\displaystyle \textbf{Question 1: }\quad \text{If }A=\begin{bmatrix} \cos\alpha & -\sin\alpha & 0\\ \sin\alpha & \cos\alpha & 0\\ 0 & 0 & 1 \end{bmatrix},\text{ find }\mathrm{adj}\ A\text{ and verify that } \\ A(\mathrm{adj}\ A)=(\mathrm{adj}\ A)A=|A|I_{3}. \hspace{2.2cm} \quad [\text{CBSE 2016}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }C_{ij}\text{ be the cofactor of }a_{ij}\text{ in }A=[a_{ij}].\text{ Then,}
\displaystyle C_{11}=(-1)^{1+1}\begin{vmatrix} \cos\alpha & 0\\ 0 & 1 \end{vmatrix}=\cos\alpha,\quad C_{12}=(-1)^{1+2}\begin{vmatrix} \sin\alpha & 0\\ 0 & 1 \end{vmatrix}=-\sin\alpha,
\displaystyle C_{13}=(-1)^{1+3}\begin{vmatrix} \sin\alpha & \cos\alpha\\ 0 & 0 \end{vmatrix}=0,\quad C_{21}=(-1)^{2+1}\begin{vmatrix} -\sin\alpha & 0\\ 0 & 1 \end{vmatrix}=\sin\alpha,
\displaystyle C_{22}=(-1)^{2+2}\begin{vmatrix} \cos\alpha & 0\\ 0 & 1 \end{vmatrix}=\cos\alpha,\quad C_{23}=(-1)^{2+3}\begin{vmatrix} \cos\alpha & -\sin\alpha\\ 0 & 0 \end{vmatrix}=0,
\displaystyle C_{31}=(-1)^{3+1}\begin{vmatrix} -\sin\alpha & 0\\ \cos\alpha & 0 \end{vmatrix}=0,\quad C_{32}=(-1)^{3+2}\begin{vmatrix} \cos\alpha & 0\\ \sin\alpha & 0 \end{vmatrix}=0,
\displaystyle C_{33}=(-1)^{3+3}\begin{vmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{vmatrix}=\cos^{2}\alpha+\sin^{2}\alpha=1.
\displaystyle \therefore\ \mathrm{adj}\ A=\begin{bmatrix} \cos\alpha & -\sin\alpha & 0\\ \sin\alpha & \cos\alpha & 0\\ 0 & 0 & 1 \end{bmatrix}^{T}=\begin{bmatrix} \cos\alpha & \sin\alpha & 0\\ -\sin\alpha & \cos\alpha & 0\\ 0 & 0 & 1 \end{bmatrix}
\displaystyle \text{and,}\quad |A|=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}
\displaystyle |A|=(\cos\alpha)(\cos\alpha)+(-\sin\alpha)(-\sin\alpha)+0\times 0=\cos^{2}\alpha+\sin^{2}\alpha=1.
\displaystyle \text{Now,}\quad A(\mathrm{adj}\ A)=\begin{bmatrix} \cos\alpha & -\sin\alpha & 0\\ \sin\alpha & \cos\alpha & 0\\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos\alpha & \sin\alpha & 0\\ -\sin\alpha & \cos\alpha & 0\\ 0 & 0 & 1 \end{bmatrix}
\displaystyle =\begin{bmatrix} \cos^{2}\alpha+\sin^{2}\alpha & \sin\alpha\cos\alpha-\cos\alpha\sin\alpha & 0\\ \cos\alpha\sin\alpha-\sin\alpha\cos\alpha & \sin^{2}\alpha+\cos^{2}\alpha & 0\\ 0 & 0 & 1 \end{bmatrix}
\displaystyle =\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}=|A|I.
\displaystyle \text{and,}\quad (\mathrm{adj}\ A)A=\begin{bmatrix} \cos\alpha & \sin\alpha & 0\\ -\sin\alpha & \cos\alpha & 0\\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos\alpha & -\sin\alpha & 0\\ \sin\alpha & \cos\alpha & 0\\ 0 & 0 & 1 \end{bmatrix}
\displaystyle =\begin{bmatrix} \cos^{2}\alpha+\sin^{2}\alpha & -\sin\alpha\cos\alpha+\cos\alpha\sin\alpha & 0\\ -\sin\alpha\cos\alpha+\cos\alpha\sin\alpha & \sin^{2}\alpha+\cos^{2}\alpha & 0\\ 0 & 0 & 1 \end{bmatrix}
\displaystyle =\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}=|A|I.
\displaystyle \text{Hence, }A(\mathrm{adj}\ A)=|A|I=(\mathrm{adj}\ A)A\text{ is verified.}
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\displaystyle \textbf{Question 2: }\quad \text{Show that }A=\begin{bmatrix} 2 & -3\\ 3 & 4 \end{bmatrix}\text{ satisfies the equation } \\ x^{2}-6x+17=0.\text{ Hence, find }A^{-1}. \hspace{2.2cm} \quad [\text{CBSE 2007}]
\displaystyle \text{Answer:}
\displaystyle \text{We have,}\quad A=\begin{bmatrix} 2 & -3\\ 3 & 4 \end{bmatrix}
\displaystyle \therefore\ A^{2}=AA=\begin{bmatrix} 2 & -3\\ 3 & 4 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 3 & 4 \end{bmatrix}=\begin{bmatrix} 4-9 & -6-12\\ 6+12 & -9+16 \end{bmatrix}=\begin{bmatrix} -5 & -18\\ 18 & 7 \end{bmatrix}
\displaystyle -6A=(-6)\begin{bmatrix} 2 & -3\\ 3 & 4 \end{bmatrix}=\begin{bmatrix} -12 & 18\\ -18 & -24 \end{bmatrix}\quad \text{and,}\quad 17I=17\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 17 & 0\\ 0 & 17 \end{bmatrix}
\displaystyle \therefore\ A^{2}-6A+17I=\begin{bmatrix} -5 & -18\\ 18 & 7 \end{bmatrix}+\begin{bmatrix} -12 & 18\\ -18 & -24 \end{bmatrix}+\begin{bmatrix} 17 & 0\\ 0 & 17 \end{bmatrix}
\displaystyle =\begin{bmatrix} -5-12+17 & -18+18+0\\ 18-18+0 & 7-24+17 \end{bmatrix}=\begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}=O.
\displaystyle \text{Hence, the matrix }A\text{ satisfies the equation }x^{2}-6x+17=0.
\displaystyle \text{Now,}\quad A^{2}-6A+17I=O
\displaystyle \Rightarrow A^{2}-6A=-17I
\displaystyle \Rightarrow A^{-1}(A^{2}-6A)=A^{-1}(-17I)\quad [\text{Pre-multiplying both sides by }A^{-1}]
\displaystyle \Rightarrow A^{-1}A^{2}-6A^{-1}A=-17(A^{-1}I)
\displaystyle \Rightarrow A-6I=-17A^{-1}
\displaystyle \Rightarrow A^{-1}=-\frac{1}{17}(A-6I)=\frac{1}{17}(6I-A)
\displaystyle =\frac{1}{17}\left(\begin{bmatrix} 6 & 0\\ 0 & 6 \end{bmatrix}-\begin{bmatrix} 2 & -3\\ 3 & 4 \end{bmatrix}\right)=\frac{1}{17}\begin{bmatrix} 4 & 3\\ -3 & 2 \end{bmatrix}.
\\

