NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 Determinants - FREE PDF Download
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NCERT solutions for exercise 4.4 class 12 chapter 4 "Determinants" deals with the key concepts such as the condition for the existence of the inverse of a matrix, adjoint and inverse of a matrix, and nonsingular matrices. This chapter provides a strong foundation for understanding how to work with matrices and their properties.
Vedantu’s NCERT Solutions offer clear and detailed explanations on these topics. Class 12 Ex 4.4 gives an insight of how to compute a given structure of linear equations with the help of determinants. The entire NCERT Class 12 Maths solutions from vedantu has been updated according to the 2024-25 syllabus by our masters.
Glance on NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.4 | Vedantu
NCERT solutions of exercise 4.4 class 12 maths of chapter determinants, deals with the topics inverse of a matrix, condition for existence of inverse of a matrix, adjoint of a matrix, singular matrix and non singular matrix.
Inverse of a matrix A is another matrix denoted by $ A^{-1} $ such that when A is multiplied by $ A^{-1} $ the result is the identity matrix ie in mathematically ,
$ A\times A^{-1}= A^{-1}\times A=I $
Conditions for Existence of an Inverse of a matrix:
A matrix must be square (same number of rows and columns) and have a non-zero determinant to have an inverse.
A singular matrix is a square matrix that does not have an inverse. This occurs when the determinant of the matrix is zero. ie,$\left | A \right |$=0.
A non-singular matrix is a square matrix that has an inverse. This occurs when the determinant of the matrix is non-zero.
The adjoint (or adjugate) of a square matrix is the transpose of its cofactor matrix.
This article Determinants ex 4.4 class 12 contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter - Determinants, which you can download as PDFs.
There are 18 fully solved questions and solutions in class 12th Maths chapter 4 Determinants.
Important Formulas Used in Class 12 Chapter 4 Exercise 4.4
Inverse of a square matrix 𝐴 :
$A^{-1}=\dfrac{1}{|A|}\times adj(A)$
Access NCERT Solutions for Class 12 Maths Chapter 4 - Determinants Exercise 4.4
1. Find the adjoint of each of the matrices. \[\mathbf{\left[ \begin{matrix} \text{1} & \text{2} \\ \text{3} & \text{4} \\ \end{matrix} \right]}\]
Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{2} \\ \text{3} & \text{4} \\ \end{matrix} \right]\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Thus,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{A}_{11}}\text{=4}\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{A}_{12}}=-3\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}-2\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=1\]
We know that adjoint of a matrix is the transpose of its cofactor matrix.
Thus,\[\text{adjA=}{{\left[ \begin{matrix} {{\text{A}}_{11}} & {{\text{A}}_{12}} \\ {{\text{A}}_{21}} & {{\text{A}}_{22}} \\ \end{matrix} \right]}^{T}}\]
\[\therefore adjA\text{=}\left[ \begin{matrix} \text{4} & \text{-2} \\ \text{-3} & \text{1} \\ \end{matrix} \right]\].
2. Find adjoint of each of the matrices \[\mathbf{\left[ \begin{matrix} \text{1} & \text{-1} & \text{2} \\ \text{2} & \text{3} & \text{5} \\ \text{-2} & \text{0} & \text{1} \\ \end{matrix} \right]}\].
Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2} \\ \text{2} & \text{3} & \text{5} \\ \text{-2} & \text{0} & \text{1} \\ \end{matrix} \right]\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Thus,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{A}_{11}}\text{=}\left| \begin{matrix} \text{3} & \text{5} \\ \text{0} & \text{1} \\ \end{matrix} \right|=3-0=3\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{A}_{12}}=\left| \begin{matrix} \text{2} & \text{5} \\ \text{-2} & \text{1} \\ \end{matrix} \right|=-\left( \text{2+10} \right)=-12\]
\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]
\[\Rightarrow {{A}_{13}}=\left| \begin{matrix} 2 & 3 \\ -2 & 0 \\ \end{matrix} \right|\text{=0+6=6}\]
Similarly,
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}\left| \begin{matrix} \text{-1} & \text{2} \\ \text{0} & \text{1} \\ \end{matrix} \right|=-\left( -1-0 \right)\text{=1}\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=\left| \begin{matrix} \text{1} & \text{2} \\ -2 & \text{1} \\ \end{matrix} \right|\text{=1+4=5}\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}\left| \begin{matrix} \text{1} & \text{-1} \\ \text{-2} & 0 \\ \end{matrix} \right|\text{=}\left( 0-2 \right)\text{=2}\]
and
\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]
\[\Rightarrow {{\text{A}}_{31}}=\left| \begin{matrix} -1 & \text{2} \\ \text{2} & \text{5} \\ \end{matrix} \right|=-5-4=-9\]
\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]
\[\Rightarrow {{\text{A}}_{32}}=\left| \begin{matrix} \text{1} & \text{2} \\ \text{2} & \text{5} \\ \end{matrix} \right|=-\left( 5-4 \right)=-1\]
\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]
\[\Rightarrow {{\text{A}}_{33}}=\left| \begin{matrix} \text{1} & \text{-1} \\ \text{2} & \text{3} \\ \end{matrix} \right|\text{=3+2=5}\].
We know that adjoint of a matrix is the transpose of its cofactor matrix.
Thus, \[\text{adjA=}{{\left[ \begin{matrix} A11 & A\text{12} & A\text{13} \\ A\text{21} & {{\text{A}}_{\text{22}}} & A23 \\ A\text{31} & {{\text{A}}_{\text{32}}} & A\text{33} \\ \end{matrix} \right]}^{T}}\text{=}\left[ \begin{matrix} \text{3} & -12 & 6 \\ 1 & \text{5} & 2 \\ -9 & -1 & \text{5} \\ \end{matrix} \right]\]
3. Verify \[\mathbf{\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I}\]. \[\mathbf{\left[ \begin{matrix} \text{2} & \text{3} \\ \text{-4} & \text{-6} \\ \end{matrix} \right]}\]
Ans: Given,\[\text{A=}\,\left[ \begin{matrix} \text{2} & \text{3} \\ \text{-4} & \text{-6} \\ \end{matrix} \right]\]
\[\therefore \left| \text{A} \right|=-12-\left( -12 \right)\]
\[\Rightarrow \left| \text{A} \right|=0\]
Hence, \[\left| A \right|I=0\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
\[\Rightarrow \left| A \right|I=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Then,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{A}_{11}}\text{=}-6\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{A}_{12}}=4\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}-3\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=2\]
Cofactor matrix is $\left[ \begin{matrix} -6 & 4 \\ -3 & 2 \\ \end{matrix} \right]$.
We know that adjoint of a matrix is the transpose of its cofactor matrix.
Thus, \[\text{adj}\,\text{A=}\,\left[ \begin{matrix} -6 & -3 \\ 4 & \text{2} \\ \end{matrix} \right]\]
Now, multiplying $A$ with its adjoint, we have:
\[\Rightarrow \text{A}\left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{2} & \text{3} \\ \text{-4} & \text{-6} \\ \end{matrix} \right]\left[ \begin{matrix} -6 & -3 \\ 4 & \text{2} \\ \end{matrix} \right]\]
\[\Rightarrow \text{A}\left( \text{adjA} \right)\text{=}\left[ \begin{matrix} -12+12 & -6+6 \\ 24-24 & 12-12 \\ \end{matrix} \right]\]
\[\therefore \text{A}\left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{0} & \text{0} \\ \text{0} & \text{0} \\ \end{matrix} \right]\]
Similarly, multiplying \[\left( adjA \right)\] with \[A\], we get:
\[\Rightarrow \left( \text{adjA} \right)\text{A=}\left[ \begin{matrix} -6 & -3 \\ 4 & \text{2} \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{3} \\ -4 & -6 \\ \end{matrix} \right]\]
\[\Rightarrow \left( \text{adjA} \right)\text{A=}\left[ \begin{matrix} -12+12 & -18+18 \\ 8-8 & 12-12 \\ \end{matrix} \right]\]
\[\therefore \left( \text{adjA} \right)\text{A=}\left[ \begin{matrix} \text{0} & \text{0} \\ \text{0} & \text{0} \\ \end{matrix} \right]\]
Thus, \[\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I\]
Hence verified.
