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# NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 4 - Determinants

Last updated date: 14th Jul 2024
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## NCERT Solutions for Maths Miscellaneous Exercise Class 12 Chapter 4 Determinants - Free PDF Download

NCERT Solution for Miscellaneous Exercise Class 12 Chapter 4 Determinants includes solutions to all Miscellaneous Exercise problems. Determinants Class 12 NCERT Solutions Miscellaneous Exercises are based on the ideas presented in Maths Chapter 4. This activity is crucial for both the CBSE Board examinations and competitive tests. To perform well on the board exam, download the NCERT Solutions for Class 12 Maths Determinants Miscellaneous Exercise in PDF format and practice them offline. Students can download the revised Class 12 Maths NCERT Solutions from our page which is prepared so that you can understand it easily.

The Class 12 Maths Chapter 4 Miscellaneous Solutions are aligned with the updated CBSE guidelines for Class 12, ensuring students are well-prepared for exams. Access the Class 12 Maths Syllabus here.

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Maths Chapter 4 Determinants

### Miscellaneous Exercise

1. Prove that the determinant $\mathbf{\left| \begin{matrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \\ \end{matrix} \right|}$ is independent of $\mathbf{\theta}$.

Ans: Let $\Delta =\left| \begin{matrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \\ \end{matrix} \right|$

To solve it, we have:

$\Rightarrow \Delta =x\left( {{x}^{2}}-1 \right)-\sin \theta \left( -x\sin \theta -\cos \theta \right)+\cos \theta \left( -\sin \theta +x\cos \theta \right)$

$\Rightarrow \Delta ={{x}^{3}}-x+x{{\sin }^{2}}\theta +\sin \theta \cos \theta -\sin \theta \cos \theta +x{{\cos }^{2}}\theta$

$\Rightarrow \Delta ={{x}^{3}}-x+x\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)$

$\Rightarrow \Delta \text{=}{{\text{x}}^{\text{3}}}\text{-x+x}$

$\therefore \Delta \text{=}{{\text{x}}^{\text{3}}}$

Hence, $\text{ }\!\!\Delta\!\!\text{ }$ is independent of $\theta$ .

2. Evaluate $\mathbf{\left| \begin{matrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\ \end{matrix} \right|}$

Ans: Let $\Delta =\left| \begin{matrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\ \end{matrix} \right|$

Expanding along column ${{\text{C}}_{3}}$

$\Rightarrow \Delta =-\sin \alpha \left( -\sin \alpha {{\sin }^{2}}\beta +{{\cos }^{2}}\beta \sin \alpha \right)+\cos \alpha \left( \cos \alpha {{\cos }^{2}}\beta +\cos \alpha {{\sin }^{2}}\beta \right)$

$\Rightarrow \Delta ={{\sin }^{2}}\alpha \left( s{{\sin }^{2}}\beta +{{\cos }^{2}}\beta \right)+{{\cos }^{2}}\alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)$

$\Rightarrow \Delta ={{\sin }^{2}}\alpha \left( 1 \right)+{{\cos }^{2}}\alpha \left( 1 \right)$

$\therefore \Delta \text{=1}$

3. If $\mathbf{{{\mathbf{A}}^{-1}}=\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \\ \end{matrix} \right]$, find ${{\left( AB \right)}^{-1}}}$.

Ans: The below result will be used for simplification,

${{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}$

Finding the inverse of matrix B. So, the determinant is

$\left| B \right|=1\left( 3-0 \right)-2\left( -1-0 \right)-2\left( 2-0 \right)$

$\Rightarrow \left| B \right|=3+2-4$

$\therefore \left| B \right|=1$

Now finding the cofactors,

${{B}_{11}}={{(-1)}^{1+1}}\left| \begin{matrix} 3 & 0 \\ -2 & 1 \\ \end{matrix} \right|\Rightarrow {{B}_{11}}=3$

${{B}_{12}}={{(-1)}^{1+2}}\left| \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right|\Rightarrow {{B}_{12}}=1$

${{B}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix} -1 & 3 \\ 0 & -2 \\ \end{matrix} \right|\Rightarrow {{B}_{13}}=2$

