NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.1 - Determinants

Exercise (4.1)

1. Evaluate the determinant: $\begin{vmatrix}2 & 4 \\ -5& -1 \\\end{vmatrix}$

Ans: Solving the determinant  $\begin{vmatrix}2 & 4 \\ -5& -1 \\\end{vmatrix}$ we have:

$\Rightarrow \begin{vmatrix} 2 & 4 \\ -5& -1 \\ \end{vmatrix}=2(-1)-4(-5)$ $\Rightarrow \begin{vmatrix} 2 & 4 \\ -5& -1 \\ \end{vmatrix}=-2+20$ $\Rightarrow \begin{vmatrix} 2 & 4 \\ -5& -1 \\ \end{vmatrix}=18$

2. Evaluate the determinants.

 i. \[\left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|\] Ans: Solving the determinant

$\left| \begin{matrix} \cos \theta  & -\sin \theta   \\ \sin \theta  & \cos \theta   \\ \end{matrix} \right|$

We have: $\Rightarrow \left| \begin{matrix} \cos \theta  & -\sin \theta   \\ \sin \theta  & \cos \theta   \\ \end{matrix} \right|=\left( \cos \theta  \right)\left( \cos \theta  \right)-\left( -\sin \theta  \right)\left( \sin \theta  \right)$

\[\Rightarrow \left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|=\text{ }{{\cos }^{2}}\theta +{{\sin }^{2}}\theta \] We know, \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] \[\therefore \left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|=1\] ii. \[\left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1} \\ \text{x+1} & \text{x+1} \\ \end{matrix} \right|\]

Ans: Solving the determinant

$\left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|$, 

We have: $\Rightarrow \left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|\text{=}\left( {{\text{x}}^{\text{2}}}\text{-x+1} \right)\left( \text{x+1} \right)\text{-}\left( \text{x-1} \right)\left( \text{x+1} \right)$ 

$\Rightarrow \left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{-}{{\text{x}}^{\text{2}}}\text{+x+}{{\text{x}}^{\text{2}}}\text{-x+1-}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)$ 

So, $\left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{+1-}{{\text{x}}^{\text{2}}}\text{+1}$

$\therefore \left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{-}{{\text{x}}^{\text{2}}}\text{+2}$.

3. If $\text{A=}$\[\left| \begin{matrix}1 & 2  \\   4 & 2  \\ \end{matrix} \right|\], then show that \[\left| \text{2A} \right|\text{=4}\left| \text{A} \right|\].

Ans: Given that,\[\text{A=}\left[ \begin{matrix} \text{1} & \text{2} \\ \text{4} & \text{2} \\ \end{matrix} \right]\] Multiplying $\text{A}$ by $2$, we have: \[\Rightarrow \text{2A= 2}\left[ \begin{matrix} \text{1} & \text{2} \\ \text{4} & \text{2} \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2} & \text{4} \\ \text{8} & \text{4} \\ \end{matrix} \right]\] \[\Rightarrow \text{2A=}\left[ \begin{matrix} \text{2} & \text{4} \\ \text{8} & \text{4} \\ \end{matrix} \right]\] \[\therefore \] L.H.S \[\text{=}\left| \text{2A} \right|\text{=}\left| \begin{matrix} \text{2} & \text{4} \\ \text{8} & \text{4} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{2A} \right|\text{=}2\times 4-4\times 8\] \[\Rightarrow \left| \text{2A} \right|\text{=}8-32\] \[\therefore \left| \text{2A} \right|\text{=}-24\] The value of determinant $\text{A}$ is \[\Rightarrow \left| \text{A} \right|\text{=}\left| \begin{matrix} \text{1} & \text{2} \\ \text{4} & \text{2} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}2-8\] \[\therefore \left| \text{A} \right|\text{=}-6\] R.H.S is given as $\text{4}\left| \text{A} \right|$. \[\therefore \text{4}\left| \text{A} \right|\text{=4 }\!\!\times\!\!\text{ }\left( \text{-6} \right)\text{=-24}\] Hence, we have L.H.S $=$ R.H.S \[\therefore \left| \text{2A} \right|\text{=4}\left| \text{A} \right|\].

4. If \[\text{A=}\left[ \begin{matrix} 1 & 0 & 1  \\  0 & 1 & 2  \\ 0 & 0 & 4  \\  \end{matrix} \right]\], then show that \[\left| {\text{3A}} \right|\text{=27}\left| \text{A} \right|\].