4. Verify \[\mathbf{\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I}\]. \[\mathbf{\left[ \begin{matrix} \text{1} & -1 & \text{2} \\ \text{3} & \text{0} & -2 \\ \text{1} & \text{0} & \text{3} \\ \end{matrix} \right]}\].
Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2} \\ \text{3} & \text{0} & \text{-2} \\ \text{1} & \text{0} & \text{3} \\ \end{matrix} \right]\]
\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{0-0} \right)\text{+1}\left( \text{9+2} \right)\text{+2}\left( \text{0-0} \right)\]
\[\therefore \left| \text{A} \right|\text{=11}\]
\[\left| \text{A} \right|\text{I=11}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{0} & \text{1} & \text{0} \\ \text{0} & \text{0} & \text{1} \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{11} & \text{0} & \text{0} \\ \text{0} & \text{11} & \text{0} \\ \text{0} & \text{0} & \text{11} \\ \end{matrix} \right]\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Thus,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{A}_{11}}\text{=0}\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{A}_{12}}=-\left( 9+2 \right)=-11\]
\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]
\[\Rightarrow {{A}_{13}}=0\]
Similarly,
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( -3+0 \right)=3\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=3-2=1\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 0+1 \right)=-1\]
and
\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]
\[\Rightarrow {{\text{A}}_{31}}=2-0=2\]
\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]
\[\Rightarrow {{\text{A}}_{32}}=-\left( -2-6 \right)=8\]
\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]
\[\Rightarrow {{\text{A}}_{33}}=0+3=3\].
Cofactor matrix is \[\left[ \begin{matrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \\ \end{matrix} \right]\].
We know that adjoint of a matrix is the transpose of its cofactor matrix.
\[\Rightarrow adjA={{\left[ \begin{matrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \\ \end{matrix} \right]}^{T}}\]
\[\therefore \text{adj}\,\text{A=}\left[ \begin{matrix} \text{0} & 3 & 2 \\ -11 & \text{1} & \text{8} \\ \text{0} & -1 & \text{3} \\ \end{matrix} \right]\]
Now, multiplying $A$ with its adjoint, we have:
\[\Rightarrow \text{A}\left( \text{adj}\,\text{A} \right)\text{=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2} \\ \text{3} & \text{0} & \text{-2} \\ \text{1} & \text{0} & \text{3} \\ \end{matrix} \right]\left[ \begin{matrix} \text{0} & \text{3} & \text{2} \\ \text{-11} & \text{1} & \text{8} \\ \text{0} & \text{-1} & \text{3} \\ \end{matrix} \right]\]
\[\Rightarrow \text{A}\left( \text{adj}\,\text{A} \right)\text{=}\left[ \begin{matrix} \text{0+11+0} & \text{3-1-2} & \text{2-8+6} \\ \text{0+0+0} & \text{9+0+2} & \text{6+0-6} \\ \text{0+0+0} & \text{3+0-3} & \text{2+0+9} \\ \end{matrix} \right]\]
\[\therefore \text{A}\left( \text{adj}\,\text{A} \right)\text{=}\left[ \begin{matrix} \text{11} & \text{0} & \text{0} \\ \text{0} & \text{11} & \text{0} \\ \text{0} & \text{0} & \text{11} \\ \end{matrix} \right]\]
Similarly, multiplying \[\left( adjA \right)\] with \[A\], we get:
\[\Rightarrow \left( \text{adj}\,\text{A} \right)\text{A=}\left[ \begin{matrix} \text{0} & \text{3} & \text{2} \\ \text{-11} & \text{1} & \text{8} \\ \text{0} & \text{-1} & \text{3} \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{-1} & \text{2} \\ \text{3} & \text{0} & \text{-2} \\ \text{1} & \text{0} & \text{3} \\ \end{matrix} \right]\]
\[\Rightarrow \left( \text{adj}\,\text{A} \right)\text{A=}\left[ \begin{matrix} \text{0+9+2} & \text{0+0+0} & \text{0-6+6} \\ \text{-11+3+8} & \text{11+0+0} & \text{-22-2+24} \\ \text{0-3+3} & \text{0+0+0} & \text{0+2+9} \\ \end{matrix} \right]\]
\[\therefore \left( \text{adj}\,\text{A} \right)\text{A=}\left[ \begin{matrix} \text{11} & \text{0} & \text{0} \\ \text{0} & \text{11} & \text{0} \\ \text{0} & \text{0} & \text{11} \\ \end{matrix} \right]\]
Thus, \[\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I\]
Hence verified.
5. Find the inverse of each of the matrices (if it exists). $\mathbf{\left[ \begin{matrix} 2 & -2 \\ 4 & 3 \\ \end{matrix} \right]}$
Ans: Let \[\text{A=}\left[ \begin{matrix} 2 & -2 \\ 4 & 3 \\ \end{matrix} \right]\]
\[\Rightarrow \left| \text{A} \right|=6+8\]
\[\therefore \left| \text{A} \right|=14\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Then,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{\text{A}}_{11}}\text{=3}\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{\text{A}}_{12}}\text{=-4}\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}=2\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=2\]
Cofactor matrix is $\left[ \begin{matrix} 3 & -4 \\ 2 & 2 \\ \end{matrix} \right]$.
We know that adjoint of a matrix is the transpose of its cofactor matrix.
\[\therefore \text{adjA=}\left[ \begin{matrix} 3 & 2 \\ -4 & 2 \\ \end{matrix} \right]\]
Hence, the inverse of the matrix $A$ is given by,
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]
\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{14}}\left[ \begin{matrix} 3 & 2 \\ \text{-4} & 2 \\ \end{matrix} \right]\].
6. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{-1} & \text{5} \\ \text{-3} & \text{2} \\ \end{matrix} \right]}\]
Ans: Let \[\text{A=}\left[ \begin{matrix} \text{-1} & \text{5} \\ \text{-3} & \text{2} \\ \end{matrix} \right]\]
\[\Rightarrow \left| \text{A} \right|=-2+15\]
\[\therefore \left| \text{A} \right|=13\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Then,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{\text{A}}_{11}}\text{=2}\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{\text{A}}_{12}}\text{=3}\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}=-5\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=-1\]
Cofactor matrix is $\left[ \begin{matrix} 2 & 3 \\ -5 & -1 \\ \end{matrix} \right]$.
We know that adjoint of a matrix is the transpose of its cofactor matrix.
\[\therefore \text{adjA=}\left[ \begin{matrix} \text{2} & \text{-5} \\ \text{3} & \text{-1} \\ \end{matrix} \right]\]
Hence, the inverse of the matrix $A$ is given by,
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]
\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{13}}\left[ \begin{matrix} \text{2} & \text{-5} \\ \text{3} & \text{-1} \\ \end{matrix} \right]\].
7. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{2} & \text{3} \\ \text{0} & \text{2} & \text{4} \\ \text{0} & \text{0} & \text{5} \\ \end{matrix} \right]}\]
Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{2} & \text{3} \\ \text{0} & \text{2} & \text{4} \\ \text{0} & \text{0} & \text{5} \\ \end{matrix} \right]\]
Then,
\[\Rightarrow \left| \text{A} \right|\text{=1}\left( 10-0 \right)-2\left( 0-0 \right)\text{+3}\left( 0-0 \right)\]
\[\therefore \left| \text{A} \right|\text{=10}\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Thus,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{A}_{11}}\text{=}10-0=10\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{A}_{12}}=-\left( 0+0 \right)=0\]
\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]
\[\Rightarrow {{A}_{13}}\text{=0}\]
Similarly,
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( 10-0 \right)\text{=}-10\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=5-0=5\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 0-0 \right)=0\]
And
\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]
\[\Rightarrow {{\text{A}}_{31}}=8-6=2\]
\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]
\[\Rightarrow {{\text{A}}_{32}}=-\left( 4-0 \right)=-4\]
\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]
\[\Rightarrow {{\text{A}}_{33}}=2-0=2\].
