their selling prices are Rs \[80\], Rs \[60\] and Rs \[40\] each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Ans: The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:

\[12\left[ \begin{matrix}    10 & 8 & 10  \\ \end{matrix} \right]\left[ \begin{matrix}   80  \\   60  \\   40  \\ \end{matrix} \right]\]

\[\Rightarrow 12\left[ 10\times 80+8\times 60+10\times 40 \right]\]

\[\Rightarrow 12\left[ 1680 \right]\]

\[\Rightarrow 20160\]

Therefore, the bookshop will receive Rs \[20160\] from the sale.

21. Assume \[X,Y,Z,W\] and \[P\] are the matrices of order \[\mathbf{2\times n,3\times k,2\times p,n\times 3}\] and \[\mathbf{p\times k}\] respectively. The restriction on \[\mathbf{n,k,p}\] so that \[\mathbf{PY+WY}\] will be defined are:

1. \[\mathbf{k=3,p=n}\]

2. \[\mathbf{k}\] is arbitrary, \[\mathbf{p=2}\]

3. \[\mathbf{p}\] is arbitrary, \[\mathbf{k=3}\]

4. \[\mathbf{k=2,p=3}\]

Ans: Matrices \[P\] and \[Y\] are of the orders \[p\times k\] and \[3\times k\] respectively.

Therefore, the matrix \[PY\] will be defined if \[k=3\] .

Also, \[PY\] will be of order \[p\times k\] .

Since the number of columns in matrix \[W\] is equal to the number of rows in matrix \[Y\] .

Therefore, matrix \[WY\] is well-defined and is of the order \[n\times k\] .

Moreover, Matrices \[PY\] and \[WY\] can be added only when their orders are the same.

But \[PY\] is of the order \[p\times k\] and \[WY\] is of the order \[n\times k\] .

Therefore, we must have \[p=n\] .

\[\therefore p=n,k=3\] are the restrictions on \[n,k,p\] so that \[PY+WY\] will be defined.

Thus, option (A) is correct.

22. Assume \[\mathbf{X,Y,Z,W}\] and \[\mathbf{P}\] are matrices of order \[\mathbf{2\times n,3\times k,2\times p,n\times 3}\] and \[\mathbf{p\times k}\] respectively. If \[\mathbf{p=n}\] , then the order of the matrix \[\mathbf{7X-5Z}\] is:

1. \[\mathbf{p\times 2}\]

2. \[\mathbf{2\times n}\]

3. \[\mathbf{n\times 3}\]

4. \[\mathbf{p\times n}\]

Ans: Matrix \[X\] is of the order \[2\times n\] .

Thus, matrix \[7X\] is also of the same order.

Since, \[p=n\]

Matrix \[Z\] is of the order \[2\times p\] or \[2\times n\] .

Thus, matrix \[5Z\] is also of the same order.

Since, both the matrices \[7X\] and \[5Z\] are of the same order \[2\times n\].

Therefore, \[7X-5Z\] is well defined and is of the order \[2\times n\] .

Thus, option (B) is correct.

Exercise 3.3

1. Find the transpose of each of the following matrices:

i. \[\left[ \begin{matrix}   5  \\   \dfrac{1}{2}  \\   -1  \\ \end{matrix} \right]\]

Ans: The transpose of a matrix is obtained by changing its rows into columns and its columns into rows.

Thus, if \[A=\left[ \begin{matrix}  5  \\   \dfrac{1}{2}  \\   -1  \\ \end{matrix} \right]\] , then \[{{A}^{T}}=\left[ \begin{matrix}   5 & \dfrac{1}{2} & -1  \\ \end{matrix} \right]\]

ii. \[\left[ \begin{matrix}   1 & -1  \\   2 & 3  \\ \end{matrix} \right]\]

Ans: The transpose of a matrix is obtained by changing its rows into columns and its columns into rows.

Thus, if \[A=\left[ \begin{matrix}  1 & -1  \\   2 & 3  \\ \end{matrix} \right]\] , then \[{{A}^{T}}=\left[ \begin{matrix}   1 & 2  \\   -1 & 3  \\ \end{matrix} \right]\]

iii. \[\left[ \begin{matrix}   -1 & 5 & 6  \\   \sqrt{3} & 5 & 6  \\    2 & 3 & -1  \\ \end{matrix} \right]\]

Ans: The transpose of a matrix is obtained by changing its rows into columns and its columns into rows.

Thus, if \[A=\left[ \begin{matrix}   -1 & 5 & 6  \\   \sqrt{3} & 5 & 6  \\   2 & 3 & -1  \\ \end{matrix} \right]\] , then \[{{A}^{T}}=\left[ \begin{matrix}   -1 & \sqrt{3} & 2  \\   5 & 5 & 3  \\   6 & 6 & -1  \\ \end{matrix} \right]\]

2. If \[A=\left[ \begin{matrix}   -1 & 2 & 3  \\   5 & 7 & 9  \\   -2 & 1 & 1  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   -4 & 1 & -5  \\   1 & 2 & 0  \\   1 & 3 & 1  \\ \end{matrix} \right]\] , then verify that

i. \[\left( A+B \right)'=A'+B'\]

Ans: Given that: \[A=\left[ \begin{matrix}   -1 & 2 & 3  \\   5 & 7 & 9  \\   -2 & 1 & 1  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   -4 & 1 & -5  \\   1 & 2 & 0  \\   1 & 3 & 1  \\ \end{matrix} \right]\] 

Thus, we have \[A'=\left[ \begin{matrix}   -1 & 5 & -2  \\   2 & 7 & 1  \\   3 & 9 & 1  \\ \end{matrix} \right]\] and \[B'=\left[ \begin{matrix}   -4 & 1 & 1  \\   1 & 2 & 3  \\   -5 & 0 & 1  \\ \end{matrix} \right]\]

\[\left( A+B \right)=\left[ \begin{matrix}   -1 & 2 & 3  \\   5 & 7 & 9  \\   -2 & 1 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   -4 & 1 & -5  \\   1 & 2 & 0  \\   1 & 3 & 1  \\ \end{matrix} \right]\]

\[\therefore \left( A+B \right)=\left[ \begin{matrix}   -5 & 3 & -2  \\   6 & 9 & 9  \\   -1 & 4 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow \left( A+B \right)'=\left[ \begin{matrix}   -5 & 6 & -1  \\   3 & 9 & 4  \\   -2 & 9 & 2  \\ \end{matrix} \right]\]                                   …(1)

And \[A'+B'=\left[ \begin{matrix}   -1 & 5 & -2  \\   2 & 7 & 1  \\   3 & 9 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   -4 & 1 & 1  \\   1 & 2 & 3  \\   -5 & 0 & 1  \\ \end{matrix} \right]\]

\[\therefore A'+B'=\left[ \begin{matrix}   -5 & 6 & -1  \\   3 & 9 & 4  \\   -2 & 9 & 2  \\ \end{matrix} \right]\]                                         …(2)

Thus, from equation (1) and (2),

\[\left( A+B \right)'=A'+B'\]

Hence proved.

ii. \[\left( A-B \right)'=A'-B'\]

Ans: Given that: \[A=\left[ \begin{matrix}   -1 & 2 & 3  \\   5 & 7 & 9  \\   -2 & 1 & 1  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   -4 & 1 & -5  \\   1 & 2 & 0  \\   1 & 3 & 1  \\ \end{matrix} \right]\] 

Thus, we have \[A'=\left[ \begin{matrix}   -1 & 5 & -2  \\   2 & 7 & 1  \\   3 & 9 & 1  \\ \end{matrix} \right]\] and \[B'=\left[ \begin{matrix}   -4 & 1 & 1  \\   1 & 2 & 3  \\   -5 & 0 & 1  \\ \end{matrix} \right]\]

\[\left( A-B \right)=\left[ \begin{matrix}   -1 & 2 & 3  \\   5 & 7 & 9  \\   -2 & 1 & 1  \\ \end{matrix} \right]-\left[ \begin{matrix}  -4 & 1 & -5  \\   1 & 2 & 0  \\   1 & 3 & 1  \\ \end{matrix} \right]\]

\[\therefore \left( A-B \right)=\left[ \begin{matrix}   3 & 1 & 8  \\   4 & 5 & 9  \\   -3 & -2 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow \left( A-B \right)'=\left[ \begin{matrix}   3 & 4 & -3  \\   1 & 5 & -2 \\   8 & 9 & 0  \\ \end{matrix} \right]\]                                   …(1)

And \[A'-B'=\left[ \begin{matrix}   -1 & 5 & -2  \\   2 & 7 & 1  \\   3 & 9 & 1  \\ \end{matrix} \right]-\left[ \begin{matrix}   -4 & 1 & 1  \\   1 & 2 & 3  \\   -5 & 0 & 1  \\ \end{matrix} \right]\]

\[\therefore A'-B'=\left[ \begin{matrix}   3 & 4 & -3  \\   1 & 5 & -2  \\   8 & 9 & 0  \\ \end{matrix} \right]\]                                         …(2)

Thus, from equation (1) and (2),

\[\left( A-B \right)'=A'-B'\]

Hence proved.