\displaystyle \textbf{Question 3: }\quad \text{Use elementary column operation }C_{2}\rightarrow C_{2}-2C_{1}\text{ in the matrix equation}
\displaystyle \begin{bmatrix} 4 & 2\\ 3 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 2\\ 0 & 3 \end{bmatrix}\begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix} \hspace{2.2cm}  \quad [\text{CBSE 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{If }A,B,C\text{ are three matrices such that }C=AB,\text{ then any elementary row (column)}
\displaystyle \text{transformation of }C\text{ can be obtained by subjecting the pre-factor (post-factor }B)\text{ to the same}
\displaystyle \text{elementary row (column) transformation. Therefore, given matrix equation after applying}
\displaystyle C_{2}\rightarrow C_{2}-2C_{1},\text{ becomes}
\displaystyle \begin{bmatrix} 4 & 2-2\times 4\\ 3 & 3-2\times 3 \end{bmatrix}=\begin{bmatrix} 1 & 2\\ 0 & 3 \end{bmatrix}\begin{bmatrix} 2 & 0-2\times 2\\ 1 & 1-2\times 1 \end{bmatrix}
\displaystyle \text{or,}\quad \begin{bmatrix} 4 & -6\\ 3 & -3 \end{bmatrix}=\begin{bmatrix} 1 & 2\\ 0 & 3 \end{bmatrix}\begin{bmatrix} 2 & -4\\ 1 & -1 \end{bmatrix}
\\

\displaystyle \textbf{Question 4: }\ \ \text{Find the inverse of the matrix }A=\begin{bmatrix} 1&2&-2\\ -1&3&0\\ 0&-2&1 \end{bmatrix}\ \text{by using elementary} \\ \text{row} \text{ transformations.} \hspace{2.2cm}  \qquad [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle \text{We know that}
\displaystyle A=IA
\displaystyle \text{or,}\quad \begin{bmatrix} 1&2&-2\\ -1&3&0\\ 0&-2&1 \end{bmatrix}=\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}A
\displaystyle \Rightarrow\ \begin{bmatrix} 1&2&-2\\ 0&5&-2\\ 0&-2&1 \end{bmatrix}=\begin{bmatrix} 1&0&0\\ 1&1&0\\ 0&0&1 \end{bmatrix}A\qquad [\text{Applying }R_{2}\rightarrow R_{2}+R_{1}]
\displaystyle \Rightarrow\ \begin{bmatrix} 1&2&-2\\ 0&1&0\\ 0&-2&1 \end{bmatrix}=\begin{bmatrix} 1&0&0\\ 1&1&2\\ 0&0&1 \end{bmatrix}A\qquad [\text{Applying }R_{2}\rightarrow R_{2}+2R_{3}]
\displaystyle \Rightarrow\ \begin{bmatrix} 1&0&-2\\ 0&1&0\\ 0&0&1 \end{bmatrix}=\begin{bmatrix} -1&-2&-4\\ 1&1&2\\ 2&2&5 \end{bmatrix}A\qquad [\text{Applying }R_{1}\rightarrow R_{1}+(-2)R_{2},\ R_{3}\rightarrow R_{3}+2R_{2}]
\displaystyle \Rightarrow\ \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}=\begin{bmatrix} 3&2&6\\ 1&1&2\\ 2&2&5 \end{bmatrix}A\qquad [\text{Applying }R_{1}\rightarrow R_{1}+2R_{3}]
\displaystyle \text{Hence,}\quad A^{-1}=\begin{bmatrix} 3&2&6\\ 1&1&2\\ 2&2&5 \end{bmatrix}
\\