Cofactor matrix is $\left[ \begin{matrix} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \\ \end{matrix} \right]$.
We know that adjoint of a matrix is the transpose of its cofactor matrix.
\[\therefore \text{adjA=}\left[ \begin{matrix} \text{10} & -10 & \text{2} \\ \text{0} & \text{5} & -4 \\ \text{0} & \text{0} & \text{2} \\ \end{matrix} \right]\]
Hence, the inverse of the matrix $A$ is given by,
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]
\[\therefore {{\text{A}}^{\text{-1}}}=\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{10} & \text{-10} & \text{2} \\ \text{0} & \text{5} & \text{-4} \\ \text{0} & \text{0} & \text{2} \\ \end{matrix} \right]\]
8. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{3} & \text{3} & \text{0} \\ \text{5} & \text{2} & \text{-1} \\ \end{matrix} \right]}\]
Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{3} & \text{3} & \text{0} \\ \text{5} & \text{2} & \text{-1} \\ \end{matrix} \right]\]
Then,
\[\Rightarrow \left| \text{A} \right|\text{=1}\left( -3-0 \right)\text{-0+0}\]
\[\therefore \left| \text{A} \right|=-3\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Thus,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{A}_{11}}\text{=}-3-0=-3\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{A}_{12}}=-\left( -3-0 \right)=3\]
\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]
\[\Rightarrow {{A}_{13}}=6-15=-9\]
Similarly,
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( 0+0 \right)=0\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=-1-0=-1\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 2-0 \right)=-2\]
and
\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]
\[\Rightarrow {{\text{A}}_{31}}=0-0=0\]
\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]
\[\Rightarrow {{\text{A}}_{32}}=-\left( 0-0 \right)=0\]
\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]
\[\Rightarrow {{\text{A}}_{33}}=3-0=3\].
Cofactor matrix is $\left[ \begin{matrix} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \\ \end{matrix} \right]$.
We know that adjoint of a matrix is the transpose of its cofactor matrix.
\[\Rightarrow adjA={{\left[ \begin{matrix} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \\ \end{matrix} \right]}^{T}}\]
\[\therefore \text{adjA=}\left[ \begin{matrix} \text{-3} & \text{0} & \text{0} \\ \text{3} & \text{-1} & \text{0} \\ \text{-9} & \text{-2} & \text{3} \\ \end{matrix} \right]\]
Hence, the inverse of the matrix $A$ is given by,
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]
\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{-3} & \text{0} & \text{0} \\ \text{3} & \text{-1} & \text{0} \\ \text{-9} & \text{-2} & \text{3} \\ \end{matrix} \right]\]
9. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{2} & \text{1} & \text{3} \\ \text{4} & \text{-1} & \text{0} \\ \text{-7} & \text{2} & \text{1} \\ \end{matrix} \right]}\]
Ans: Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{1} & \text{3} \\ \text{4} & \text{-1} & \text{0} \\ \text{-7} & \text{2} & \text{1} \\ \end{matrix} \right]\]
Thus,
\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{-1-0} \right)\text{-1}\left( \text{4-0} \right)\text{+3}\left( \text{8-7} \right)\]
\[\Rightarrow \left| A \right|=2\left( -1 \right)-1\left( 4 \right)+3\left( 1 \right)\]
\[\therefore \left| \text{A} \right|=-3\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Thus,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{A}_{11}}\text{=}-1-0=-1\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{A}_{12}}=-\left( 4-0 \right)=-4\]
\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]
\[\Rightarrow {{A}_{13}}=8-7=1\]
Similarly,
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( 1-6 \right)=5\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=2+21=23\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 4+7 \right)=-11\]
and
\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]
\[\Rightarrow {{\text{A}}_{31}}=0+3=3\]
\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]
\[\Rightarrow {{\text{A}}_{32}}=-\left( 0-12 \right)=12\]
\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]
\[\Rightarrow {{\text{A}}_{33}}=-2-4=-6\].
Cofactor matrix is $\left[ \begin{matrix} -1 & -4 & 1 \\ 5 & 23 & -11 \\ 3 & 12 & -6 \\ \end{matrix} \right]$.
We know that adjoint of a matrix is the transpose of its cofactor matrix.
$\Rightarrow adjA={{\left[ \begin{matrix} -1 & -4 & 1 \\ 5 & 23 & -11 \\ 3 & 12 & -6 \\ \end{matrix} \right]}^{T}}$
\[\therefore \text{adjA=}\left[ \begin{matrix} \text{-1} & \text{5} & \text{3} \\ \text{-4} & \text{23} & \text{12} \\ \text{1} & \text{-11} & \text{-6} \\ \end{matrix} \right]\]
Hence, the inverse of the matrix $A$ is given by,
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]
\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{-1} & \text{5} & \text{3} \\ \text{-4} & \text{23} & \text{12} \\ \text{1} & \text{-11} & \text{-6} \\ \end{matrix} \right]\]
10. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{-1} & \text{2} \\ \text{0} & \text{2} & \text{-3} \\ \text{3} & \text{-2} & \text{4} \\ \end{matrix} \right]}\]
Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2} \\ \text{0} & \text{2} & \text{-3} \\ \text{3} & \text{-2} & \text{4} \\ \end{matrix} \right]\]
Expanding along column \[{{\text{C}}_{1}}\],
\[\Rightarrow \left| \text{A} \right|\text{=1}\left( 8-6 \right)\text{-0+3}\left( 3-4 \right)\]
\[\therefore \left| \text{A} \right|=-1\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Thus,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{A}_{11}}\text{=8-6=2}\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{A}_{12}}=-\left( 0+9 \right)=-9\]
\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]
\[\Rightarrow {{A}_{13}}=0-6=-6\]
Similarly,
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( -4+4 \right)=0\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=4-6=-2\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( -2+3 \right)=-1\]
and
\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]
\[\Rightarrow {{\text{A}}_{31}}=3-4=-1\]
\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]
\[\Rightarrow {{\text{A}}_{32}}=-\left( -3-0 \right)=3\]
\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]
\[\Rightarrow {{\text{A}}_{33}}=2-0=2\].
Cofactor matrix is \[\left[ \begin{matrix} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \\ \end{matrix} \right]\].
We know that adjoint of a matrix is the transpose of its cofactor matrix.
\[\Rightarrow adjA={{\left[ \begin{matrix} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \\ \end{matrix} \right]}^{T}}\]
\[\Rightarrow adjA=\left[ \begin{matrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \\ \end{matrix} \right]\]
The inverse of the matrix $A$ is given by,
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]
$\therefore A^{-1}=-1\begin{bmatrix}2 &0 &-1 \\ -9 &-2 &3 \\ -6 &-1 &2 \end{bmatrix}$
$\text{Hence},A^{-1}=\begin{bmatrix}-2 &0 &1 \\ 9 &2 &-3 \\ 6 &1 &-2 \end{bmatrix}$
11. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{0} & \text{cos}\,\text{a} & \text{sin}\,\text{a} \\ \text{0} & \text{sin}\,\text{a} & \text{-cos}\,\text{a} \\ \end{matrix} \right]}\]
Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{0} & \text{cos a} & \text{sin a} \\ \text{0} & \text{sin a} & \text{-cos a} \\ \end{matrix} \right]\]
Expanding along column, \[{{C}_{1}}\]
\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{-co}{{\text{s}}^{\text{2}}}\text{a-si}{{\text{n}}^{\text{2}}}\text{a} \right)\]
\[\Rightarrow \left| \text{A} \right|=-\left( \text{co}{{\text{s}}^{\text{2}}}\text{a+si}{{\text{n}}^{\text{2}}}\text{a} \right)\]
\[\therefore \left| \text{A} \right|=-1\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Thus,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{A}_{11}}\text{=}-\text{co}{{\text{s}}^{\text{2}}}\text{a}-\text{si}{{\text{n}}^{\text{2}}}\text{a=}-1\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{A}_{12}}=0\]
\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]
\[\Rightarrow {{A}_{13}}=0\]
Similarly,
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}\text{=0}\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=-\cos a\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]
\[\Rightarrow {{\text{A}}_{23}}\text{=}-\sin a\]
and
\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]
\[\Rightarrow {{\text{A}}_{31}}=0\]
\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]
\[\Rightarrow {{\text{A}}_{32}}=-\sin a\]
\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]
\[\Rightarrow {{\text{A}}_{33}}=\cos a\].