3. If \[A'=\left[ \begin{matrix}   3 & 4  \\   -1 & 2  \\   0 & 1  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   -1 & 2 & 1  \\   1 & 2 & 3  \\ \end{matrix} \right]\] , then verify that

i. \[\left( A+B \right)'=A'+B'\]

Ans: We know that, \[A=\left( A' \right)'\]

Thus, we have: \[A=\left[ \begin{matrix}   3 & -1 & 0  \\   4 & 2 & 1  \\ \end{matrix} \right]\]

And \[B'=\left[ \begin{matrix}   -1 & 1  \\   2 & 2  \\   1 & 3  \\ \end{matrix} \right]\]

\[\therefore A+B=\left[ \begin{matrix}   3 & -1 & 0  \\   4 & 2 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   -1 & 2 & 1  \\   1 & 2 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow A+B=\left[ \begin{matrix}   2 & 1 & 1  \\   5 & 4 & 4  \\ \end{matrix} \right]\]

\[\therefore \left( A+B \right)'=\left[ \begin{matrix}   2 & 5  \\   1 & 4  \\   1 & 4  \\ \end{matrix} \right]\]                                …(1)

\[A'+B'=\left[ \begin{matrix}   3 & 4  \\   -1 & 2  \\   0 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   -1 & 1  \\   2 & 2  \\   1 & 3  \\ \end{matrix} \right]\]

\[\therefore A'+B'=\left[ \begin{matrix}   2 & 5  \\   1 & 4  \\   1 & 4  \\ \end{matrix} \right]\]                                      …(2)

Thus, from equation (1) and (2),

\[\left( A+B \right)'=A'+B'\]

Hence proved.

ii. \[\left( A-B \right)'=A'-B'\]

Ans: We know that, \[A=\left( A' \right)'\]

Thus, we have: \[A=\left[ \begin{matrix}  3 & -1 & 0  \\   4 & 2 & 1  \\ \end{matrix} \right]\]

And \[B'=\left[ \begin{matrix}   -1 & 1  \\   2 & 2  \\   1 & 3  \\ \end{matrix} \right]\]

\[\therefore A-B=\left[ \begin{matrix}   3 & -1 & 0  \\   4 & 2 & 1  \\ \end{matrix} \right]-\left[ \begin{matrix}   -1 & 2 & 1  \\   1 & 2 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow A-B=\left[ \begin{matrix}   4 & -3 & -1  \\   3 & 0 & -2  \\ \end{matrix} \right]\]

\[\therefore \left( A-B \right)'=\left[ \begin{matrix}   4 & 3  \\   -3 & 0  \\   -1 & -2  \\ \end{matrix} \right]\]                                …(1)

\[A'-B'=\left[ \begin{matrix}   3 & 4  \\   -1 & 2  \\   0 & 1  \\ \end{matrix} \right]-\left[ \begin{matrix}   -1 & 1  \\   2 & 2  \\   1 & 3  \\ \end{matrix} \right]\]

\[\therefore A'-B'=\left[ \begin{matrix}   4 & 3  \\   -3 & 0  \\   -1 & -2  \\ \end{matrix} \right]\]                                      …(2)

Thus, from equation (1) and (2),

\[\left( A-B \right)'=A'-B'\]

Hence proved.

4. If \[A'=\left[ \begin{matrix}   -2 & 3  \\   1 & 2  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   -1 & 0  \\   1 & 2  \\ \end{matrix} \right]\] , then ( A+2B \right)' .

Ans: We know that, \[A=\left( A' \right)'\]

\[\therefore A=\left[ \begin{matrix}   -2 & 1  \\   3 & 2  \\ \end{matrix} \right]\]

\[\therefore A+2B=\left[ \begin{matrix}   -2 & 1  \\   3 & 2  \\ \end{matrix} \right]+2\left[ \begin{matrix}   -1 & 0  \\   1 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow A+2B=\left[ \begin{matrix}   -2 & 1  \\   3 & 2  \\ \end{matrix} \right]+\left[ \begin{matrix}   -2 & 0  \\   2 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow A+2B=\left[ \begin{matrix}   -4 & 1  \\   5 & 6  \\ \end{matrix} \right]\]

\[\therefore \left( A+2B \right)'=\left[ \begin{matrix}   -4 & 5  \\   1 & 6  \\ \end{matrix} \right]\]

5. For the matrices \[A\] and \[B\] , verify that \[\left( AB \right)'=B'A'\] where

i. \[A=\left[ \begin{matrix}   1  \\   -4  \\   3  \\ \end{matrix} \right],B=\left[ \begin{matrix}    -1 & 2 & 1  \\ \end{matrix} \right]\]

Ans: LHS: \[AB=\left[ \begin{matrix}   1  \\   -4  \\   3  \\ \end{matrix} \right]\left[ \begin{matrix}   -1 & 2 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow AB=\left[ \begin{matrix}   -1 & 2 & 1  \\   4 & -8 & -4  \\   -3 & 6 & 3  \\ \end{matrix} \right]\]

\[\therefore \left( AB \right)'=\left[ \begin{matrix}   -1 & 4 & -3  \\   2 & -8 & 6 \\   1 & -4 & 3  \\ \end{matrix} \right]\]                                 …(1)

Now, \[A'=\left[ \begin{matrix}   1 & -4 & 3  \\ \end{matrix} \right]\]

And \[B'=\left[ \begin{matrix}   -1  \\   2  \\   1  \\ \end{matrix} \right]\]

RHS: \[\therefore B'A'=\left[ \begin{matrix}   -1  \\   2  \\   1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & -4 & 3  \\ \end{matrix} \right]\]                      

 \[\therefore B'A'=\left[ \begin{matrix}   -1 & 4 & -3  \\   2 & -8 & 6  \\   1 & -4 & 3  \\ \end{matrix} \right]\]                                     …(2)

Thus, from equation (1) and (2),

\[\left( AB \right)'=B'A'\]

Hence proved.

ii. \[A=\left[ \begin{matrix}   0  \\   1  \\ 2  \\ \end{matrix} \right], B=\left[ \begin{matrix}   1 & 5 & 7  \\ \end{matrix} \right]\]

Ans: LHS: \[AB=\left[ \begin{matrix}   0  \\   1  \\   2  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 5 & 7  \\ \end{matrix} \right]\]

\[\Rightarrow AB=\left[ \begin{matrix}   0 & 1 & 2  \\   0 & 5 & 10  \\   0 & 7 & 14  \\ \end{matrix} \right]\]

\[\therefore \left( AB \right)'=\left[ \begin{matrix}   0 & 1 & 2  \\   0 & 5 & 10  \\   0 & 7 & 14  \\ \end{matrix} \right]\]                                 …(1)

Now, \[A'=\left[ \begin{matrix}   0 & 1 & 2  \\ \end{matrix} \right]\]

And \[B'=\left[ \begin{matrix}   1  \\   5  \\   7  \\ \end{matrix} \right]\]

RHS: \[\therefore B'A'=\left[ \begin{matrix}   1  \\   5  \\   7  \\ \end{matrix} \right]\left[ \begin{matrix}   0 & 1 & 2  \\ \end{matrix} \right]\]

\[\therefore B'A'=\left[ \begin{matrix}   0 & 1 & 2  \\   0 & 5 & 10  \\   0 & 7 & 14  \\ \end{matrix} \right]\]                                  …(2)

Thus, from equation (1) and (2),

\[\left( AB \right)'=B'A'\]

Hence proved.