Ans: Given, 

\[\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{1} \\ \text{0} & \text{1} & \text{2} \\ \text{0} & \text{0} & \text{4} \\ \end{matrix} \right]\] Determining the value of determinant $\text{A}$, by expanding along the first column, i.e., \[\text{C1}\], we get: \[\Rightarrow \left| \text{A} \right|\text{=1}\left| \begin{matrix} \text{1} & \text{2} \\ \text{0} & \text{4} \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{0} & \text{1} \\ \text{0} & \text{4} \\ \end{matrix} \right|\text{+0}\left| \begin{matrix} \text{0} & \text{1} \\ \text{1} & \text{2} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}1\left( 4-0 \right)-0+0\] \[\Rightarrow \left| \text{A} \right|\text{=4}\] \[\therefore 27\left| \text{A} \right|\text{=27}\times \text{4}=\text{108}\] ……(1) The value of $\left| 3\text{A} \right|$ is obtained as: \[\begin{align} & \Rightarrow \text{3A=3}\left[ \begin{matrix} \text{1} & \text{0} & \text{1} \\ \text{0} & \text{1} & \text{2} \\ \text{0} & \text{0} & \text{4} \\ \end{matrix} \right] \\ & \Rightarrow \text{3A=}\left[ \begin{matrix} \text{3} & \text{0} & \text{3} \\ \text{0} & \text{3} & \text{6} \\ \text{0} & \text{0} & \text{12} \\ \end{matrix} \right] \\ \end{align}\] \[\therefore \left| \text{3A} \right|\text{=3}\left| \begin{matrix} \text{3} & \text{6} \\ \text{0} & \text{12} \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{0} & \text{3} \\ \text{0} & \text{12} \\ \end{matrix} \right|\text{+0}\left| \begin{matrix} \text{0} & \text{3} \\ \text{3} & \text{6} \\ \end{matrix} \right|\] \[\Rightarrow \text{3}\left( \text{36-0} \right)\text{+0+0}\] \[\Rightarrow \left| \text{3A} \right|\text{=3}\times \text{36}\] Thus, \[\left| \text{3A} \right|\text{=}108\] ……(2) From equations (1) and (2), we have: \[\left| \text{3A} \right|\text{=27}\left| \text{A} \right|\]

Hence proved.

5. Evaluate the determinants

i. \[\left| \begin{matrix} \text{3} & \text{-1} & \text{-2} \\ \text{0} & \text{0} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|\]

Ans:Let \[\text{A=}\left| \begin{matrix} \text{3} & \text{-1} & \text{-2} \\ \text{0} & \text{0} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|\] Determining the value of $\text{A}$ by expanding along the second row, we have: \[\Rightarrow \left| \text{A} \right|\text{=}-0\left| \begin{matrix} -1 & -2 \\ -5 & 0 \\ \end{matrix} \right|+0\left| \begin{matrix} 3 & -2 \\ 3 & 0 \\ \end{matrix} \right|-\left( -1 \right)\left| \begin{matrix} 3 & -1 \\ 3 & -5 \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}\left( \text{-15+3} \right)\] \[\therefore \left| \text{A} \right|\text{=-12}\]

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

Ans: Let \[\text{A=}\left| \begin{matrix} \text{3} & \text{-4} & \text{5} \\ \text{1} & \text{1} & \text{-2} \\ \text{2} & \text{3} & \text{1} \\ \end{matrix} \right|\] Determining the value of $\text{A}$ by expanding along the first row, we have: \[\Rightarrow \left| \text{A} \right|\text{=3}\left| \begin{matrix} \text{1} & \text{-2} \\ \text{3} & \text{1} \\ \end{matrix} \right|\text{+4}\left| \begin{matrix} \text{1} & \text{-2} \\ \text{2} & \text{1} \\ \end{matrix} \right|\text{+5}\left| \begin{matrix} \text{1} & \text{1} \\ \text{2} & \text{3} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=3}\left( \text{1+6} \right)\text{+4}\left( \text{1+4} \right)\text{+5}\left( \text{3-2} \right)\] \[\Rightarrow \left| \text{A} \right|\text{=21+20+5}\] \[\therefore \left| \text{A} \right|\text{=}46\]

iii. \[\left| \begin{matrix} \text{0} & \text{1} & \text{2} \\ \text{-1} & \text{0} & \text{-3} \\ \text{-2} & \text{3} & \text{0} \\ \end{matrix} \right|\]

Ans: Let \[\text{A=}\left| \begin{matrix} \text{0} & \text{1} & \text{2} \\ \text{-1} & \text{0} & \text{-3} \\ \text{-2} & \text{3} & \text{0} \\ \end{matrix} \right|\] Determining the value of $\text{A}$ by expanding along the first row, we have: \[\Rightarrow \left| \text{A} \right|\text{=0}\left| \begin{matrix} \text{0} & \text{-3} \\ \text{3} & \text{0} \\ \end{matrix} \right|\text{-1}\left| \begin{matrix} \text{-1} & \text{-3} \\ \text{-2} & \text{0} \\ \end{matrix} \right|\text{+2}\left| \begin{matrix} \text{-1} & \text{0} \\ \text{-2} & \text{3} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=0}\left( 9 \right)-\left( -6 \right)+2\left( -3 \right)\] \[\therefore \left| \text{A} \right|\text{=0}\]

iv. $\left| \begin{matrix} \text{2} & \text{-1} & \text{-2} \\ \text{0} & \text{2} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|$

Ans: Let $\text{A=}\left| \begin{matrix} \text{2} & \text{-1} & \text{-2} \\ \text{0} & \text{2} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|$ Determining the value of $\text{A}$ by expanding along the first column, we have: \[\Rightarrow \left| \text{A} \right|\text{=2}\left| \begin{matrix} \text{2} & \text{-1} \\ \text{-5} & \text{0} \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{-1} & \text{-2} \\ \text{-5} & \text{0} \\ \end{matrix} \right|\text{+3}\left| \begin{matrix} \text{-1} & \text{-2} \\ \text{2} & \text{-1} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}2\left( -5 \right)-0+3\left( 5 \right)\] \[\Rightarrow \left| \text{A} \right|\text{=}-\text{10+15}\] \[\therefore \left| \text{A} \right|\text{=5}\]

6. If \[\text{A=}\left[ \begin{matrix} 1 & 1 & -2  \\   2 & 1 & -3  \\ 5 & 4 & -9  \\ \end{matrix} \right]\], find \[\left| \text{A} \right|\].