Cofactor matrix is \[\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -\cos a & -\sin a \\ 0 & -\sin a & \cos a \\ \end{matrix} \right]\].
We know that adjoint of a matrix is the transpose of its cofactor matrix.
\[\Rightarrow adjA={{\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -\cos a & -\sin a \\ 0 & -\sin a & \cos a \\ \end{matrix} \right]}^{T}}\]
\[\therefore \text{adjA=}\left[ \begin{matrix} \text{-1} & \text{0} & \text{0} \\ \text{0} & \text{-cos}\,\text{a} & \text{-sin}\,\text{a} \\ \text{0} & \text{-sin}\,\text{a} & \text{cos}\,\text{a} \\ \end{matrix} \right]\]
The inverse of the matrix $A$ is given by,
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]
\[\therefore {{\text{A}}^{\text{-1}}}\text{=}-\text{1}\left[ \begin{matrix} \text{-1} & \text{0} & \text{0} \\ \text{0} & \text{-cos}\,\text{a} & \text{-sin}\,\text{a} \\ \text{0} & \text{-sin}\,\text{a} & \text{cos}\,\text{a} \\ \end{matrix} \right]\]
Hence, \[{{\text{A}}^{-1}}\text{=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{0} & \text{cos}\,\text{a} & \text{sin}\,\text{a} \\ \text{0} & \text{sin}\,\text{a} & \text{-cos}\,\text{a} \\ \end{matrix} \right]\].
12. Let \[\mathbf{\text{A=}\left[ \begin{matrix} \text{3} & \text{7} \\ \text{2} & \text{5} \\ \end{matrix} \right]}\] and \[\mathbf{\text{B=}\left[ \begin{matrix} \text{6} & \text{8} \\ \text{7} & \text{9} \\ \end{matrix} \right]}\] . Verify that \[\mathbf{{{\left( \text{AB} \right)}^{\text{-1}}}\text{=}{{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}}\].
Ans: Let \[\text{A=}\left[ \begin{matrix} \text{3} & \text{7} \\ \text{2} & \text{5} \\ \end{matrix} \right]\]
Thus, determining the value of \[\left| \text{A} \right|\],
\[\Rightarrow \left| \text{A} \right|\text{=}15-14\]
\[\therefore \left| \text{A} \right|=1\]
Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].
Thus,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{\text{A}}_{11}}\text{=5}\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}-2\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}=-7\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=3\]
We know that adjoint of a matrix is the transpose of its cofactor matrix.
\[\therefore \text{adjA=}{{\left[ \begin{matrix} \text{5} & -2 \\ -7 & \text{3} \\ \end{matrix} \right]}^{T}}\]
\[\Rightarrow \text{adjA=}\left[ \begin{matrix} \text{5} & -7 \\ -2 & \text{3} \\ \end{matrix} \right]\]
The inverse of a matrix is given by, \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]
Hence, \[{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{5} & \text{-7} \\ \text{-2} & \text{3} \\ \end{matrix} \right]\]
For \[\text{B=}\left[ \begin{matrix} \text{6} & \text{8} \\ \text{7} & \text{9} \\ \end{matrix} \right]\]
\[\Rightarrow \left| \text{B} \right|\text{=54}-\text{56}\]
\[\therefore \left| \text{B} \right|\text{=}-\text{2}\]
Thus,
\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]
\[\Rightarrow {{\text{A}}_{11}}\text{=9}\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]
\[\Rightarrow {{\text{A}}_{12}}\text{=}-7\]
\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]
\[\Rightarrow {{\text{A}}_{21}}=-8\]
\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]
\[\Rightarrow {{\text{A}}_{22}}=6\]
We know that adjoint of a matrix is the transpose of its cofactor matrix.
\[\therefore \text{adjA=}{{\left[ \begin{matrix} 9 & -7 \\ -8 & 6 \\ \end{matrix} \right]}^{T}}\]
\[\Rightarrow \text{adjA=}\left[ \begin{matrix} 9 & -8 \\ -7 & 6 \\ \end{matrix} \right]\]
Hence, \[\text{adjB=}\left[ \begin{matrix} \text{9} & \text{-8} \\ \text{-7} & \text{6} \\ \end{matrix} \right]\]
\[\therefore {{\text{B}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{B} \right|}\text{adjB=}-\dfrac{\text{1}}{\text{2}}\left[ \begin{matrix} \text{9} & -\text{8} \\ -\text{7} & \text{6} \\ \end{matrix} \right]\]
Thus, \[{{\text{B}}^{\text{-1}}}=\left[ \begin{matrix} -\dfrac{\text{9}}{\text{2}} & \text{4} \\ \dfrac{\text{7}}{\text{2}} & -\text{3} \\ \end{matrix} \right]\].
Now, multiplying ${{B}^{-1}}$ and ${{A}^{-1}}$, we get:
\[\Rightarrow {{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{9}}{\text{2}} & \text{4} \\ \dfrac{\text{7}}{\text{2}} & \text{-3} \\ \end{matrix} \right]\left[ \begin{matrix} \text{5} & \text{-7} \\ \text{-2} & \text{3} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{45}}{\text{2}}\text{-8} & \dfrac{\text{63}}{\text{2}}\text{+12} \\ \dfrac{\text{35}}{\text{2}}\text{+6} & \text{-}\dfrac{\text{49}}{\text{2}}\text{-9} \\ \end{matrix} \right]\]
\[\therefore {{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{61}}{\text{2}} & \dfrac{\text{87}}{\text{2}} \\ \dfrac{\text{47}}{\text{2}} & \text{-}\dfrac{\text{67}}{\text{2}} \\ \end{matrix} \right]\] ……(1)
Similarly, multiplying the matrices $A$ and $B$, we get:
\[\Rightarrow \text{AB=}\left[ \begin{matrix} \text{3} & \text{7} \\ \text{2} & \text{5} \\ \end{matrix} \right]\left[ \begin{matrix} \text{6} & \text{8} \\ \text{7} & \text{9} \\ \end{matrix} \right]\]
\[\Rightarrow \text{AB=}\left[ \begin{matrix} \text{18+49} & \text{24+63} \\ \text{12+35} & \text{16+45} \\ \end{matrix} \right]\]
\[\therefore \text{AB=}\left[ \begin{matrix} \text{67} & \text{87} \\ \text{47} & \text{61} \\ \end{matrix} \right]\]
The value of \[\left| \text{AB} \right|\] is
\[\Rightarrow \left| \text{AB} \right|\text{=67 }\!\!\times\!\!\text{ 61-87 }\!\!\times\!\!\text{ 47}\]
\[\Rightarrow \left| \text{AB} \right|\text{=4087-4089}\]
\[\therefore \left| \text{AB} \right|=-2\]
The adjoint of $\left( AB \right)$ is given by,
\[\Rightarrow \text{adj}\left( \text{AB} \right)\text{=}\left[ \begin{matrix} \text{61} & \text{-87} \\\text{-47} & \text{67} \\\end{matrix} \right]\]
Thus, the inverse is,
\[\Rightarrow {{\left( \text{AB} \right)}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{AB} \right|}\text{adj}\left( \text{AB} \right)\]
\[\Rightarrow {{\left( \text{AB} \right)}^{\text{-1}}}-\dfrac{\text{1}}{\text{2}}\left[ \begin{matrix} \text{61} & \text{-87} \\ \text{-47} & \text{67} \\ \end{matrix} \right]\]
\[\therefore {{\left( \text{AB} \right)}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{61}}{\text{2}} & \dfrac{\text{87}}{\text{2}} \\ \dfrac{\text{47}}{\text{2}} & \text{-}\dfrac{\text{67}}{\text{2}} \\ \end{matrix} \right]\] ……. (2)
From (1) and (2), we have:
\[{{\left( \text{AB} \right)}^{\text{-1}}}\text{=}{{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\]
Hence proved.