6. If 

i. \[A=\left[ \begin{matrix}   \cos \alpha  & \sin \alpha   \\   -\sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\] , then verify that \[A'A=I\]

Ans: Here, \[A'=\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

\[\therefore A'A=\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\left[ \begin{matrix}   \cos \alpha  & \sin \alpha   \\   -\sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow A'A=\left[ \begin{matrix}   {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha  & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha   \\   \sin \alpha \cos \alpha -\sin \alpha \cos \alpha  & {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha   \\ \end{matrix} \right]\]

\[\therefore A'A=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

\[\therefore A'A=I\]

Hence proved.

ii. \[A=\left[ \begin{matrix}   \sin \alpha  & \cos \alpha   \\   -\cos \alpha  & \sin \alpha   \\ \end{matrix} \right]\] , then verify that \[A'A=I\]

Ans: Here, \[A'=\left[ \begin{matrix}   \sin \alpha  & -\cos \alpha   \\   \cos \alpha  & \sin \alpha   \\ \end{matrix} \right]\]

\[\therefore A'A=\left[ \begin{matrix}   \sin \alpha  & -\cos \alpha   \\    \cos \alpha  & \sin \alpha   \\ \end{matrix} \right]\left[ \begin{matrix}   \sin \alpha  & \cos \alpha   \\   -\cos \alpha  & \sin \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow A'A=\left[ \begin{matrix}   {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha  & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha   \\   \sin \alpha \cos \alpha -\sin \alpha \cos \alpha  & {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha   \\ \end{matrix} \right]\]

\[\therefore A'A=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

\[\therefore A'A=I\]

Hence proved.

7. Show that

i. The matrix \[A=\left[ \begin{matrix}   1 & -1 & 5  \\   -1 & 2 & 1  \\   5 & 1 & 3  \\ \end{matrix} \right]\] is a symmetric matrix.

Ans: Given \[A=\left[ \begin{matrix}   1 & -1 & 5  \\   -1 & 2 & 1  \\   5 & 1 & 3  \\ \end{matrix} \right]\]

\[\therefore A'=\left[ \begin{matrix}   1 & -1 & 5  \\   -1 & 2 & 1  \\   5 & 1 & 3  \\ \end{matrix} \right]\]

Here, we have \[A'=A\]

Thus, \[A\] is a symmetric matrix.

ii. The matrix \[A=\left[ \begin{matrix}   0 & 1 & -1  \\   -1 & 0 & 1  \\   1 & -1 & 0  \\ \end{matrix} \right]\] is a skew symmetric matrix.

Ans: Given \[A=\left[ \begin{matrix}   0 & 1 & -1  \\   -1 & 0 & 1  \\   1 & -1 & 0  \\ \end{matrix} \right]\]

\[\therefore A'=\left[ \begin{matrix}   0 & -1 & 1  \\   1 & 0 & -1  \\   -1 & 1 & 0 \\ \end{matrix} \right]\]

\[\Rightarrow A'=-\left[ \begin{matrix}   0 & 1 & -1  \\   -1 & 0 & 1  \\   1 & -1 & 0  \\ \end{matrix} \right]\]

Here, we have \[A'=-A\]

Thus, \[A\] is a skew symmetric matrix.

8. For the matrix \[A=\left[ \begin{matrix}   1 & 5  \\   6 & 7  \\ \end{matrix} \right]\] , verify that

i. \[\left( A+A' \right)\] is a symmetric matrix.

Ans: Given \[A=\left[ \begin{matrix}   1 & 5  \\   6 & 7  \\ \end{matrix} \right]\]

\[\therefore A'=\left[ \begin{matrix}   1 & 6  \\   5 & 7  \\ \end{matrix} \right]\]

\[A+A'=\left[ \begin{matrix}   1 & 5  \\   6 & 7  \\ \end{matrix} \right]+\left[ \begin{matrix}   1 & 6  \\   5 & 7  \\ \end{matrix} \right]\]

\[\Rightarrow A+A'=\left[ \begin{matrix}   2 & 11  \\   11 & 14  \\ \end{matrix} \right]\]

\[\left( A+A' \right)'=\left[ \begin{matrix}   2 & 11  \\   11 & 14  \\ \end{matrix} \right]\]

Here, we have \[A+A'=\left( A+A' \right)'\]

Thus, \[A+A'\] is a symmetric matrix.

ii. \[\left( A-A' \right)\] is a skew symmetric matrix.

Ans: Given \[A=\left[ \begin{matrix}   1 & 5  \\   6 & 7  \\ \end{matrix} \right]\]

\[\therefore A'=\left[ \begin{matrix}   1 & 6  \\   5 & 7  \\ \end{matrix} \right]\]

\[A-A'=\left[ \begin{matrix}   1 & 5  \\   6 & 7  \\ \end{matrix} \right]-\left[ \begin{matrix}   1 & 6  \\   5 & 7  \\ \end{matrix} \right]\]

\[\Rightarrow A-A'=\left[ \begin{matrix}   0 & -1  \\   1 & 0  \\ \end{matrix} \right]\]

\[\left( A-A' \right)'=\left[ \begin{matrix}   0 & 1  \\   -1 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow \left( A-A' \right)'=-\left[ \begin{matrix}   0 & -1  \\   1 & 0  \\ \end{matrix} \right]\]

Here, we have \[A-A'=-\left( A-A' \right)'\]

Thus, \[A-A'\] is a skew symmetric matrix.

9. Find \[\dfrac{1}{2}\left( A+A' \right)\] and \[\dfrac{1}{2}\left( A-A' \right)\] , when \[A=\left[ \begin{matrix}   0 & a & b  \\   -a & 0 & c  \\   -b & -c & 0  \\ \end{matrix} \right]\] .

Ans: Given \[A=\left[ \begin{matrix}   0 & a & b  \\   -a & 0 & c  \\   -b & -c & 0  \\ \end{matrix} \right]\]

\[\therefore A'=\left[ \begin{matrix}   0 & -a & -b  \\   a & 0 & -c  \\   b & c & 0  \\ \end{matrix} \right]\]

\[A+A'=\left[ \begin{matrix}   0 & a & b  \\   -a & 0 & c  \\   -b & -c & 0  \\ \end{matrix} \right]+\left[ \begin{matrix}   0 & -a & -b  \\   a & 0 & -c  \\   b & c & 0 \\ \end{matrix} \right]\]

\[\Rightarrow A+A'=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

\[\therefore \dfrac{1}{2}\left( A+A' \right)=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

Now, \[A-A'=\left[ \begin{matrix}   0 & a & b  \\   -a & 0 & c  \\   -b & -c & 0  \\ \end{matrix} \right]-\left[ \begin{matrix}   0 & -a & -b  \\   a & 0 & -c  \\   b & c & 0  \\ \end{matrix} \right]\]

\[\left( A-A' \right)=\left[ \begin{matrix}   0 & 2a & 2b  \\  -2a & 0 & 2c  \\   -2b & -2c & 0  \\ \end{matrix} \right]\]

\[\therefore \dfrac{1}{2}\left( A-A' \right)=\left[ \begin{matrix}   0 & a & b  \\   -a & 0 & c  \\   -b & -c & 0  \\ \end{matrix} \right]\]

10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

i. \[\left[ \begin{matrix}   3 & 5  \\   1 & -1  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}  3 & 5  \\   1 & -1  \\ \end{matrix} \right]\] and \[A'=\left[ \begin{matrix}   3 & 1  \\   5 & -1  \\ \end{matrix} \right]\]

Now, \[A+A'=\left[ \begin{matrix}   3 & 5  \\   1 & -1  \\ \end{matrix} \right]+\left[ \begin{matrix}   3 & 1  \\   5 & -1  \\ \end{matrix} \right]\]

\[\Rightarrow A+A'=\left[ \begin{matrix}   6 & 6  \\   6 & -2  \\ \end{matrix} \right]\]

Let \[P=\dfrac{1}{2}\left( A+A' \right)\]

\[\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix}  6 & 6  \\   6 & -2  \\ \end{matrix} \right]\]

\[\Rightarrow P=\left[ \begin{matrix}   3 & 3  \\   3 & -1  \\ \end{matrix} \right]\]

Now, \[P'=\left[ \begin{matrix}   3 & 3  \\   3 & -1  \\ \end{matrix} \right]\]

Here, \[P=P'\]

\[\therefore P=\dfrac{1}{2}\left( A+A' \right)\] is a symmetric matrix.