Ans: Given,\[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{-2} \\ \text{2} & \text{1} & \text{-3} \\ \text{5} & \text{4} & \text{-9} \\ \end{matrix} \right]\] Determining the value of $\text{A}$ by expanding along the first row, we have: \[\Rightarrow \left| \text{A} \right|\text{=1}\left| \begin{matrix} \text{1} & \text{-3} \\ \text{4} & \text{-9} \\ \end{matrix} \right|\text{-1}\left| \begin{matrix} \text{2} & \text{-3} \\ \text{5} & \text{-9} \\ \end{matrix} \right|\text{-2}\left| \begin{matrix} \text{2} & \text{1} \\ \text{5} & \text{4} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}1\left( -9+12 \right)-1\left( -18+15 \right)-2\left( 8-5 \right)\] \[\Rightarrow \left| \text{A} \right|\text{=}3+3-6\] \[\therefore \left| \text{A} \right|\text{=}0\]

7. Find values of \[x\] , if

i. \[\left| \begin{matrix} \text{2} & \text{4} \\ \text{5} & \text{1} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{2x} & \text{4} \\ \text{6} & \text{x} \\ \end{matrix} \right|\] Ans: Given, \[\left| \begin{matrix} \text{2} & \text{4} \\ \text{5} & \text{1} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{2x} & \text{4} \\ \text{6} & \text{x} \\ \end{matrix} \right|\] Solving it, we have: \[\Rightarrow \left( 2\times 1 \right)-\left( 5\times 4 \right)=\left( 2x\times x \right)-\left( 6\times 4 \right)\] \[\Rightarrow 2-20=2{{x}^{2}}-24\] \[\Rightarrow -18+24=2{{x}^{2}}\] \[\Rightarrow 3={{x}^{2}}\] Applying square root on both the sides, we obtain: \[\Rightarrow \text{x = }\!\!\pm\!\!\text{ }\sqrt{\text{3}}\] ii. \[\left| \begin{matrix} \text{2} & \text{3} \\ \text{4} & \text{5} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{3} \\ \text{2x} & \text{5} \\ \end{matrix} \right|\] Ans: Given, \[\left| \begin{matrix} \text{2} & \text{3} \\ \text{4} & \text{5} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{3} \\ \text{2x} & \text{5} \\ \end{matrix} \right|\] Solving it, we have: \[\Rightarrow \left( \text{2 }\!\!\times\!\!\text{ 5} \right)-\left( \text{3 }\!\!\times\!\!\text{ 4} \right)\text{=}\left( \text{x }\!\!\times\!\!\text{ 5} \right)-\left( \text{3 }\!\!\times\!\!\text{ 2x} \right)\] \[\Rightarrow \text{10}-\text{12=5x}-\text{6x}\] \[\Rightarrow -\text{2=}-\text{x}\] Multiplying by $\left( -1 \right)$ on both the sides, we obtain: \[\Rightarrow \text{x = 2}\]

8. If \[\left| \begin{matrix} x & 2  \\ 18 & x  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} 6 & 2 \\  18 & 6  \\ \end{matrix} \right|\], then \[x\] is equal to

1. \[\text{6}\]

2. \[\text{ }\!\!\pm\!\!\text{ 6}\]

3. \[\text{-6}\]

4. \[\text{0}\]

Ans: Given,\[\left| \begin{matrix} \text{x} & \text{2} \\ \text{18} & \text{x} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{6} & \text{2} \\ \text{18} & \text{6} \\ \end{matrix} \right|\] Solving it, we have: \[\Rightarrow {{\text{x}}^{\text{2}}}-\text{36 = 36}-\text{36}\] \[\Rightarrow {{\text{x}}^{\text{2}}}-\text{36=0}\] \[\Rightarrow {{\text{x}}^{\text{2}}}\text{=36}\] Applying square root on both the sides, we obtain: \[\Rightarrow \text{x= }\!\!\pm\!\!\text{ 6}\] Hence, B. \[\text{ }\!\!\pm\!\!\text{ 6}\] is the correct answer.

Conclusion

Vedantu's NCERT Solutions for Class 12 Maths Chapter 4 - Determinants, Exercise 4.1, provides a complete understanding of determinants and their properties. It's important to focus on the basic concepts and methods of calculating determinants, as these are fundamental for solving more complex problems. Practice each question thoroughly to build a strong foundation. Pay special attention to the different types of determinant problems to ensure you're well-prepared for the exams.

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