13. If $A = \begin{bmatrix} 3&1 \\ -1 &2 \end{bmatrix} , \text{show that}\;\; A^2-5A+7I=0. \text{Hence find} A^{-1}$.
Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{3} & \text{1} \\ \text{-1} & \text{2} \\ \end{matrix} \right]\]
We can write, \[{{\text{A}}^{\text{2}}}\text{=A}\text{.A}\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{3} & \text{1} \\ \text{-1} & \text{2} \\ \end{matrix} \right]\left[ \begin{matrix} \text{3} & \text{1} \\ \text{-1} & \text{2} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{9-1} & \text{3+2} \\ \text{-3-2} & \text{-1+4} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{8} & \text{5} \\ \text{-5} & \text{3} \\ \end{matrix} \right]\]
\[\therefore \] The value of \[{{\text{A}}^{\text{2}}}\text{-5A+7I}\] is:
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-5A+7I=}\left[ \begin{matrix} \text{8} & \text{5} \\ \text{-5} & \text{3} \\ \end{matrix} \right]\text{-5}\left[ \begin{matrix} \text{3} & \text{1} \\ \text{-1} & \text{2} \\ \end{matrix} \right]\text{+7}\left[ \begin{matrix} \text{1} & \text{0} \\ \text{0} & \text{1} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-5A+7I=}\left[ \begin{matrix} \text{-7} & \text{0} \\ \text{0} & \text{-7} \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{7} & \text{0} \\ \text{0} & \text{7} \\ \end{matrix} \right]\]
\[\therefore {{\text{A}}^{\text{2}}}\text{-5A+7I=}\left[ \begin{matrix} \text{0} & \text{0} \\ \text{0} & \text{0} \\ \end{matrix} \right]\]
Hence, \[{{\text{A}}^{\text{2}}}\text{-5A+7I=0}\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-5A=-7I}\]
Multiplying by \[{{\text{A}}^{\text{-1}}}\] on both the sides, we have:
\[\Rightarrow \text{AA}\left( {{\text{A}}^{\text{-1}}} \right)-\text{5A}{{\text{A}}^{\text{-1}}}\text{=}-\text{7I}{{\text{A}}^{\text{-1}}}\]
\[\Rightarrow \text{A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)-\text{5I=}-\text{7I}{{\text{A}}^{\text{-1}}}\]
\[\Rightarrow \text{AI}-\text{5I=}-\text{7I}{{\text{A}}^{\text{-1}}}\]
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}-\dfrac{\text{1}}{\text{7}}\left( \text{A}-\text{5I} \right)\]
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{7}}\left( \text{5I}-\text{A} \right)\]
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{7}}\left( \left[ \begin{matrix} \text{5} & \text{0} \\ \text{0} & \text{5} \\ \end{matrix} \right]-\left[ \begin{matrix} \text{3} & \text{1} \\ \text{-1} & \text{2} \\ \end{matrix} \right] \right)\]
\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{7}}\left[ \begin{matrix} \text{2} & \text{-1} \\ \text{1} & \text{3} \\ \end{matrix} \right]\]
14. For the matrix\[\mathbf{\text{A=}\left[ \begin{matrix} \text{3} & \text{2} \\ \text{1} & \text{1} \\ \end{matrix} \right]}\] .find the number \[\mathbf{\text{a}}\] and \[\mathbf{\text{b}}\] such that. \[\mathbf{{{\text{A}}^{\text{2}}}\text{+aA+bI=0}}\].
Ans: Given \[\text{A=}\left[ \begin{matrix} \text{3} & \text{2} \\ \text{1} & \text{1} \\ \end{matrix} \right]\]
We can write, \[{{\text{A}}^{\text{2}}}\text{=A}\text{.A}\]
\[\therefore {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{3} & \text{2} \\ \text{1} & \text{1} \\ \end{matrix} \right]\left[ \begin{matrix} \text{3} & \text{2} \\ \text{1} & \text{1} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{9+2} & \text{6+2} \\ \text{3+1} & \text{2+1} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{11} & \text{8} \\ \text{4} & \text{3} \\ \end{matrix} \right]\]
Solving \[{{\text{A}}^{\text{2}}}\text{+aA+bI=0}\] by multiplying the whole equation by \[{{A}^{-1}}\].
\[\Rightarrow \left( \text{AA} \right){{\text{A}}^{\text{-1}}}\text{+aA}{{\text{A}}^{\text{-1}}}\text{+bI}{{\text{A}}^{\text{-1}}}\text{=0}\]
\[\Rightarrow \text{A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{+aI+b}\left( \text{I}{{\text{A}}^{\text{-1}}} \right)\text{=0}\]
\[\Rightarrow \text{AI+aI+b}{{\text{A}}^{\text{-1}}}\text{=0}\]
\[\Rightarrow \text{A+aI=-b}{{\text{A}}^{\text{-1}}}\]
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{b}}\left( \text{A+aI} \right)\]
Now, determining the value of \[{{\text{A}}^{\text{-1}}}\].
We know that the adjoint of a square matrix is the transpose of its cofactor matrix.
Hence, the adjoint of matrix $A$ is:
$\therefore adjA=\left[ \begin{matrix} 1 & -2 \\ -1 & 3 \\ \end{matrix} \right]$
The inverse is given by, \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}adjA\].
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{1}}\left[ \begin{matrix} \text{1} & \text{-2} \\ \text{-1} & \text{3} \\ \end{matrix} \right]\]
\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{1} & \text{-2} \\ \text{-1} & \text{3} \\ \end{matrix} \right]\]
Thus,
\[\Rightarrow \left[ \begin{matrix} \text{1} & \text{-2} \\ \text{-1} & \text{3} \\ \end{matrix} \right]\text{=-}\dfrac{\text{1}}{\text{b}}\left( \left[ \begin{matrix} \text{3} & \text{2} \\ \text{1} & \text{1} \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{a} & \text{0} \\ \text{0} & \text{a} \\ \end{matrix} \right] \right)\]
\[\Rightarrow \left[ \begin{matrix} \text{1} & \text{-2} \\ \text{-1} & \text{3} \\ \end{matrix} \right]\text{=-}\dfrac{\text{1}}{\text{b}}\left[ \begin{matrix} \text{3+a} & \text{2} \\ \text{1} & \text{1+a} \\ \end{matrix} \right]\]
\[\Rightarrow \left[ \begin{matrix} \text{1} & \text{-2} \\ \text{-1} & \text{3} \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \dfrac{\text{-3-a}}{\text{b}} & \text{-}\dfrac{\text{2}}{\text{b}} \\ \text{-}\dfrac{\text{1}}{\text{b}} & \dfrac{\text{-1-a}}{\text{b}} \\ \end{matrix} \right]\]
Equating the corresponding elements of the two matrices, we get:
\[\Rightarrow \text{-}\dfrac{\text{1}}{\text{b}}\text{=-1}\]
\[\therefore \text{b=1}\]
\[\Rightarrow \dfrac{\text{-3-a}}{\text{b}}=1\]
\[\therefore \text{a=}-4\]
Thus, \[-4\] and \[1\] are the required values of \[\text{a}\] and \[\text{b}\] respectively.
15. For the matrix \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{1} & \text{2} & \text{-3} \\ \text{2} & \text{-1} & \text{3} \\ \end{matrix} \right]}\] show that \[\mathbf{{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=0}}\] . Hence, \[\mathbf{{{\text{A}}^{\text{-1}}}}\].
Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{1} & \text{2} & \text{-3} \\ \text{2} & \text{-1} & \text{3} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{1} & \text{2} & \text{-3} \\ \text{2} & \text{-1} & \text{3} \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{1} & \text{2} & \text{-3} \\ \text{2} & \text{-1} & \text{3} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{1+1+1} & \text{1+2-1} & \text{1-3+3} \\ \text{1+2-6} & \text{1+4+3} & \text{1-6-9} \\ \text{2-1+6} & \text{2-2-3} & \text{2+3+9} \\ \end{matrix} \right]\]
\[\therefore {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1} \\ \text{-3} & \text{8} & \text{-14} \\ \text{7} & \text{-3} & \text{14} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}{{\text{A}}^{\text{2}}}\text{.A=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1} \\ \text{-3} & \text{8} & \text{-14} \\ \text{7} & \text{-3} & \text{14} \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{1} & \text{2} & \text{-3} \\ \text{2} & \text{-1} & \text{3} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{4+2+2} & \text{4+4-1} & \text{4-6+3} \\ \text{-3+8-28} & \text{-3+16+14} & \text{-3-24-42} \\ \text{7-3+28} & \text{7-6-14} & \text{7+9+42} \\ \end{matrix} \right]\]
\[\therefore {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{8} & \text{7} & \text{1} \\ \text{-23} & \text{27} & \text{-69} \\ \text{32} & \text{-13} & \text{58} \\ \end{matrix} \right]\]
Substituting the values for \[{{\text{A}}^{\text{3}}}\], \[{{\text{A}}^{2}}\] and \[\text{A}\] in \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I}\], we have:
\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{8} & \text{7} & \text{1} \\ \text{-23} & \text{27} & \text{-69} \\ \text{32} & \text{-13} & \text{58} \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{4} & \text{2} & \text{1} \\ \text{-3} & \text{8} & \text{-14} \\ \text{7} & \text{-3} & \text{14} \\ \end{matrix} \right]\text{+5}\left[ \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{1} & \text{2} & \text{-3} \\ \text{2} & \text{-1} & \text{3} \\ \end{matrix} \right]\text{+11}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{0} & \text{1} & \text{0} \\ \text{0} & \text{0} & \text{1} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{8} & \text{7} & \text{1} \\ \text{-23} & \text{27} & \text{-69} \\ \text{32} & \text{-13} & \text{58} \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{24} & \text{12} & \text{6} \\ \text{-18} & \text{48} & \text{-84} \\ \text{42} & \text{-18} & \text{84} \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{5} & \text{5} & \text{5} \\ \text{5} & \text{10} & \text{-15} \\ \text{2} & \text{-5} & \text{15} \\ \end{matrix} \right]\text{+11}\left[ \begin{matrix} \text{11} & \text{0} & \text{0} \\ \text{0} & \text{11} & \text{0} \\ \text{0} & \text{0} & \text{11} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{24} & \text{12} & \text{6} \\ \text{-18} & \text{48} & \text{-84} \\ \text{42} & \text{-18} & \text{84} \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{24} & \text{12} & \text{6} \\ \text{-18} & \text{48} & \text{-84} \\ \text{42} & \text{-18} & \text{84} \\ \end{matrix} \right]\]
\[\therefore {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{0} & \text{0} & \text{0} \\ \text{0} & \text{0} & \text{0} \\ \text{0} & \text{0} & \text{0} \\ \end{matrix} \right]\text{=0}\]
Thus, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=0}\]
Since, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=0}\].
Multiplying the whole equation by \[{{A}^{-1}}\], we have:
\[\Rightarrow \left( \text{AAA} \right){{\text{A}}^{\text{-1}}}\text{-6}\left( \text{AA} \right){{\text{A}}^{\text{-1}}}\text{+5A}{{\text{A}}^{\text{-1}}}\text{+11I}{{\text{A}}^{\text{-1}}}\text{=0}\]
\[\Rightarrow \text{AA}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{-6A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{+5}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{=11}\left( \text{I}{{\text{A}}^{\text{-1}}} \right)\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=-11}{{\text{A}}^{\text{-1}}}\]
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{\text{11}}\left( {{\text{A}}^{\text{2}}}\text{-6A+5I} \right)\] …. (1)
Now, \[{{\text{A}}^{\text{2}}}\text{-6A+5I}\] is given by:
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1} \\ \text{-3} & \text{8} & \text{-14} \\ \text{7} & \text{-3} & \text{14} \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{1} & \text{1} & \text{1} \\ \text{1} & \text{2} & \text{-3} \\ \text{2} & \text{-1} & \text{3} \\ \end{matrix} \right]\text{+5}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{0} & \text{1} & \text{0} \\ \text{0} & \text{0} & \text{1} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1} \\ \text{-3} & \text{8} & \text{-14} \\ \text{7} & \text{-3} & \text{14} \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{6} & \text{6} & \text{6} \\ \text{6} & \text{12} & \text{-18} \\ \text{12} & \text{6} & \text{18} \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{5} & \text{0} & \text{0} \\ \text{0} & \text{5} & \text{0} \\ \text{0} & \text{0} & \text{5} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1} \\ \text{-3} & \text{13} & \text{-14} \\ \text{7} & \text{-3} & \text{19} \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{6} & \text{6} & \text{6} \\ \text{6} & \text{12} & \text{-18} \\ \text{12} & \text{-6} & \text{18} \\ \end{matrix} \right]\]
\[\therefore {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{3} & \text{-4} & \text{-5} \\ \text{-9} & \text{1} & \text{4} \\ \text{-5} & \text{3} & \text{1} \\ \end{matrix} \right]\]
Substituting for \[{{\text{A}}^{\text{2}}}\text{-6A+5I}\] equation (1), we get
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{3} & \text{-4} & \text{-5} \\ \text{-9} & \text{1} & \text{4} \\ \text{-5} & \text{3} & \text{1} \\ \end{matrix} \right]\]
\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{-3} & \text{4} & \text{5} \\ \text{9} & \text{-1} & \text{-4} \\ \text{5} & \text{-3} & \text{-1} \\ \end{matrix} \right]\]
16. If \[\mathbf{\text{A=}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1} \\ \text{-1} & \text{2} & \text{-1} \\ \text{1} & \text{-1} & \text{2} \\ \end{matrix} \right]}\] verify that \[\mathbf{{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A+4I=0}}\] and hence find \[\mathbf{{{\text{A}}^{\text{-1}}}}\].