Now, \[A-A'=\left[ \begin{matrix}   3 & 5  \\   1 & -1  \\ \end{matrix} \right]-\left[ \begin{matrix}   3 & 1  \\   5 & -1  \\ \end{matrix} \right]\]

\[\Rightarrow A-A'=\left[ \begin{matrix}   0 & 4  \\   -4 & 0  \\ \end{matrix} \right]\]

Let \[Q=\dfrac{1}{2}\left( A-A' \right)\]

\[\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix}   0 & 4  \\   -4 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow Q=\left[ \begin{matrix}   0 & 2  \\   -2 & 0  \\ \end{matrix} \right]\]

Now, \[Q'=\left[ \begin{matrix}   0 & 2  \\   -2 & 0  \\ \end{matrix} \right]\]

Here, \[Q=-Q'\]

\[\therefore Q=\dfrac{1}{2}\left( A-A' \right)\] is a skew symmetric matrix.

Thus, \[A\]is the sum of matrices \[P=\left[ \begin{matrix}   3 & 3  \\   3 & -1  \\ \end{matrix} \right]\] and \[Q=\left[ \begin{matrix}   0 & 2  \\   -2 & 0  \\ \end{matrix} \right]\].

ii. \[\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\] and \[A'=\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1 \\   2 & -1 & 3  \\ \end{matrix} \right]\]

Now, \[A+A'=\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]+\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\]

Let \[P=\dfrac{1}{2}\left( A+A' \right)\]

\[\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix}   12 & -4 & 4  \\   -4 & 6 & -2 \\   4 & -2 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow P=\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\]

Now, \[P'=\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\]

Here, \[P=P'\]

\[\therefore P=\dfrac{1}{2}\left( A+A' \right)\] is a symmetric matrix.

Now, \[A-A'=\left[ \begin{matrix}  6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]-\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow A-A'=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

Let \[Q=\dfrac{1}{2}\left( A-A' \right)\]

\[\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow Q=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0 \\ \end{matrix} \right]\]

Now, \[Q'=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

Here, \[Q=-Q'\]

\[\therefore Q=\dfrac{1}{2}\left( A-A' \right)\] is a skew symmetric matrix.

Thus, \[A\]is the sum of matrices \[P=\left[ \begin{matrix}  6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\] and \[Q=\left[ \begin{matrix}   0 & 0 & 0  \\  0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\].

iii. \[\left[ \begin{matrix}   3 & 3 & -1  \\   -2 & -2 & 1  \\   -4 & -5 & 2  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   3 & 3 & -1  \\   -2 & -2 & 1  \\   -4 & -5 & 2  \\ \end{matrix} \right]\] and \[A'=\left[ \begin{matrix}   3 & -2 & -4  \\   3 & -2 & -5  \\   -1 & 1 & 2  \\ \end{matrix} \right]\]

Now, \[A+A'=\left[ \begin{matrix}   3 & 3 & -1  \\   -2 & -2 & 1  \\   -4 & -5 & 2  \\ \end{matrix} \right]+\left[ \begin{matrix}   3 & -2 & -4  \\   3 & -2 & -5  \\   -1 & 1 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow A+A'=\left[ \begin{matrix}   6 & 1 & -5  \\   1 & -4 & -4  \\   -5 & -4 & 4  \\ \end{matrix} \right]\]

Let \[P=\dfrac{1}{2}\left( A+A' \right)\]

\[\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix}   6 & 1 & -5  \\   1 & -4 & -4 \\   -5 & -4 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow P=\left[ \begin{matrix}   3 & \dfrac{1}{2} & \dfrac{-5}{2}  \\   \dfrac{1}{2} & -2 & -2  \\   \dfrac{-5}{2} & -2 & 2  \\ \end{matrix} \right]\]

Now, \[P'=\left[ \begin{matrix}   3 & \dfrac{1}{2} & \dfrac{-5}{2}  \\  \dfrac{1}{2} & -2 & -2  \\   \dfrac{-5}{2} & -2 & 2  \\ \end{matrix} \right]\]

Here, \[P=P'\]

\[\therefore P=\dfrac{1}{2}\left( A+A' \right)\] is a symmetric matrix.

Now, \[A-A'=\left[ \begin{matrix}   3 & 3 & -1  \\   -2 & -2 & 1  \\   -4 & -5 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix}   3 & -2 & -4  \\   3 & -2 & -5  \\   -1 & 1 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow A-A'=\left[ \begin{matrix}   0 & 5 & 3  \\   -5 & 0 & 6  \\   -3 & -6 & 0  \\ \end{matrix} \right]\]

Let \[Q=\dfrac{1}{2}\left( A-A' \right)\]

\[\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix}   0 & 5 & 3  \\   -5 & 0 & 6  \\   -3 & -6 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow Q=\left[ \begin{matrix}   0 & \dfrac{5}{2} & \dfrac{3}{2}  \\   \dfrac{-5}{2} & 0 & 3  \\   \dfrac{-3}{2} & -3 & 0  \\ \end{matrix} \right]\]

Now, \[Q'=\left[ \begin{matrix}   0 & -\dfrac{5}{2} & -\dfrac{3}{2}  \\ \dfrac{5}{2} & 0 & -3  \\   \dfrac{3}{2} & 3 & 0  \\ \end{matrix} \right]\]

Here, \[Q=-Q'\]

\[\therefore Q=\dfrac{1}{2}\left( A-A' \right)\] is a skew symmetric matrix.

Thus, \[A\]is the sum of matrices \[P=\left[ \begin{matrix}   3 & \dfrac{1}{2} & \dfrac{-5}{2}  \\   \dfrac{1}{2} & -2 & -2  \\  \dfrac{-5}{2} & -2 & 2  \\ \end{matrix} \right]\] and \[Q=\left[ \begin{matrix}  0 & \dfrac{5}{2} & \dfrac{3}{2}  \\    \dfrac{-5}{2} & 0 & 3  \\  \dfrac{-3}{2} & -3 & 0  \\ \end{matrix} \right]\].

iv. \[\left[ \begin{matrix}   1 & 5  \\   -1 & 2  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   1 & 5  \\   -1 & 2  \\ \end{matrix} \right]\] and \[A'=\left[ \begin{matrix}   1 & -1  \\   5 & 2  \\ \end{matrix} \right]\]

Now, \[A+A'=\left[ \begin{matrix}   1 & 5  \\   -1 & 2  \\ \end{matrix} \right]+\left[ \begin{matrix}   1 & -1  \\   5 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow A+A'=\left[ \begin{matrix}   2 & 4  \\   4 & 4  \\ \end{matrix} \right]\]

Let \[P=\dfrac{1}{2}\left( A+A' \right)\]

\[\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix}   2 & 4  \\   4 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow P=\left[ \begin{matrix}   1 & 2  \\   2 & 2  \\ \end{matrix} \right]\]

Now, \[P'=\left[ \begin{matrix}   1 & 2  \\   2 & 2  \\ \end{matrix} \right]\]

Here, \[P=P'\]

\[\therefore P=\dfrac{1}{2}\left( A+A' \right)\] is a symmetric matrix.

Now, \[A-A'=\left[ \begin{matrix}   1 & 5  \\   -1 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix}   1 & -1  \\   5 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow A-A'=\left[ \begin{matrix}   0 & 6  \\   -6 & 0  \\ \end{matrix} \right]\]

Let \[Q=\dfrac{1}{2}\left( A-A' \right)\]

\[\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix}   0 & 6  \\   -6 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow Q=\left[ \begin{matrix}   0 & 3  \\   -3 & 0  \\ \end{matrix} \right]\]

Now, \[Q'=\left[ \begin{matrix}    0 & -3  \\   3 & 0  \\ \end{matrix} \right]\]

Here, \[Q=-Q'\]

\[\therefore Q=\dfrac{1}{2}\left( A-A' \right)\] is a skew symmetric matrix.

