Master Class 12 Determinants Exercise 4.4 Solutions for Better Exam Performance
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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 2025-26 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 :
$A^{-1}=\dfrac{1}{|A|}\times adj(A)$
Master Class 12 Determinants Exercise 4.4 Solutions for Better Exam Performance
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.
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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
1. How do you find the minor of an element in a 3x3 determinant?
Delete the row and column containing the specific element to find its minor. Then, calculate the determinant of the remaining 2x2 sub-matrix. For an element aᵢⱼ, its minor Mᵢⱼ is the value of the determinant of the sub-matrix formed by removing the i-th row and j-th column.
2. How can I download the Class 12 Maths Chapter 4 Exercise 4.4 solutions PDF?
Click the “Download PDF” button on this page to save the complete solutions. This Free PDF allows you to access all the solved problems from Class 12 Maths Determinants Exercise 4.4 offline, making it easy to revise formulas and check answers on any device without an internet connection.
3. What is the formula for finding the inverse of a matrix?
Use the formula A⁻¹ = (1/|A|) adj(A) to find the inverse of a square matrix A. First, ensure the determinant |A| is non-zero (meaning the matrix is non-singular). Then, multiply the adjoint of A by the reciprocal of the determinant to get the final inverse matrix.
4. How do you calculate the cofactor of an element?
Calculate the cofactor Aᵢⱼ using the formula Aᵢⱼ = (-1)ⁱ⁺ʲ Mᵢⱼ, where Mᵢⱼ is the minor of the element aᵢⱼ. The sign depends on the element's position. If the sum of the row and column indices (i+j) is even, Aᵢⱼ = Mᵢⱼ; if it is odd, Aᵢⱼ = -Mᵢⱼ.
5. How can these NCERT solutions help with self-assessment?
Solve the problems from Exercise 4.4 on your own first. Afterwards, compare your steps and final answers with the detailed solutions provided here. This process helps you identify mistakes in your calculations of minors, cofactors, or the final inverse, improving your accuracy for exams.
6. How do you find the adjoint of a 3x3 matrix as required in Exercise 4.4?
To find the adjoint of a 3x3 matrix, first calculate the cofactor of every element and then find the transpose of the resulting cofactor matrix.
The adjoint is a crucial step for finding the inverse of a matrix, which is a key concept for solving systems of linear equations and a frequent question in Class 12 Maths Chapter 4.
7. What is the process to verify the property A(adj A) = |A|I?
Verify the property A(adj A) = |A|I by calculating each side of the equation separately and confirming they are identical. This is a common verification problem in the class 12 determinants exercise 4.4 solutions.
Steps:
- Calculate the Determinant |A|. Find the determinant of the given square matrix A.
- Find the Adjoint (adj A). Calculate all minors and cofactors, form the cofactor matrix, and then find its transpose to get adj(A).
- Multiply A by adj(A). Perform the matrix multiplication of the original matrix A with its adjoint.
- Calculate |A|I. Multiply the scalar determinant value |A| with the identity matrix I of the same order as A.
8. How can a student effectively use the step-by-step NCERT Solutions for Exercise 4.4?
Use Vedantu's NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 as a learning tool to master the method, not just to find the final answers.
Simply copying answers does not build problem-solving skills. Understanding the logic behind calculating determinants, adjoints, and inverses is crucial for performing well in exams.
9. What is the shortcut for finding the inverse of a 2x2 matrix?
Find the inverse of a 2x2 matrix, A = [[a, b], [c, d]], by swapping the main diagonal elements, changing the signs of the off-diagonal elements, and dividing the resulting matrix by the determinant.
Formula: The inverse is given by A⁻¹ = (1 / (ad - bc)) * [[d, -b], [-c, a]].
10. Under what condition does the inverse of a square matrix not exist?
The inverse of a square matrix A does not exist if its determinant is zero (|A| = 0). A matrix with a determinant of zero is known as a singular matrix.
The formula for the inverse is A⁻¹ = (1/|A|) (adj A), which involves division by the determinant. Since division by zero is undefined, the inverse cannot be calculated if |A| = 0. This is the first and most important check to perform in any problem that asks for a matrix inverse.