Ans: Given,\[\text{A=}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1} \\ \text{-1} & \text{2} & \text{-1} \\ \text{1} & \text{-1} & \text{2} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1} \\ \text{-1} & \text{2} & \text{-1} \\ \text{1} & \text{-1} & \text{2} \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{-1} & \text{1} \\ \text{-1} & \text{2} & \text{-1} \\ \text{1} & \text{-1} & \text{2} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{4+1+1} & \text{-2-2-1} & \text{2+1+2} \\ \text{-2-2-1} & \text{1+4+1} & \text{-1-2-2} \\ \text{2+1+2} & \text{-1-2-2} & \text{1+1+4} \\ \end{matrix} \right]\]
\[\therefore {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5} \\ \text{-5} & \text{6} & \text{-5} \\ \text{5} & \text{-5} & \text{6} \\ \end{matrix} \right]\]
Similarly,
\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}{{\text{A}}^{\text{2}}}\text{A=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5} \\ \text{-5} & \text{6} & \text{-5} \\ \text{5} & \text{-5} & \text{6} \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{-1} & \text{1} \\ \text{-1} & \text{2} & \text{-1} \\ \text{1} & \text{-1} & \text{2} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{12+5+5} & \text{-6-10-5} & \text{6+5+10} \\ \text{-10-6-5} & \text{5+12+5} & \text{-5-6-10} \\ \text{10+5+6} & \text{-5-10-6} & \text{5+5+12} \\ \end{matrix} \right]\]
\[\therefore {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{22} & \text{-21} & \text{21} \\ \text{-21} & \text{22} & \text{-21} \\ \text{21} & \text{-21} & \text{22} \\ \end{matrix} \right]\]
Now, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I}\] is given by:
\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=}\left[ \begin{matrix} \text{22} & \text{-21} & \text{21} \\ \text{-21} & \text{22} & \text{-21} \\ \text{21} & \text{-21} & \text{22} \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5} \\ \text{-5} & \text{6} & \text{-5} \\ \text{5} & \text{-5} & \text{6} \\ \end{matrix} \right]\text{+9}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1} \\ \text{-1} & \text{2} & \text{-1} \\ \text{1} & \text{-1} & \text{2} \\ \end{matrix} \right]\text{-4}\left[ \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{0} & \text{1} & \text{0} \\ \text{0} & \text{0} & \text{1} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=}\left[ \begin{matrix} \text{22} & \text{-21} & \text{21} \\ \text{-21} & \text{22} & \text{-21} \\ \text{21} & \text{-21} & \text{22} \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{36} & \text{-30} & \text{30} \\ \text{-30} & \text{36} & \text{-30} \\ \text{30} & \text{-30} & \text{36} \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{18} & \text{-9} & \text{9} \\ \text{-9} & \text{18} & \text{-9} \\ \text{9} & \text{-9} & \text{18} \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{4} & \text{0} & \text{0} \\ \text{0} & \text{4} & \text{0} \\ \text{0} & \text{0} & \text{4} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{3}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=}\left[ \begin{matrix} \text{40} & \text{-30} & \text{30} \\ \text{-30} & \text{40} & \text{-30} \\ \text{30} & \text{-30} & \text{40} \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{40} & \text{-30} & \text{30} \\ \text{-30} & \text{40} & \text{-30} \\ \text{30} & \text{-30} & \text{40} \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{0} & \text{0} & \text{0} \\ \text{0} & \text{0} & \text{0} \\ \text{0} & \text{0} & \text{0} \\ \end{matrix} \right]\]
\[\therefore {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=0}\]
Since, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=0}\].
Multiplying the whole equation by \[{{A}^{-1}}\], we have:
\[\Rightarrow \left( \text{AAA} \right){{\text{A}}^{\text{-1}}}\text{-6}\left( \text{AA} \right){{\text{A}}^{\text{-1}}}\text{+9A}{{\text{A}}^{\text{-1}}}\text{-4I}{{\text{A}}^{\text{-1}}}\text{=0}\]
\[\Rightarrow \text{AA}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{-6A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{+9}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{=4}\left( \text{I}{{\text{A}}^{\text{-1}}} \right)\]
\[\Rightarrow \text{AAI-6AI+9I=4}{{\text{A}}^{\text{-1}}}\]
\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+9I=4}{{\text{A}}^{\text{-1}}}\]
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{4}}\left( {{\text{A}}^{\text{2}}}\text{-6A+9I} \right)\] …... (1)
Now, \[{{\text{A}}^{\text{2}}}\text{-6A+9I}\] is given by:
\[{{\text{A}}^{\text{2}}}\text{-6A+9I=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5} \\ \text{-5} & \text{6} & \text{-5} \\ \text{5} & \text{-5} & \text{6} \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1} \\ \text{-1} & \text{2} & \text{-1} \\ \text{1} & \text{-1} & \text{2} \\ \end{matrix} \right]\text{+9}\left[ \begin{matrix} \text{0} & \text{0} & \text{0} \\ \text{0} & \text{0} & \text{0} \\ \text{0} & \text{0} & \text{0} \\ \end{matrix} \right]\]
\[{{\text{A}}^{\text{2}}}\text{-6A+9I=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5} \\ \text{-5} & \text{6} & \text{-5} \\ \text{5} & \text{-5} & \text{6} \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{12} & \text{-6} & \text{6} \\ \text{-6} & \text{12} & \text{-6} \\ \text{6} & \text{-6} & \text{12} \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{9} & \text{0} & \text{0} \\ \text{0} & \text{9} & \text{0} \\ \text{0} & \text{0} & \text{9} \\ \end{matrix} \right]\]
\[\therefore {{\text{A}}^{\text{2}}}\text{-6A+9I=}\left[ \begin{matrix} \text{3} & \text{1} & \text{-1} \\ \text{1} & \text{3} & \text{1} \\ \text{-1} & \text{3} & \text{3} \\ \end{matrix} \right]\]
Substituting for \[{{\text{A}}^{\text{2}}}\text{-6A+9I}\] equation (1), we get
\[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{4}}\left[ \begin{matrix} \text{3} & \text{1} & \text{-1} \\ \text{1} & \text{3} & \text{1} \\ \text{-1} & \text{3} & \text{3} \\ \end{matrix} \right]\].
17. Let \[\mathbf{\text{A}}\] be nonsingular square matrix of order \[\mathbf{\text{3 }\!\!\times\!\!\text{ 3}}\] . Then \[\mathbf{\left| \text{adjA} \right|}\] is equal to
\[\mathbf{\left| \text{A} \right|}\]
\[\mathbf{{{\left| \text{A} \right|}^{\text{2}}}}\]
\[\mathbf{{{\left| \text{A} \right|}^{\text{3}}}}\]
\[\mathbf{\text{3}\left| \text{A} \right|}\]
Ans: Given $A$ is a nonsingular square matrix, i.e., it is a square matrix whose determinant is not equal to zero.
The inverse of a matrix is given as \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}adjA\].
$ \Rightarrow {{A}^{-1}}A=\dfrac{1}{\left| A \right|}adjA$
$\Rightarrow \left| A \right|I=adjA$
The adjoint of matrix $A$ is given by,
\[\Rightarrow \left( \text{adjA} \right)\text{=A=}\left| \text{A} \right|\text{I=}\left[ \begin{matrix} \left| \text{A} \right| & \text{0} & \text{0} \\ \text{0} & \left| \text{A} \right| & \text{0} \\ \text{0} & \text{0} & \left| \text{A} \right| \\ \end{matrix} \right]\]
\[\Rightarrow \left| \left( \text{adjA} \right)\text{A} \right|\text{=}\left[ \begin{matrix} \left| \text{A} \right| & \text{0} & \text{0} \\ \text{0} & \left| \text{A} \right| & \text{0} \\ \text{0} & \text{0} & \left| \text{A} \right| \\ \end{matrix} \right]\]
\[\Rightarrow \left| \text{adjA} \right|\left| \text{A} \right|\text{=}{{\left| \text{A} \right|}^{\text{3}}}\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{0} & \text{1} & \text{0} \\ \text{0} & \text{0} & \text{1} \\ \end{matrix} \right|\text{=}{{\left| \text{A} \right|}^{\text{3}}}\left( \text{I} \right)\]
\[\therefore \left| \text{adjA} \right|\text{=}{{\left| \text{A} \right|}^{\text{3}}}\]
Hence, B. \[{{\left| \text{A} \right|}^{\text{2}}}\] is the correct answer.
18. If \[\mathbf{\text{A}}\] is an invertible matrix of order \[\mathbf{\text{2}}\] , then \[\mathbf{\text{det}\left( {{\text{A}}^{\text{-1}}} \right)}\] is equal to
\[\mathbf{\text{det}\left( \text{A} \right)}\]
\[\mathbf{\dfrac{\text{1}}{\text{det}\left( \text{A} \right)}}\]
\[\mathbf{\text{1}}\]
\[\mathbf{\text{0}}\]
Ans: Since \[\text{A}\] is an invertible matrix, thus \[{{\text{A}}^{\text{-1}}}\] exists and it is given by: \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\].
As matrix \[\text{A}\] is of order \[\text{2}\],
$\therefore $ Let \[\text{A=}\left[ \begin{matrix} \text{a} & \text{b} \\ \text{c} & \text{d} \\ \end{matrix} \right]\] .