Thus, \[A\]is the sum of matrices \[P=\left[ \begin{matrix}   1 & 2  \\   2 & 2  \\ \end{matrix} \right]\] and \[Q=\left[ \begin{matrix}   0 & 3  \\   -3 & 0  \\ \end{matrix} \right]\].

11. If \[A,B\] are symmetric matrices of same order, then \[AB-BA\] is a:

1. Skew symmetric matrix

2. Symmetric matrix

3. Zero matrix

4. Identity matrix

Ans: \[A\] and \[B\] are symmetric matrix, therefore, we have:

\[A'=A\] and \[B'=B\]                                                    …(1)

Here, \[\left( AB-BA \right)'=\left( AB \right)'-\left( BA \right)'\]

\[\Rightarrow \left( AB-BA \right)'=B'A'-A'B'\]

From (1),

\[\Rightarrow \left( AB-BA \right)'=BA-AB\]

\[\Rightarrow \left( AB-BA \right)'=-\left( AB-BA \right)\]                           …(2)

From equation (2),

\[AB-BA\] is a skew symmetric matrix.

12. If \[A=\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\] , then \[\mathbf{A+A'=I}\] , if the value of \[\mathbf{\alpha}\] is

1. \[\dfrac{\pi }{6}\]

2. \[\dfrac{\pi }{3}\]

3. \[n\]

4. \[\dfrac{3\pi }{2}\]

Ans: Given \[A=\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow A'=\left[ \begin{matrix}   \cos \alpha  & \sin \alpha   \\   -\sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

Now, \[\because A+A'=I\]

\[\therefore \left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]+\left[ \begin{matrix}   \cos \alpha  & \sin \alpha   \\    -\sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2\cos \alpha  & 0  \\   0 & 2\cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

Equating the corresponding elements of the two matrices, we have:

\[\cos \alpha =\dfrac{1}{2}\]

\[\Rightarrow \alpha ={{\cos }^{-1}}\left( \dfrac{1}{2} \right)\]

\[\Rightarrow \alpha =\dfrac{\pi }{3}\]

Thus, option (B) is correct.

Miscellaneous Solutions

1. Let \[A=\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\] , show that \[{{\left( aI+bA \right)}^{n}}={{a}^{n}}I+n{{a}^{n-1}}bA\] , where \[I\] is the identity matrix of order \[2}\] and \[n\in N\].

Ans: Given \[A=\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\]

By using the principle of mathematical induction.

For \[n=1\] ,

\[P\left( 1 \right):{{\left( aI+bA \right)}^{1}}={{a}^{1}}I+{{a}^{0}}bA\]

\[\Rightarrow P\left( 1 \right):\left( aI+bA \right)=aI+bA\]

Therefore, the result is true for \[n=1\] .

Let the result be true for \[n=k\] .

That is, \[P\left( k \right):{{\left( aI+bA \right)}^{k}}={{a}^{k}}I+k{{a}^{k-1}}bA\]

Now, we have to prove that the result is true for \[n=k+1\] .

\[{{\left( aI+bA \right)}^{k-1}}={{\left( aI+bA \right)}^{k}}\left( aI+bA \right)\]

\[\Rightarrow {{\left( aI+bA \right)}^{k-1}}=\left( {{a}^{k}}I+k{{a}^{k-1}}bA \right)\left( aI+bA \right)\]

\[\Rightarrow {{\left( aI+bA \right)}^{k-1}}={{a}^{k}}I+\left( k+1 \right){{a}^{k}}bA+k{{a}^{k-1}}{{b}^{2}}{{A}^{2}}\]                        …(1)

Now, 

\[{{A}^{2}}=\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\]

\[{{A}^{2}}=\left[ \begin{matrix}   0 & 0  \\  0 & 0  \\ \end{matrix} \right]\]

\[{{A}^{2}}=0\]

\[\Rightarrow {{\left( aI+bA \right)}^{k-1}}={{a}^{k}}I+\left( k+1 \right){{a}^{k}}bA\]

Thus, the result is true for \[n=k+1\]

Therefore, by the principal of mathematical induction, we have:

\[{{\left( aI+bA \right)}^{n}}={{a}^{n}}I+n{{a}^{n-1}}bA\] where \[A=\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\] , \[n\in N\].

2. If \[A=\left[ \begin{matrix}  1 & 1 & 1  \\   1 & 1 & 1  \\   1 & 1 & 1  \\ \end{matrix} \right]\] , prove that \[{{A}^{n}}=\left[ \begin{matrix}   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\ \end{matrix} \right]\] , \[n\in N\].

Ans: Given \[A=\left[ \begin{matrix}   1 & 1 & 1  \\   1 & 1 & 1  \\   1 & 1 & 1  \\ \end{matrix} \right]\]

By using the principles of mathematical induction.

For \[n=1\] , we have

\[P\left( 1 \right)=\left[ \begin{matrix}   {{3}^{1-1}} & {{3}^{1-1}} & {{3}^{1-1}}  \\   {{3}^{1-1}} & {{3}^{1-1}} & {{3}^{1-1}}  \\   {{3}^{1-1}} & {{3}^{1-1}} & {{3}^{1-1}}  \\ \end{matrix} \right]\]

\[\Rightarrow P\left( 1 \right)=\left[ \begin{matrix}   {{3}^{0}} & {{3}^{0}} & {{3}^{0}}  \\   {{3}^{0}} & {{3}^{0}} & {{3}^{0}}  \\   {{3}^{0}} & {{3}^{0}} & {{3}^{0}}  \\ \end{matrix} \right]\]

\[\Rightarrow P\left( 1 \right)=\left[ \begin{matrix}   1 & 1 & 1  \\   1 & 1 & 1  \\   1 & 1 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow P\left( 1 \right)=A\]

Therefore, the result is true for \[n=1\] .

Let the result be true for \[n=k\] .

\[P\left( k \right):{{A}^{k}}=\left[ \begin{matrix}   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\ \end{matrix} \right]\]

Now, we have to prove that the result is true for \[n=k+1\] .

Now, \[{{A}^{k+1}}=A\cdot {{A}^{k}}\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   1 & 1 & 1  \\   1 & 1 & 1  \\   1 & 1 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}}  \\   3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}}  \\   3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}}  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}}  \\   {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}}  \\   {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}}  \\ \end{matrix} \right]\]

Thus, the result is true for \[n=k+1\]

Therefore, by the principal of mathematical induction, we have:

 \[{{A}^{n}}=\left[ \begin{matrix}   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\  {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\ \end{matrix} \right]\] where \[A=\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\] , \[n\in N\].

3. If \[A=\left[ \begin{matrix}   3 & -4  \\   1 & -1  \\ \end{matrix} \right]\] , then prove \[{{A}^{n}}=\left[ \begin{matrix}   1+2n & -4n  \\   n & 1-2n  \\ \end{matrix} \right]\] where \[n\] is any positive integer.

Ans: Given \[A=\left[ \begin{matrix}   3 & -4  \\   1 & -1  \\ \end{matrix} \right]\]

By using the principle of mathematical induction.

For \[n=1\] ,

\[P\left( 1 \right):A=\left[ \begin{matrix}   3 & -4  \\   1 & -1  \\ \end{matrix} \right]\]

Therefore, the result is true for \[n=1\] .

Let the result be true for \[n=k\] .