${{B}_{21}}={{(-1)}^{2+1}}\left| \begin{matrix} 2 & -2 \\ -2 & 1 \\ \end{matrix} \right|\Rightarrow {{B}_{21}}=-\left( 2-4 \right)=2$

${{B}_{22}}={{(-1)}^{2+2}}\left| \begin{matrix} 1 & -2 \\ 0 & 1 \\ \end{matrix} \right|\Rightarrow {{B}_{22}}=1$

${{B}_{23}}={{(-1)}^{2+3}}\left| \begin{matrix} 1 & 2 \\ 0 & -2 \\ \end{matrix} \right|\Rightarrow {{B}_{23}}=2$

${{B}_{31}}={{(-1)}^{3+1}}\left| \begin{matrix} 2 & -2 \\ 3 & 0 \\ \end{matrix} \right|\Rightarrow {{B}_{31}}=6$

${{B}_{32}}={{(-1)}^{3+2}}\left| \begin{matrix} 1 & -2 \\ -1 & 0 \\ \end{matrix} \right|\Rightarrow {{B}_{32}}=2$

${{B}_{33}}={{(-1)}^{3+3}}\left| \begin{matrix} 1 & 2 \\ -1 & 3 \\ \end{matrix} \right|\Rightarrow {{B}_{33}}=\left( 3+2 \right)=5$

The cofactor matrix is $\left[ \begin{matrix} 3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5 \\ \end{matrix} \right]$. The adjoint will be the transpose of the cofactor matrix.

$adj\left( B \right)=\left[ \begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \\ \end{matrix} \right]$

The inverse is given by ${{B}^{-1}}=\frac{adj\left( B \right)}{\left| B \right|}$. So,

${{B}^{-1}}=\left[ \begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \\ \end{matrix} \right]$

Now, it is already given that ${{A}^{-1}}=\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \\ \end{matrix} \right]$.

So, we can compute ${{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}$ as below,

${{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \\ \end{matrix} \right]$

$\Rightarrow {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10 \\ \end{matrix} \right]$

$\therefore {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \\ \end{matrix} \right]$

4. Let $\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{2} & \text{1} \\ \text{2} & \text{3} & \text{1} \\ \text{1} & \text{1} & \text{5} \\ \end{matrix} \right]}$ verify that

(i) $\mathbf{{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)}$

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

$\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{15-1} \right)\text{-2}\left( \text{10-1} \right)\text{+1}\left( \text{2-3} \right)$

$\Rightarrow \left| A \right|\text{=14-18-1}$

$\therefore \left| A \right|\text{=}-5$

Thus,

$\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}$

$\Rightarrow {{A}_{11}}\text{=14}$

$\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}$

$\Rightarrow {{A}_{12}}=-9$

$\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}$

$\Rightarrow {{A}_{13}}=-1$

Similarly,

$\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}$

$\Rightarrow {{\text{A}}_{21}}\text{=-9}$

$\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}$

$\Rightarrow {{\text{A}}_{22}}=4$

$\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}$

$\Rightarrow {{\text{A}}_{23}}\text{=1}$

And

$\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}$

$\Rightarrow {{\text{A}}_{31}}=-1$

$\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}$

$\Rightarrow {{\text{A}}_{32}}=1$

$\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}$

$\Rightarrow {{\text{A}}_{33}}=-1$.

Cofactor matrix is $\left[ \begin{matrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \\ \end{matrix} \right]$.

We know that the adjoint of a matrix is the transpose of its cofactor matrix.

$\Rightarrow adjA={{\left[ \begin{matrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \\ \end{matrix} \right]}^{T}}$ $\therefore \text{adjA=}\left[ \begin{matrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \\ \end{matrix} \right]$

Let us denote the adjoint of A as B. So, $B\text{=}\left[ \begin{matrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \\ \end{matrix} \right]$.

The inverse of A is given by

${{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)$

$\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{5}\left[ \begin{matrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \\ \end{matrix} \right]$

$\therefore {{A}^{\text{-1}}}\text{=}\dfrac{\text{1}}{5}\left[ \begin{matrix} -14 & 9 & 1 \\ 9 & -4 & -1 \\ 1 & -1 & 1 \\ \end{matrix} \right]$

Now, we have to verify ${{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)$.