Hence, \[\left| \text{A} \right|\text{=ad-bc}\].
The adjoint of \[\text{A}\] would be, \[\text{adjA=}\left[ \begin{matrix} \text{d} & \text{-b} \\ \text{-c} & \text{a} \\ \end{matrix} \right]\] .
Now, the inverse of the matrix is given by:
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{d}}{\left| \text{A} \right|} & \dfrac{\text{-b}}{\left| \text{A} \right|} \\ \dfrac{\text{-c}}{\left| \text{A} \right|} & \dfrac{\text{a}}{\left| \text{A} \right|} \\ \end{matrix} \right]\]
\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{d}}{\left| \text{A} \right|} & \dfrac{\text{-b}}{\left| \text{A} \right|} \\ \dfrac{\text{-c}}{\left| \text{A} \right|} & \dfrac{\text{a}}{\left| \text{A} \right|} \\ \end{matrix} \right]\]
\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\dfrac{\text{1}}{\left| {{\text{A}}^{\text{2}}} \right|}\left| \begin{matrix} \text{d} & \text{-b} \\ \text{-c} & \text{a} \\ \end{matrix} \right|\]
\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|=\dfrac{\text{1}}{\left| {{\text{A}}^{\text{2}}} \right|}\left( \text{ad-bc} \right)\]
\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\dfrac{\text{1}}{\left| {{\text{A}}^{\text{2}}} \right|}\text{.}\left| \text{A} \right|\]
\[\therefore \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\]
Thus, \[\text{det}\left( {{\text{A}}^{\text{-1}}} \right)\text{=}\dfrac{\text{1}}{\text{det}\left( \text{A} \right)}\].
Hence, B. \[\dfrac{\text{1}}{\text{det}\left( \text{A} \right)}\] is the correct answer.
Conclusion
The NCERT Solutions for Exercise 4.4 Class 12 Maths by Vedantu provide a comprehensive understanding of important concepts such as the condition for the existence of the inverse of a matrix, the adjoint and inverse of a matrix, and nonsingular matrices. These exercise 4.4 class 12 maths solutions are crucial for mastering the topic of determinants and their applications. Students should focus on understanding the conditions under which a matrix is invertible, how to calculate the adjoint, and the properties of nonsingular matrices. Practicing 4.4 maths class 12 solutions will enhance your problem-solving skills and help you perform better in exams. For a better understanding, Vedantu’s detailed explanations and step-by-step solutions and resources are available for free.
Class 12 Maths Chapter 4: Exercises Breakdown
S.No. | Chapter 4 - Determinants Exercises in PDF Format | |
1 | Class 12 Maths Chapter 4 Exercise 4.1 - 8 Questions & Solutions (3 Short Answers, 5 Long Answers) | |
2 | Class 12 Maths Chapter 4 Exercise 4.2 - 10 Questions & Solutions (4 Short Answers, 10 Long Answers) | |
3 | Class 12 Maths Chapter 4 Exercise 4.3 - 5 Questions & Solutions (2 Short Answers, 3 Long Answers) | |
4 | Class 12 Maths Chapter 4 Exercise 4.5 - 18 Questions & Solutions (4 Short Answers, 14 Long Answers) | |
CBSE Class 12 Maths Chapter 4 Other Study Materials
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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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Related Links for NCERT Class 12 Maths in Hindi
Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.
S.No. | Related NCERT Solutions for Class 12 Maths |
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Important Related Links for NCERT Class 12 Maths
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FAQs on NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4
1. What is the correct step-by-step method to find the inverse of a square matrix in Class 12 Maths Chapter 4 as per NCERT Solutions?
To find the inverse of a square matrix as per NCERT Solutions, follow these steps:
- Check if the matrix is square (equal rows and columns) and its determinant is not zero.
- Calculate the matrix of cofactors (each element is replaced with its cofactor).
- Find the adjoint (adjugate) by transposing the cofactor matrix.
- Divide the adjoint by the determinant: A-1 = 1/|A| × adj(A).
2. Why is a matrix with a zero determinant called singular and non-invertible, according to NCERT?
A matrix is called singular when its determinant equals zero. Such a matrix does not have an inverse because dividing by zero is undefined in mathematics. This concept is critical for CBSE since only non-singular matrices (with non-zero determinant) can be inverted as per the syllabus and exam requirements.
3. How do NCERT Solutions for Determinants Exercise 4.4 ensure stepwise clarity for CBSE board exams?
The solutions provide step-by-step calculations, explaining each determinant, cofactor, and adjoint computation. Every answer follows the NCERT textbook structure, with each manipulation justified. This approach ensures you demonstrate every required step in the CBSE marking scheme for full marks.
4. What are common mistakes students should avoid when solving inverse and adjoint problems in Class 12 Maths NCERT Solutions?
To avoid errors:
- Always check if the determinant is zero before finding an inverse.
- Calculate minors and cofactors carefully for each element.
- Transpose the cofactor matrix correctly to obtain the adjoint.
- Show the division by the determinant explicitly.
- Justify each row/column operation as per CBSE pattern.
5. What is the official NCERT method for verifying the property A(adjA) = |A|I for a square matrix?
According to NCERT, for any square matrix A, you can verify the property by multiplying the matrix with its adjoint: A × adj(A) = adj(A) × A = |A| × I, where I is the identity matrix. Show this equality stepwise to substantiate your solution, as expected in CBSE board exams.
6. How are the solutions to miscellaneous exercises in Class 12 Determinants beneficial for deeper understanding?
Miscellaneous exercises include application-based and higher-order questions that test conceptual understanding beyond rote steps. Solving these with stepwise NCERT Solutions strengthens the grasp of properties, proofs, and exceptions, which is important for CBSE and competitive exam readiness.
7. What role does the adjoint of a matrix play in finding its inverse as per NCERT Solutions?
The adjoint of a matrix is essential for computing the inverse. Once the adjoint is determined by transposing the cofactor matrix, it is multiplied by the reciprocal of the determinant. This gives the inverse using the formula A-1 = 1/|A| × adj(A), a required CBSE method.
8. How does the NCERT Solutions approach differ when expanding a determinant along different rows or columns?
NCERT Solutions recommend expanding along rows or columns that contain zeros to simplify the calculation. This reduces the number of non-zero terms in the expansion, making the process quicker and aligning with official CBSE guidelines for efficiency.
9. In what situations can you not find the inverse of a matrix according to CBSE Class 12 Maths Chapter 4?
If the matrix is not square (unequal rows and columns) or its determinant is zero, then it is singular and the inverse does not exist. Always check these conditions before proceeding with finding the inverse in NCERT-style solutions.
10. How do NCERT Solutions for Chapter 4 Determinants help with board and state exams beyond CBSE?
These solutions provide stepwise, CBSE-approved answers in both Hindi and English medium, making them suitable for state boards (like UP Board) that follow the NCERT pattern. They ensure clarity, correctness, and compliance with the official marking scheme for all board exams aligned to NCERT.
11. What is the significance of verifying matrix properties such as (AB)-1 = B-1A-1 in NCERT Solutions for Class 12 Maths?
Proving properties like (AB)-1 = B-1A-1 demonstrates understanding of matrix algebra and reinforces conceptual knowledge, which are often tested in higher-mark questions and competitive exams per the latest CBSE trends.
12. Why do NCERT Solutions for Determinants Exercise 4.4 emphasize CBSE-marking compliant intermediate steps?
Including all intermediate steps—such as minor, cofactor, adjoint, and verification—ensures you meet CBSE’s marking scheme and avoid score loss due to skipped or unclear working. Each step demonstrates methodical reasoning essential for full credit in board exams.
13. How can mastering Class 12 NCERT Solutions for matrices and determinants aid in future entrance exams?
A strong understanding of matrix and determinant operations, as taught through stepwise NCERT Solutions, is foundational for JEE, NEET, and other entrances. The ability to systematically solve, justify, and prove properties prepares you for high-difficulty problems common in such exams.