That is, \[P\left( k \right):{{A}^{k}}=\left[ \begin{matrix}   1+2k & -4k  \\   k & 1-2k  \\ \end{matrix} \right]\] , \[n\in N\]

Now, we have to prove that the result is true for \[n=k+1\] .

\[{{A}^{k+1}}=A\cdot {{A}^{k}}\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   1+2k & -4k  \\   k & 1-2k  \\ \end{matrix} \right]\left[ \begin{matrix}   3 & -4  \\   1 & -1  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   3+2k & -4-4k  \\  1+k & -1-2k  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   1+2\left( k+1 \right) & -4\left( k+1 \right)  \\   1+k & 1-2\left( k+1 \right)  \\ \end{matrix} \right]\]

Thus, the result is true for \[n=k+1\]

Therefore, by the principal of mathematical induction, we have:

\[{{A}^{n}}=\left[ \begin{matrix}   1+2n & -4n  \\   n & 1-2n  \\ \end{matrix} \right]\] , \[n\in N\].

4. If \[A\] and \[B\] are symmetric matrices, prove that \[AB-BA\] is a skew symmetric matrix.

Ans: \[A\] and \[B\] are symmetric matrix, therefore, we have:

\[A'=A\] and \[B'=B\]                                                    …(1)

Here, \[\left( AB-BA \right)'=\left( AB \right)'-\left( BA \right)'\]

\[\Rightarrow \left( AB-BA \right)'=B'A'-A'B'\]

From (1),

\[\Rightarrow \left( AB-BA \right)'=BA-AB\]

\[\Rightarrow \left( AB-BA \right)'=-\left( AB-BA \right)\]                           …(2)

From equation (2),

\[AB-BA\] is a skew symmetric matrix.

5. Show that the matrix \[B'AB\] is symmetric or skew symmetric accordingly as \[A\] is symmetric or skew symmetric.

Ans: Let \[A\] be a symmetric matrix, then \[A'=A\]               …(1)

\[\left( B'AB \right)'=\left\{ B'\left( AB \right) \right\}'\]

\[\Rightarrow \left( B'AB \right)'=\left( AB \right)'B'\]

\[\Rightarrow \left( B'AB \right)'=B'\left( A'B \right)\]

From (1),

\[\Rightarrow \left( B'AB \right)'=B'\left( AB \right)\]

Thus, if \[A\] is a symmetric matrix, then \[B'AB\] is a symmetric matrix.

Let \[A\] be a skew symmetric matrix, then \[A'=-A\]               …(2)

\[\left( B'AB \right)'=\left\{ B'\left( AB \right) \right\}'\]

\[\Rightarrow \left( B'AB \right)'=\left( AB \right)'B'\]

\[\Rightarrow \left( B'AB \right)'=B'\left( A'B \right)\]

From (2),

\[\Rightarrow \left( B'AB \right)'=B'\left( -AB \right)\]

\[\Rightarrow \left( B'AB \right)'=-B'AB\]

Thus, if \[A\] is a skew-symmetric matrix, then \[B'AB\] is a skew-symmetric matrix.

Therefore, the matrix \[B'AB\] is symmetric or skew symmetric accordingly as \[A\] is symmetric or skew symmetric.

6. Solve the system of linear equations, using matrix method.

\[2x-y=-2\]

\[3x+4y=3\]

Ans: The given system of equation can be written in the form of \[AX=B\] where,

\[A=\left[ \begin{matrix}   2 & -1  \\   3 & 4  \\ \end{matrix} \right],X=\left[ \begin{matrix}   x  \\   y  \\ \end{matrix} \right],B=\left[ \begin{matrix}   -2  \\  3  \\ \end{matrix} \right]\]

Now, \[\left| A \right|=8+3=11\]

\[\because \left| A \right|\ne 0\] ,therefore its inverse exists.

We know that,

\[{{A}^{-1}}=\dfrac{1}{\left| A \right|}adjA\]

\[\therefore {{A}^{-1}}=\dfrac{1}{11}adj\left[ \begin{matrix}   4 & 1  \\   -3 & 2  \\ \end{matrix} \right]\]

We can write \[X={{A}^{-1}}B\]

\[\therefore X=\dfrac{1}{11}\left[ \begin{matrix}   4 & 1  \\   -3 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   -2  \\   3  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   x  \\   y  \\ \end{matrix} \right]=\dfrac{1}{11}\left[ \begin{matrix}   -5  \\   12  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   x  \\   y  \\ \end{matrix} \right]=\left[ \begin{matrix}   \dfrac{-5}{11}  \\   \dfrac{12}{11}  \\ \end{matrix} \right]\]

Thus, \[x=\dfrac{-5}{11}\] and \[y=\dfrac{12}{11}\].

7. For what values of \[x\] , \[\left[ \begin{matrix}   1 & 2 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 0  \\   2 & 0 & 1  \\   1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   0  \\   2  \\   x  \\ \end{matrix} \right]=O\] ?

Ans: Given \[\left[ \begin{matrix}   1 & 2 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 0  \\   2 & 0 & 1  \\   1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   0  \\   2  \\   x  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ \begin{matrix}   1+4+1 & 2+0+0 & 0+2+2  \\ \end{matrix} \right]\left[ \begin{matrix}   0  \\   2  \\   x  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ \begin{matrix}   6 & 2 & 4  \\\end{matrix} \right]\left[ \begin{matrix}   0  \\   2  \\   x  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ 6\left( 0 \right)+2\left( 2 \right)+4\left( x \right) \right]=O\]

\[\Rightarrow \left[ 4+4x \right]=\left[ 0 \right]\]

\[\Rightarrow 4+4x=0\]

\[\therefore x=-1\]

Thus, the required value of \[x\] is \[-1\] .

8. If \[A=\left[ \begin{matrix}   3 & 1  \\   -1 & 2  \\ \end{matrix} \right]\] , show that \[{{A}^{2}}-5A+7I=O\] .

Ans: Given \[A=\left[ \begin{matrix}   3 & 1  \\   -1 & 2  \\ \end{matrix} \right]\]

\[\therefore {{A}^{2}}=A\cdot A\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   3 & 1  \\   -1 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   3 & 1  \\   -1 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   3\left( 3 \right)+1\left( -1 \right) & 3\left( 1 \right)+1\left( 2 \right)  \\   -1\left( 3 \right)+2\left( -1 \right) & -1\left( 1 \right)+2\left( 2 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   9-1 & 3+2  \\   -3-2 & -1+4  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   8 & 5  \\   -5 & 3  \\ \end{matrix} \right]\]

LHS: \[{{A}^{2}}-5A+7I\]

\[\therefore \left[ \begin{matrix}   8 & 5  \\   -5 & 3  \\ \end{matrix} \right]-5\left[ \begin{matrix}  3 & 1  \\   -1 & 2  \\ \end{matrix} \right]+7\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   8 & 5  \\   -5 & 3  \\ \end{matrix} \right]-\left[ \begin{matrix}   15 & 5  \\   -5 & 10  \\ \end{matrix} \right]+\left[ \begin{matrix}   7 & 0  \\   0 & 7  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   0 & 0  \\   0 & 0  \\ \end{matrix} \right]\] 

\[\Rightarrow O\]

RHS: \[\Rightarrow O\]

\[LHS=RHS\]

\[\therefore {{A}^{2}}-5A+7I=O\] hence proved.

9. Find \[X\] , if \[\left[ \begin{matrix}   x & -5 & -1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   x  \\   4  \\   1  \\ \end{matrix} \right]=O\] .

Ans: Given \[\left[ \begin{matrix}   x & -5 & -1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   x  \\   4  \\   1  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ \begin{matrix}   x-2 & -10 & 2x-8  \\ \end{matrix} \right]\left[ \begin{matrix}   x  \\   4  \\   1  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ x\left( x-2 \right)-40+2x-8 \right]=O\]

\[\Rightarrow \left[ {{x}^{2}}-48 \right]=\left[ 0 \right]\]

\[\Rightarrow {{x}^{2}}-48=0\]

\[\therefore x=\pm 4\sqrt{3}\]

Thus, the required value of \[x\] is \[\pm 4\sqrt{3}\] .

10. A manufacture produces three products \[X,Y,Z\] which he sells in two markets.

Annual sales are indicated below:

1. If unit sale prices of \[X,Y,Z\] are Rs \[2.50\] , Rs \[1.50\] and Rs \[1.00\] , respectively, find the total revenue in each market with the help of matrix algebra.

