NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 3 - Matrices

Miscellaneous Exercise

1. 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.

2. 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.

3. Find the values of x, y, z if the matrix A =$\begin{pmatrix}0 & 2y& z \\x&y& -z \\ x& -y& z\\\end{pmatrix}$ satisfy the equation ${A}'A=I$

Ans: Given matrix A=$\begin{bmatrix} 0 & 2y& z \\ x & y & -z \\ x& -y& z \\ \end{bmatrix}$

Transpose of A = ${A}'=\begin{bmatrix} 0 &x &x \\ 2y& y &-y \\ z& -z &z \\ \end{bmatrix}$

Given ${A}'A=I$

= $\begin{bmatrix} 0 & 2y &z \\ x& y&-z \\ x& -y &z \\ \end{bmatrix}\begin{bmatrix} 0 &x &x \\ 2y& y &-y \\ z &-z &z \\ \end{bmatrix}=\begin{bmatrix} 1 & 0 &0 \\ 0& 1& 0 \\ 0& 0 &1 \\ \end{bmatrix}$

${A}'$=$\begin{bmatrix} 0+x^{2}+x^{2} &0+xy-xy &0-xz+xz \\ 0+xy-xy & 4y^{2}+y^{2}+y^{2} &2yz-yz-yz \\ 0-zx+zx& 2yz-yz-yz &z^{2}+z^{2}+z^{2} \\ \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 0& 1& 0\\ 0& 0 &1 \\ \end{bmatrix}$

=$\begin{bmatrix} 2x^{2} &0 &0 \\ 0& 6y^{2}&0 \\ 0& 0 & 3z^{2} \\ \end{bmatrix}$ =$\begin{pmatrix} 1 &0 &0 \\ 0& 1& 0\\ 0& 0 &1 \\ \end{pmatrix}$

To find the values of x, y, z we have to equate the entries with corresponding matrix

$2x^{2}=1$

$\Rightarrow x^{2}$=$\dfrac{1}{2}$

$\Rightarrow$ x= $\pm \dfrac{1}{\sqrt{2}}$

$6y^{2}=1$

$\Rightarrow y^{2}$ = $\dfrac{1}{6}$

$\Rightarrow$ y = $\pm \dfrac{1}{\sqrt{6}}$ 

And

$3z^{2} = 1$

$\Rightarrow z^{2} = \dfrac{1}{3}$

$\Rightarrow$ z= $\pm \dfrac{1}{\sqrt{3}}$

4. 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 \[-1\] .

5. 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.

6. 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 \[\pm 4\sqrt{3}\] .

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

Annual sales are indicated below:

(a) 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 \[53000\].

(b) 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 \[Rs53000-Rs36000=Rs17000\] .

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

9. 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:

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

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

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

10. 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.

11. 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.

Conclusion

Matrices Class 12 Miscellaneous Exercise Solutions are important for understanding various concepts thoroughly. Class 12 Maths Chapter 3 Miscellaneous Exercise Solutions covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to master this exercise effectively.

Class 12 Maths Chapter 3: Exercises Breakdown

CBSE Class 12 Maths Chapter 3 Other Study Materials

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.