Let us compute the RHS first, i.e. the adjoint of ${{A}^{\text{-1}}}\text{=}\dfrac{\text{1}}{5}\left[ \begin{matrix} -14 & 9 & 1 \\ 9 & -4 & -1 \\ 1 & -1 & 1 \\ \end{matrix} \right]$ or ${{A}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{14}{5} & \dfrac{9}{5} & \dfrac{\text{1}}{5} \\ \dfrac{9}{5} & -\dfrac{4}{5} & -\dfrac{\text{1}}{5} \\ \dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & \dfrac{\text{1}}{5} \\ \end{matrix} \right]$.

Thus,

$\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}$

$\Rightarrow {{A}_{11}}\text{=-}\dfrac{\text{1}}{5}$

$\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}$

$\Rightarrow {{A}_{12}}=-\dfrac{2}{5}$

$\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}$

$\Rightarrow {{A}_{13}}=-\dfrac{\text{1}}{5}$

Similarly,

$\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}$

$\Rightarrow {{\text{A}}_{21}}\text{=-}\dfrac{2}{5}$

$\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}$

$\Rightarrow {{\text{A}}_{22}}=-\dfrac{3}{5}$

$\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}$

$\Rightarrow {{\text{A}}_{23}}\text{=-}\dfrac{\text{1}}{5}$

And

$\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}$

$\Rightarrow {{\text{A}}_{31}}=-\dfrac{\text{1}}{5}$

$\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}$

$\Rightarrow {{\text{A}}_{32}}=-\dfrac{\text{1}}{5}$

$\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}$

$\Rightarrow {{\text{A}}_{33}}=-1$.

Cofactor matrix is $\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1 \\ \end{matrix} \right]$.  We know that adjoint of a matrix is the transpose of its cofactor matrix.

$\Rightarrow adj{{A}^{-1}}={{\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1 \\ \end{matrix} \right]}^{T}}$

$\therefore \text{adj}{{A}^{-1}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1 \\ \end{matrix} \right]$

Next, moving to the LHS, i.e. the inverse of the adjoint of A. We have an adjoint of A as matrix B,

$B\text{=}\left[ \begin{matrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \\ \end{matrix} \right]$

The Determinant of B is

$\left| B \right|=\left[ 14\left( -4-1 \right)+9\left( 9+1 \right)-1\left( -9+4 \right) \right]$

$\Rightarrow \left| B \right|=\left[ -70+90+5 \right]$

$\therefore \left| B \right|=25$

Now the cofactors,

Thus,

$\Rightarrow {{B}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}$

$\Rightarrow {{B}_{11}}\text{=-5}$

$\Rightarrow {{B}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}$

$\Rightarrow {{B}_{12}}=-10$

$\Rightarrow {{B}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}$

$\Rightarrow {{B}_{13}}=-5$

Similarly,

$\Rightarrow {{B}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}$

$\Rightarrow {{B}_{21}}\text{=-10}$

$\Rightarrow {{B}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}$

$\Rightarrow {{B}_{22}}=-15$

$\Rightarrow {{B}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}$

$\Rightarrow {{B}_{23}}\text{=-5}$

And

$\Rightarrow {{B}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}$

$\Rightarrow {{B}_{31}}=-5$

$\Rightarrow {{B}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}$

$\Rightarrow {{B}_{32}}=-5$

$\Rightarrow {{B}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}$

$\Rightarrow {{B}_{33}}=-25$.