Ans: Here the total revenue in market \[I\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   10000 & 2000 & 18000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.50  \\   1.50  \\   1.00  \\ \end{matrix} \right]\]

\[\Rightarrow 1000\times 2.50+2000\times 1.50+18000\times 1.00\]

\[\Rightarrow 46000\]

And, the total revenue in market \[II\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   6000 & 20000 & 8000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.50  \\   1.50  \\   1.00  \\ \end{matrix} \right]\]

\[\Rightarrow 6000\times 2.50+20000\times 1.50+8000\times 1.00\]

\[\Rightarrow 53000\]

Thus, the total revenue in market \[I\] is Rs \[46000\] and the same in market \[II\] is Rs \[53000\].

2. If the unit costs of the above three commodities are Rs \[2.00\] , Rs \[1.00\] and \[50\] paise respectively. Find the gross profit.

Ans: Here, the total cost prices of all the products in the market \[I\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   10000 & 2000 & 18000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.00  \\   1.00  \\   0.50  \\ \end{matrix} \right]\]

\[\Rightarrow 1000\times 2.00+2000\times 1.00+18000\times 0.50\]

\[\Rightarrow 31000\]

As the total revenue in market \[I\] is Rs \[46000\] , the gross profit in this market is \[Rs46000-Rs31000=Rs15000\] .

And, the total revenue in market \[II\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   6000 & 20000 & 8000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.00  \\   1.00  \\   0.50  \\ \end{matrix} \right]\]

\[\Rightarrow 6000\times 2.00+20000\times 1.00+8000\times 0.50\]

\[\Rightarrow 36000\]

As the total revenue in market \[II\] is Rs \[53000\] , the gross profit in this market is \[Rs53000-Rs36000=Rs17000\] .

11. Find the matrix \[X\] so that \[X\left[ \begin{matrix}   1 & 2 & 3  \\   4 & 5 & 6  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & -8 & -9  \\   2 & 4 & 6  \\ \end{matrix} \right]\].

Ans: Given \[X\left[ \begin{matrix}   1 & 2 & 3  \\   4 & 5 & 6  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & -8 & -9  \\   2 & 4 & 6  \\ \end{matrix} \right]\]

Here, \[X\] has to be a \[2\times 2\] matrix.

Now, let \[X=\left[ \begin{matrix}   a & c  \\   b & d  \\ \end{matrix} \right]\]

Thus, we have:

\[\left[ \begin{matrix}   a & c  \\   b & d  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   4 & 5 & 6  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & -8 & -9 \\   2 & 4 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   a+4c & 2a+5c & 3a+6c  \\   b+4d & 2b+5d & 3b+6d  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & -8 & -9  \\   2 & 4 & 6  \\ \end{matrix} \right]\]

Comparing the corresponding elements of two matrices, we have:

\[a+4c=-7\]

\[2a+5c=-8\]

\[3a+6c=-9\]

\[b+4d=2\]

\[2b+5d=4\]

\[3b+6d=6\]

Now, solving the above equations we get,

\[\therefore a=1,b=2,c=-2,d=0\]

Thus, the required matrix \[X\] is \[\left[ \begin{matrix}   1 & -2  \\   2 & 0  \\ \end{matrix} \right]\].

12. If \[A\] and \[B\] are square matrices of the same order such that \[AB=BA\] , then prove by induction that \[A{{B}^{n}}={{B}^{n}}A\] . Further, prove that \[{{\left( AB \right)}^{n}}={{B}^{n}}{{A}^{n}}\] for all \[n\in N\].

Ans: Given, \[A\] and \[B\] are square matrices of the same order such that \[AB=BA\] .

For \[n=1\] , we have:

\[P\left( 1 \right):AB=BA\]                          (Given)

\[\Rightarrow A{{B}^{1}}={{B}^{1}}A\]

Therefore, the result is true for \[n=1\] .

Let the result be true for \[n=k\] .

\[P\left( k \right):A{{B}^{k}}={{B}^{k}}A\]                                        …(1)

Now, we have to prove that the result is true for \[n=k+1\] .

\[A{{B}^{k+1}}=A{{B}^{k}}\cdot B\]

\[\Rightarrow A{{B}^{k+1}}=\left( {{B}^{k}}A \right)B\]

\[\Rightarrow A{{B}^{k+1}}={{B}^{k}}\left( AB \right)\]

\[\Rightarrow A{{B}^{k+1}}={{B}^{k}}\left( BA \right)\]

\[\Rightarrow A{{B}^{k+1}}=\left( {{B}^{k}}B \right)A\]

\[\Rightarrow A{{B}^{k+1}}={{B}^{k+1}}A\]

Therefore, the result is true for \[n=k+1\] .

By the principle of mathematical induction, we have \[A{{B}^{n}}={{B}^{n}}A\] , \[n\in N\].

13. Choose the correct answer in the following questions:

If \[A=\left[ \begin{matrix}   \alpha  & \beta   \\   \gamma  & -\alpha   \\ \end{matrix} \right]\] is such that \[{{A}^{2}}=I\] then 

1. \[1+{{\alpha }^{2}}+\beta \gamma =0\]

2. \[1-{{\alpha }^{2}}+\beta \gamma =0\]

3. \[1-{{\alpha }^{2}}-\beta \gamma =0\]

4. \[1+{{\alpha }^{2}}-\beta \gamma =0\]

Ans: Given \[A=\left[ \begin{matrix}   \alpha  & \beta   \\   \gamma  & -\alpha  \\ \end{matrix} \right]\] 

\[\therefore {{A}^{2}}=A\cdot A\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   \alpha  & \beta   \\   \gamma  & -\alpha   \\ \end{matrix} \right]\left[ \begin{matrix}   \alpha  & \beta   \\   \gamma  & -\alpha   \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   {{\alpha }^{2}}+\beta \gamma  & \alpha \beta -\alpha \beta   \\   \alpha \gamma -\alpha \gamma  & \beta \gamma +{{\alpha }^{2}}  \\ \end{matrix} \right]\]

Now, \[{{A}^{2}}=I\]

\[\Rightarrow \left[ \begin{matrix}   {{\alpha }^{2}}+\gamma  & 0  \\   0 & \beta \gamma +{{\alpha }^{2}}  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

Equating the corresponding elements, we get:

\[\beta \gamma +{{\alpha }^{2}}=1\]

\[\Rightarrow {{\alpha }^{2}}+\beta \gamma -1=0\]

\[\Rightarrow 1-{{\alpha }^{2}}-\beta \gamma =0\]

14. If the matrix \[A\] is both symmetric and skew symmetric, then

1. \[A\] is a diagonal matrix

2. \[A\] is a zero matrix

3. \[A\] is a square matrix

4. None of these

Ans: If a matrix \[A\] is both symmetric and skew symmetric, then

\[A'=A\] and \[A'=-A\]

\[\Rightarrow A=-A\]

\[\Rightarrow A+A=O\]

\[\Rightarrow A=O\]

Thus, option (B) is correct.

15. If \[A\] is a square matrix such that \[{{A}^{2}}=A\] , then \[{{\left( I+A \right)}^{3}}-7A\] is equal to

1. \[A\]

2. \[I-A\]

3. \[I\]

4. \[3A\]

Ans: \[{{\left( I+A \right)}^{3}}-7A={{I}^{3}}+{{A}^{3}}+3A+3{{A}^{2}}-7A\]

\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+{{A}^{3}}+3A+3{{A}^{2}}-7A\]

Given that \[{{A}^{2}}=A\] ,

\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+{{A}^{2}}\cdot A+3A+3{{A}^{2}}-7A\]

\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+A\cdot A-A\]

\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I\]

Thus, option (C) is correct.

Details of Class 12 Matrices NCERT Solutions – Free PDF Download

3.1 Introduction

In NCERT Matrices Class 12 Maths Chapter 3 PDF, students will explore concepts of matrices. The learners can clearly understand the necessary application of matrices in various branches of mathematics. This mathematical tool simplifies the work to a great extent as compared to other straight forward methods. 

The concept of matrices in Class 12 develops from the need to simplify solving systems of linear equations. This chapter introduces students to the basics of matrices and matrix algebra. Matrices are incredibly useful, not just for solving equations but also in various real-world applications. They are used in electronic spreadsheets for budgeting, sales projections, and cost estimates, as well as in science for analyzing experimental results. Matrices also help in physical operations like magnification and reflection and are used in cryptography. Understanding matrices is essential across different fields such as genetics, economics, sociology, psychology, and industrial management.

3.2 Matrix

In this section of the chapter,  the fundamental principles of matrices are discussed. With the help of examples cited by established academicians, students can understand the whole concept. Get the proper definition of Matrix Class 12 and learn the order of a matrix considering only those matrices whose elements are real numbers or functions taking actual values. Students will get to know that the horizontal and vertical lines of entries in a matrix are called rows and columns respectively. This section also talks about the size of a matrix by the number of rows and columns it contains.

3.3 Types of Matrices

This section of the chapter expalins various types of matrices, such as column, row, square, diagonal, scalar, identity, and zero matrices, using clear examples. It also examines when two matrices are considered equal, providing examples and solutions to illustrate this. 

The questions included are designed to deepen students' understanding of these concepts and encourage a thorough exploration of the Matrix Class 12 NCERT Solutions. This part helps students get a better grasp of different matrix forms and their properties, enhancing their mathematical skills and knowledge.

3.4 Operations on Matrices

Matrix Class 12 NCERT Solutions introduces certain operations on matrices, namely, the addition of matrices, multiplication of a matrix by a scalar, differences and multiplication of matrices. Highlighting properties of matrix addition, scalar multiplication of a matrix, multiplication of matrices, etc., students can get a profound understanding of how matrices operate.

3.5. Transpose of a Matrix

This section will focus on transposing a matrix and unique types of matrices such as symmetric and skew-symmetric matrices. It will establish the foundations of the transpose of the matrix with compelling examples and properties. This chapter teaches the basics using clear examples and solutions.

3.6 Symmetric and Skew Symmetric Matrices

In this section, the chapter will assist in learning symmetric and skew-symmetric matrices. One can also comprehend the difference between a symmetric matrix and a skew-symmetric matrix. It is imperative to know about the transpose of a matrix and how to find it. Moreover, it is essential to remember all the parts of Chapter 3 that have been covered up until now. This section also explores theorems proving the nature of the symmetric matrix and skew-symmetric matrix.

Some Important Points to remember

1. A matrix is said to be an ordered rectangular array of numbers or functions. These numbers or functions in the array are called the elements or the entries of the matrix.

2. Order of a Matrix

The order of the matrix determines the dimension of the matrix and the number of rows and columns in the matrix. The general representation of matrix order is Amxn, where m is the number of rows and n is the number of columns in the matrix.

3. Types of Matrices

• Column Matrix: A column matrix is a matrix with only one column.

• Row Matrix: A row matrix is a matrix with only one row.

• Square Matrix: A square matrix is one that has an equal number of rows and columns.

• Diagonal Matrix: A square matrix where only the diagonal elements are non-zero and all other elements are zero.

• Scalar Matrix: A diagonal matrix where all the diagonal elements are the same non-zero number.

• Identity Matrix: A diagonal matrix where all diagonal elements are '1' and all other elements are zero, often symbolized as 'I'.

• Zero or Null Matrix: A matrix where all elements are zero.

4. Equality of Matrices: Let A and B be two matrices. These matrices will be equal, if

(i) orders of A and B will be the same

(ii) corresponding elements of two matrices are the same

5. Operations on Matrices

• Addition of Matrices: You can add two matrices by adding the numbers that are in the same position in each matrix, as long as both matrices are the same size.

• Subtraction of Matrices: Subtracting matrices works like addition; you subtract the numbers in the same positions, but only if the matrices are the same size.

• Multiplication of Matrices: To multiply matrices, the number of columns in the first matrix must match the number of rows in the second; you then multiply and sum specific pairs of numbers.

6. Properties of Multiplication of Matrices

• Non-commutativity: 

Matrix multiplication will be not commutative i.e. if AB & BA are both defined, then it is not mandatory that AB ≠ BA.

• Associative law:

For three matrices A, B, and C, if multiplication is defined, then we can write it as A (BC) = (AB) C.

• Multiplicative identity: 

For any square matrix A, there will be an identity matrix of the same order in which  IA = AI = A.

Summary of NCERT Solutions Class 12 Matrices 

• A matrix is an ordered rectangular array of numbers or functions.

• A matrix having $m$ rows and $n$ columns is called a matrix of order $m \times n$.

• $\left[a_y\right]_{m \times 1}$ is a column matrix.

• $\left[a_h\right]_{\mathrm{b} x}$ is a row matrix.

• An $m \times n$ matrix is a square matrix if $m=n$.

• $A=\left[a_{i j}\right]_{\text {nex }}$ is a diagonal matrix if $a_{i j}=0$, when $i \neq j$.

• $A_y=\left[a_{i j}\right]_{n \times n}$ is a scalar matrix if $a_{i j}=0$, when $i \neq j, a_{i j}=k,(k$ is some constant $)$, when $i=j$.

• $A=\left[a_y\right]_{m \times n}$ is an identity matrix, if $a_{i j}=1$. when $i=j, a_{i j}=0$, when $i \neq j$.

• A zero matrix has all its elements as zero.

• $A=\left[a_i\right]-\left[b_y\right]-B$ if

1. A and B are of asme order,

2. $a_{i j}-b_{i j}$ for all possible values of $i$ and $j$.

• $k a=k\left[a_y\right]_{m \times n}-\left[k\left(a_y\right)\right]_{m \times n}$

• $-A=(-1) A$

• $A-B=A+(-1) B$

• $A+B=B+A$

• $(A+B)+C=A+(B+C)$, where $A, B$ and $C$ are of aame order.

• $k(A+B)=k A+k B$, where $\mathrm{A}$ and $\mathrm{B}$ of asme order, $k$ is constant.

• $(k+l) A=k A+l A$, where $k$ and $l$ are constant.

• If $A=\left[a_{j j}\right]_{m \beta n}$ and $B=\left[b_{i j}\right]_{N \times p}$, then $A B=C-\left[C_{j k}\right]_{w \times p}$, where $c_{j k}=\sum_{j=1}^n a_j b_{j k}$

1. $A(B C)=(A B) C$

2. $A(B+C)=A B+A C$

3. $(A+B) C=A C+B C$

•  If $A=\left[a_W\right]_{n \times N}$, then $A$ or $A^T=\left[a_N\right]_{N \times m}$

1. $\left(A^{\prime}\right)^{\prime}=A$,

2. $(k t)^{\prime}=k A^{\prime}$,

3. $(A+B)^{\prime}=A^{\prime}+B^{\prime}$;

4. $(A B)^{\prime}=B \cdot A^{\prime}$

• $A$ is a symmetric matrix if $A^{\prime}=A$.

• A is a skew symmetric matrix if $A^{\prime}=-A^{\prime}$.

• Any aquare matrix can be represented as the sum of a symmetric and a skew symmetric matrix.

• Elementary operations of a matrix are as follows:

1. $R_i \leftrightarrow R_j$ or $C_i \leftrightarrow C_i$

2. $R_l \rightarrow k R_i$ or $C_l \rightarrow k C_i$

3. $R_i \rightarrow R_i+k R_j$ or $C_i \rightarrow C_i+k C_j$

• If $A$ and $B$ re two square matrices such that $A B-B A=I$, then $B$ is the inverse matrix of $A$ and is denoted by $A^{-1}$ and $A$ is the inverse of $\mathrm{B}$.

Overview of Deleted Syllabus for Class 12 Maths Chapter 3

Class 12 Maths Chapter 3: Exercise Breakdown

Conclusion

NCERT Solutions for Class 12 Chapter 4 - Matrices offer a comprehensive understanding of the topic. It's crucial to focus on understanding the fundamental concepts of matrices, such as types of matrices, operations, and applications. Previous year question papers have typically included around 5-7 questions from this chapter, covering various aspects like matrix multiplication, determinants, and their properties. Therefore, it's important to practice solving different types of problems to gain proficiency. Vedantu's solutions provide clear explanations and step-by-step approaches, aiding in grasping the concepts effectively. By utilizing these resources and practicing regularly, students can enhance their understanding and excel in this topic.

Other Study Materials of CBSE Class 12 Maths Chapter 3

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

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