Cofactor matrix is $\left[ \begin{matrix} -5 & -10 & -5 \\ -10 & -15 & -5 \\ -5 & -5 & -25 \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

$\Rightarrow adjB={{\left[ \begin{matrix} -5 & -10 & -5 \\ -10 & -15 & -5 \\ -5 & -5 & -25 \\ \end{matrix} \right]}^{T}}$

$\therefore \text{adjB=}\left[ \begin{matrix} -5 & -10 & -5 \\ -10 & -15 & -5 \\ -5 & -5 & -25 \\ \end{matrix} \right]$

Now the inverse is found as

${{B}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| B \right|}\left( \text{adjB} \right)$

$\Rightarrow {{B}^{\text{-1}}}\text{=}\dfrac{\text{1}}{25}\left[ \begin{matrix} -5 & -10 & -5 \\ -10 & -15 & -5 \\ -5 & -5 & -25 \\ \end{matrix} \right]$

$\therefore {{B}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1 \\ \end{matrix} \right]$

So, LHS is ${{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1 \\ \end{matrix} \right]$.

Since LHS=RHS, it is verified that ${{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)$.

(ii) $\mathbf{{{\left( {{\text{A}}^{\text{-1}}} \right)}^{\text{-1}}}\text{=A}}$

Ans: Since ${{A}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{14}{5} & \dfrac{9}{5} & \dfrac{\text{1}}{5} \\ \dfrac{9}{5} & -\dfrac{4}{5} & -\dfrac{\text{1}}{5} \\ \dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & \dfrac{\text{1}}{5} \\ \end{matrix} \right]$  and $\text{adj}{{A}^{-1}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1 \\ \end{matrix} \right]$, thus,

$\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\left[ -\dfrac{14}{5}\left( -\dfrac{4}{25}-\dfrac{1}{25} \right)-\dfrac{9}{5}\left( \dfrac{9}{25}+\dfrac{1}{25} \right)+\dfrac{1}{5}\left( -\dfrac{9}{25}+\dfrac{4}{25} \right) \right]$

$\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\left[ \dfrac{70}{125}-\dfrac{90}{125}-\dfrac{5}{125} \right]$

$\therefore \left| {{\text{A}}^{\text{-1}}} \right|\text{=-}\dfrac{1}{5}$

Also, we know that an inverse of a matrix is given by:

$\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=}\dfrac{\text{adj}{{\text{A}}^{\text{-1}}}}{\left| \text{A} \right|}$

$\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=-5}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5} \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1 \\ \end{matrix} \right]$

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

$\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=A}$

Hence verified that ${{\left( {{\text{A}}^{\text{-1}}} \right)}^{\text{-1}}}\text{=A}$.

5. Evaluate $\mathbf{\left| \begin{matrix} \text{x} & \text{y} & \text{x+y} \\ \text{y} & \text{x+y} & \text{x} \\ \text{x+y} & \text{x} & \text{y} \\ \end{matrix} \right|}$

Ans: Given, $\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x} & \text{y} & \text{x+y} \\ \text{y} & \text{x+y} & \text{x} \\ \text{x+y} & \text{x} & \text{y} \\ \end{matrix} \right|$

Applying ${{R}_{1}}\to {{R}_{1}}\text{+}{{\text{R}}_{2}}\text{+}{{\text{R}}_{3}}$

$\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{2}\left( \text{x+y} \right) & \text{2}\left( \text{x+y} \right) & \text{2}\left( \text{x+y} \right) \\ \text{y} & \text{x+y} & \text{x} \\ \text{x+y} & \text{x} & \text{y} \\ \end{matrix} \right|$

Taking $\text{2}\left( \text{x+y} \right)$ common from ${{\text{R}}_{1}}$

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

Applying the row operations ${{\text{C}}_{2}}\to {{C}_{2}}-{{C}_{1}}$ and ${{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$

$\text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left| \begin{matrix} \text{1} & \text{0} & 0 \\ \text{y} & \text{x} & \text{x-y} \\ \text{x+y} & \text{-y} & \text{-x} \\ \end{matrix} \right|$

Expanding along ${{\text{R}}_{1}}$

$\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ 1\left( \left( x\times \left( \text{-x} \right) \right)-\left( -y\left( x-y \right) \right) \right)\text{+0+}0 \right]$

$\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ 1\left( \left( \text{-}{{\text{x}}^{2}} \right)-\left( -yx+{{y}^{2}} \right) \right) \right]$

$\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ -{{\text{x}}^{\text{2}}}\text{+xy}-{{\text{y}}^{\text{2}}} \right]$

$\Rightarrow \text{ }\!\!\Delta\!\!\text{ =-2}\left( \text{x+y} \right)\left[ {{\text{x}}^{\text{2}}}-xy+{{\text{y}}^{\text{2}}} \right]$

Applying the identity,

$\therefore \text{ }\!\!\Delta\!\!\text{ =-2}\left[ {{\text{x}}^{3}}\text{+}{{\text{y}}^{3}} \right]$

6. Evaluate $\mathbf{\left| \begin{matrix} \text{1} & \text{x} & \text{y} \\ \text{1} & \text{x+y} & \text{y} \\ \text{1} & \text{x} & \text{x+y} \\ \end{matrix} \right|}$

Ans: Given, $\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & \text{y} \\ \text{1} & \text{x+y} & \text{y} \\ \text{1} & \text{x} & \text{x+y} \\ \end{matrix} \right|$

Applying the row operations ${{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}$ and ${{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}$

$\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & \text{y} \\ \text{0} & \text{y} & \text{0} \\ \text{0} & \text{0} & \text{x} \\ \end{matrix} \right|$

Expanding along ${{\text{C}}_{1}}$

$\Rightarrow \text{ }\!\!\Delta\!\!\text{ =1}\left( \text{xy-0} \right)$

$\therefore \text{ }\!\!\Delta\!\!\text{ }=\text{xy}$

7. Solve the system of the following equations

$\mathbf{\dfrac{\text{2}}{\text{x}}\text{+}\dfrac{\text{3}}{\text{y}}\text{+}\dfrac{\text{10}}{\text{z}}\text{=4}}$

$\mathbf{\dfrac{\text{4}}{\text{x}}\text{+}\dfrac{\text{6}}{\text{y}}\text{+}\dfrac{\text{5}}{\text{z}}\text{=1}}$

$\mathbf{\dfrac{\text{6}}{\text{x}}\text{+}\dfrac{\text{9}}{\text{y}}\text{+}\dfrac{\text{20}}{\text{z}}\text{=2}}$

Ans: Given equations,

$\dfrac{\text{2}}{\text{x}}\text{+}\dfrac{\text{3}}{\text{y}}\text{+}\dfrac{\text{10}}{\text{z}}\text{=4}$

$\dfrac{\text{4}}{\text{x}}\text{+}\dfrac{\text{6}}{\text{y}}\text{+}\dfrac{\text{5}}{\text{z}}\text{=1}$

$\dfrac{\text{6}}{\text{x}}\text{+}\dfrac{\text{9}}{\text{y}}\text{+}\dfrac{\text{20}}{\text{z}}\text{=2}$

Let $\dfrac{\text{1}}{\text{x}}\text{=p}$ , $\dfrac{\text{1}}{\text{y}}\text{=q}$ and $\dfrac{\text{1}}{\text{z}}\text{=r}$

Then the given system of equations becomes:

$\text{2p+3q+10r=4}$

$\text{4p-6q+5r=1}$

$\text{6p+9q+20r=2}$

Let $\text{A=}\left[ \begin{matrix} \text{2} & \text{3} & \text{10} \\ \text{4} & \text{-6} & \text{5} \\ \text{6} & \text{9} & \text{-20} \\ \end{matrix} \right]$ , $\text{X=}\left[ \begin{matrix} \text{p} \\ \text{q} \\ \text{r} \\ \end{matrix} \right]$ and $\text{B=}\left[ \begin{matrix} \text{4} \\ \text{1} \\ \text{2} \\ \end{matrix} \right]$ such that, this system can be written in the form of $\text{AX=B}$.

Now,

$\Rightarrow \left| A \right|\text{=2}\left( \text{120-45} \right)\text{-3}\left( \text{-80-30} \right)\text{+10}\left( \text{36+36} \right)$

$\Rightarrow \left| A \right|\text{=150+330+720}$

$\Rightarrow \left| A \right|\text{=1200}$

Thus, $\text{A}$ is a non-singular matrix.

$\therefore$ Its inverse exists.

So,

$\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}\left| \begin{matrix} -6 & 5 \\ 9 & -20 \\ \end{matrix} \right|$

$\Rightarrow {{A}_{11}}\text{=75}$

$\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}\left| \begin{matrix} 4 & 5 \\ 6 & -20 \\ \end{matrix} \right|$

$\Rightarrow {{A}_{12}}=110$

$\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 4 & -6 \\ 6 & 9 \\ \end{matrix} \right|$

$\Rightarrow {{A}_{13}}=72$

Similarly,

$\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}\left| \begin{matrix} 3 & 10 \\ 9 & -20 \\ \end{matrix} \right|$

$\Rightarrow {{\text{A}}_{21}}\text{=150}$

$\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 2 & 10 \\ 6 & -20 \\ \end{matrix} \right|$

$\Rightarrow {{\text{A}}_{22}}=100$

$\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}\left| \begin{matrix} 2 & 3 \\ 6 & 9 \\ \end{matrix} \right|$

$\Rightarrow {{\text{A}}_{23}}\text{=0}$

and

$\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 3 & 10 \\ -6 & 5 \\ \end{matrix} \right|$

$\Rightarrow {{\text{A}}_{31}}=75$

$\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}\left| \begin{matrix} 2 & 10 \\ 4 & 5 \\ \end{matrix} \right|$

$\Rightarrow {{\text{A}}_{32}}=30$

$\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}\left| \begin{matrix} 2 & 3 \\ 4 & -6 \\ \end{matrix} \right|$

$\Rightarrow {{\text{A}}_{33}}=-24$.

The inverse of a matrix is given by:

$\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)$

$\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{75} & \text{150} & \text{75} \\ \text{110} & \text{-100} & \text{30} \\ \text{72} & \text{0} & \text{-24} \\ \end{matrix} \right]$

Now, $\text{X=}{{\text{A}}^{\text{-1}}}\text{B}$

$\Rightarrow \left[ \begin{matrix} \text{p} \\ \text{q} \\ \text{r} \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{75} & \text{150} & \text{75} \\ \text{110} & \text{-100} & \text{30} \\ \text{72} & \text{0} & \text{-24} \\ \end{matrix} \right]\left[ \begin{matrix} \text{4} \\ \text{1} \\ \text{2} \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} \text{p} \\ \text{q} \\ \text{r} \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{300+150+150} \\ \text{440-100+60} \\ \text{288+0-48} \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} \text{p} \\ \text{q} \\ \text{r} \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{600} \\ \text{400} \\ \text{240} \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} \text{p} \\ \text{q} \\ \text{r} \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \dfrac{\text{1}}{\text{2}} \\ \dfrac{\text{1}}{\text{3}} \\ \dfrac{\text{1}}{\text{5}} \\ \end{matrix} \right]$

$\therefore \text{P=}\dfrac{\text{1}}{\text{2}}$, $\text{q=}\dfrac{\text{1}}{\text{3}}$ and $\text{r=}\dfrac{\text{1}}{\text{5}}$

Thus $\text{x=2}$ , $\text{y=3}$ and $\text{z=5}$.

8. Choose the correct answer.If $\mathbf{\text{X,Y,Z}}$ are nonzero real numbers, then the inverse of matrix $\mathbf{\text{A=}\left[ \begin{matrix} \text{x} & \text{0} & \text{0} \\ \text{0} & \text{y} & \text{0} \\ \text{0} & \text{0} & \text{z} \\ \end{matrix} \right]}$

1. $\mathbf{\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0} \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0} \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}} \\ \end{matrix} \right]}$

2. $\mathbf{\text{xyz}\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0} \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0} \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}} \\ \end{matrix} \right]}$

3. $\mathbf{\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{x} & \text{0} & \text{0} \\ \text{0} & \text{y} & \text{0} \\ \text{0} & \text{0} & \text{z} \\ \end{matrix} \right]}$

4. $\mathbf{\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{0} & \text{1} & \text{0} \\ \text{0} & \text{0} & \text{1} \\ \end{matrix} \right]}$

Ans: Given, $\text{A=}\left[ \begin{matrix} \text{x} & \text{0} & \text{0} \\ \text{0} & \text{y} & \text{0} \\ \text{0} & \text{0} & \text{z} \\ \end{matrix} \right]$

$\Rightarrow \left| \text{A} \right|\text{=x}\left( \text{yz-0} \right)$

$\therefore \left| \text{A} \right|\text{=xyz}\ne \text{0}$

Thus,

$\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}$

$\Rightarrow {{A}_{11}}\text{=yz}$

$\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}$

$\Rightarrow {{A}_{12}}=0$

$\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}$

$\Rightarrow {{A}_{13}}=0$

Similarly,

$\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}$

$\Rightarrow {{\text{A}}_{21}}\text{=0}$

$\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}$

$\Rightarrow {{\text{A}}_{22}}=xz$

$\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}$

$\Rightarrow {{\text{A}}_{23}}\text{=0}$

and

$\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}$

$\Rightarrow {{\text{A}}_{31}}=0$

$\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}$

$\Rightarrow {{\text{A}}_{32}}=0$

$\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}$

$\Rightarrow {{\text{A}}_{33}}=xy$.

Cofactor matrix is $\left[ \begin{matrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

$\Rightarrow adjA={{\left[ \begin{matrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \\ \end{matrix} \right]}^{T}}$

$\therefore \text{adjA=}\left[ \begin{matrix} \text{yz} & \text{0} & \text{0} \\ \text{0} & \text{xz} & \text{0} \\ \text{0} & \text{0} & \text{xy} \\ \end{matrix} \right]$

The inverse of a matrix is given by:

$\Rightarrow {{\text{A}}^{\text{-1}}}=\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)$

$\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{yz} & \text{0} & \text{0} \\ \text{0} & \text{xz} & \text{0} \\ \text{0} & \text{0} & \text{xy} \\ \end{matrix} \right]$

$\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{yz}}{\text{xyz}} & \text{0} & \text{0} \\ \text{0} & \dfrac{\text{xz}}{\text{xyz}} & \text{0} \\ \text{0} & \text{0} & \dfrac{\text{xy}}{\text{xyz}} \\ \end{matrix} \right]$

$\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{1}}{\text{x}} & \text{0} & \text{0} \\ \text{0} & \dfrac{\text{1}}{\text{y}} & \text{0} \\ \text{0} & \text{0} & \dfrac{\text{1}}{\text{z}} \\ \end{matrix} \right]$

$\therefore {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0} \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0} \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}} \\ \end{matrix} \right]$

Thus, A. $\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0} \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0} \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}} \\ \end{matrix} \right]$ is the correct answer.

Let $\mathbf{\text{A}=\left[ \begin{matrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \\ \end{matrix} \right]}$,

where $\mathbf{\text{0}\le \text{ }\!\!\theta\!\!\text{ }\le \text{2n}}$ , then

1. $\mathbf{\text{Det}\left( \text{A} \right)\text{=0}}$

2. $\mathbf{text{Det}\left( \text{A} \right)\in \left( \text{2,}\infty \right)}$

3. $\mathbf{\text{Det}\left( \text{A} \right)\in \left( \text{2,4} \right)}$

4. $\mathbf{\text{Det}\left( \text{A} \right)\left[ \text{2,4} \right]}$

Ans: Given, $A=\left[ \begin{matrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \\ \end{matrix} \right]$

$\therefore \left| A \right|=1\left( 1+{{\sin }^{2}}\theta \right)-\sin \theta \left( -\sin \theta +\sin \theta \right)+1\left( {{\sin }^{2}}\theta +1 \right)$

$\Rightarrow \left| A \right|=1+{{\sin }^{2}}\theta +{{\sin }^{2}}\theta +1$

$\Rightarrow \left| A \right|=2+2{{\sin }^{2}}\theta$

$\Rightarrow \left| A \right|=2\left( 1+{{\sin }^{2}}\theta \right)$

We know, $0\le \sin \theta \le 1$

$\Rightarrow 1\le 1+{{\sin }^{2}}\theta \le 2$

$\Rightarrow 2\le 2\left( 1+{{\sin }^{2}}\theta \right)\le 4$

Thus, D. $\text{Det}\left( \text{A} \right)\left[ \text{2,4} \right]$ is the correct answer.

## Conclusion

Miscellaneous Exercise in Class 12 Maths Chapter 4 is important for understanding various concepts thoroughly. Class 12 Maths Chapter 4 Miscellaneous 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 excel in this exercise effectively.

## Class 12 Maths Chapter 4: Exercises Breakdown

 Exercise Number of Questions Exercise 4.1 8 Questions and Solutions Exercise 4.2 5 Questions and Solutions Exercise 4.3 5 Questions and Solutions Exercise 4.4 18 Questions and Solutions Exercise 4.5 16 Questions and Solutions Miscellaneous Exercises 9 Questions and Solutions

## 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.

## FAQs on NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 4 - Determinants

1. How do I solve problems involving properties of determinants in Class 12 Maths Chapter 4 Miscellaneous Solutions?

The solutions would provide a clear explanation of the relevant properties of determinants, such as the effect of row/column operations or the product of diagonal elements. They would then demonstrate how to use these properties to simplify expressions or solve problems involving determinants.

2. Are there solutions for problems involving minors and cofactors in the Miscellaneous Exercise Class 12 Chapter 4?

Yes, the NCERT solutions might cover problems that involve finding minors and cofactors of elements in a determinant. They would explain how to calculate these values and potentially use them to solve related problems like finding the adjoint or inverse of a matrix (if the matrix is square).

3. How can I improve my problem-solving skills for miscellaneous exercise questions on determinants?

Here are some tips:

• Master the basics: Ensure you have a strong understanding of the concepts covered in the chapter, including finding determinants, properties of determinants, and minors/cofactors.

• Practice with different methods: Practice using various methods for calculating determinants (expansion, Cramer's rule) to gain comfort and choose the most efficient approach for a given problem.

• Focus on understanding: While referring to the NCERT solutions, pay close attention to the reasoning behind each step, not just memorizing the final answer.

4. What is covered in the Class 12 Chapter 4 Determinants Miscellaneous Exercise?

The Miscellaneous Exercise in Chapter 4 of Class 12 Maths NCERT covers a comprehensive set of problems that encapsulate the entire chapter. Chapter 4 deals with "Determinants," and the exercise includes problems on properties of determinants, applications of determinants in finding the area of triangles, solving a system of linear equations using determinants (Cramer's rule), and more.

5. How important is the Miscellaneous Exercise for board exams?

The Miscellaneous Exercise is crucial for board exams, as it includes a variety of problems that cover all the concepts taught in the chapter. Practising these problems helps students gain a deeper understanding and prepares them for different types of questions that might appear in the exams.

6. Are the solutions to the Miscellaneous Exercise available in the NCERT Solutions for Class 12 Maths?

Yes, the NCERT Solutions for Class 12 Maths include detailed solutions to the Miscellaneous Exercise problems. These solutions provide step-by-step explanations, which are very helpful for students in understanding the methodology and the application of various concepts. Miscellaneous Exercise can be downloaded online on the Vedantu website.

7. Where can I find the NCERT Solutions for Chapter 4 Miscellaneous Exercise?

The NCERT Solutions for Chapter 4 Miscellaneous Exercise can be downloaded online on the Vedantu website.

8. How can NCERT Solutions help in understanding the Miscellaneous Exercise problems better?

NCERT Solutions offers detailed, step-by-step explanations for each problem, which can help students understand the underlying concepts and methods used to solve them. This can be especially beneficial for complex problems that require a deeper understanding of the subject matter.

9. Can practising the Miscellaneous Exercise improve problem-solving skills?

Yes, practising the Miscellaneous Exercise can significantly improve problem-solving skills. It exposes students to a wide range of problems, enhances their analytical thinking, and helps them learn different approaches to solving complex problems.

10. Is it necessary to solve all the problems in the Class 12 Maths Chapter 4 Miscellaneous Solutions?

While it is beneficial to solve all the problems in the Miscellaneous Exercise for thorough preparation, students should prioritize based on their understanding and the difficulty level of each problem. Focus on mastering key concepts and solving a variety of problems to ensure a well-rounded understanding.