\displaystyle \textbf{Question 5. }\text{If }A({adj \ } A)=\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix},\text{ then the value of }|A|+| {adj \ } A|
\displaystyle \text{is equal to}\hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }12\qquad \text{(b) }9\qquad \text{(c) }3\qquad \text{(d) }27
\displaystyle \text{Answer:}
\displaystyle (a)\ \text{Given, }A(\ {adj \ }A)=\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}=3\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=3I
\displaystyle \text{We know that,}
\displaystyle A(\ {adj \ }A)=|A|I
\displaystyle \Rightarrow |A|=3\text{ and }|\ {adj \ }A|=|A|^{n-1}=(3)^{3-1}\qquad[\text{where, }n=3]
\displaystyle \Rightarrow |\ {adj \ }A|=3^2=9
\displaystyle \therefore |A|+|\ {adj \ }A|=3+9=12
\\

\displaystyle \textbf{Question 6. }\text{Let }A\text{ be a }3\times3\text{ matrix such that }|{adj \ } A|=64.\text{ Then, }|A|
\displaystyle \text{is equal to}\hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }8\text{ only}\qquad \text{(b) }-8\text{ only}
\displaystyle \text{(c) }64\qquad \text{(d) }8\text{ or }-8
\displaystyle \text{Answer:}
\displaystyle (d)\ \text{We know that, }|\ {adj \ }A|=|A|^{n-1},\text{ where }n\text{ is the order of matrix }A.
\displaystyle |\ {adj \ }A|=|A|^{3-1}
\displaystyle \Rightarrow 64=|A|^2
\displaystyle \Rightarrow |A|=\pm 8
\\

\displaystyle \textbf{Question 7. }\text{If for a square matrix }A,\ A^{2}-3A+I=O\text{ and}
\displaystyle A^{-1}=xA+yI,\text{ then the value of }x+y\text{ is}\hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }-2\qquad \text{(b) }2\qquad \text{(c) }3\qquad \text{(d) }-3
\displaystyle \text{Answer:}
\displaystyle (b)\ \text{Given, }A^2-3A+I=O
\displaystyle \Rightarrow A^2-3A=-I
\displaystyle \Rightarrow A\cdot A-3A=-I
\displaystyle \Rightarrow A\cdot A\cdot A^{-1}-3A\cdot A^{-1}=-I\cdot A^{-1}
\displaystyle \Rightarrow A\cdot I-3I=-A^{-1}
\displaystyle \Rightarrow A^{-1}=-A+3I
\displaystyle \text{On comparing with }A^{-1}=xA+yI,\text{ we get}
\displaystyle x=-1\text{ and }y=3
\displaystyle \therefore x+y=-1+3=2
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\displaystyle \textbf{Question 8. }\text{If for a square matrix }A,\ A^{2}-A+I=O,\text{ then }A^{-1}\text{ equals}
\displaystyle \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }A\qquad \text{(b) }A+I\qquad \text{(c) }I-A\qquad \text{(d) }A-I
\displaystyle \text{Answer:}
\displaystyle (c)\ \text{Given, }A^2-A+I=O
\displaystyle \Rightarrow A\cdot A-A+I=O
\displaystyle \Rightarrow A\cdot A-A=-I
\displaystyle \Rightarrow A\cdot A\cdot A^{-1}-A\cdot A^{-1}=-IA^{-1}
\displaystyle \Rightarrow A(AA^{-1})-I=-A^{-1}
\displaystyle \Rightarrow AI-I=-A^{-1}
\displaystyle \Rightarrow A^{-1}=I-A
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\displaystyle \textbf{Question 9. }\text{If }A\text{ is a square matrix of order }3,\text{ such that}
\displaystyle A({adj \ } A)=10I,\text{ then }|{adj \ } A|\text{ is equal to}\hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }1\qquad \text{(b) }10\qquad \text{(c) }100\qquad \text{(d) }10I
\displaystyle \text{Answer:}
\displaystyle (c)\ \text{We know that,}
\displaystyle 4(\ {adj \ }A)=|A|I
\displaystyle \text{We have,}
\displaystyle A(\ {adj \ }A)=10I
\displaystyle \therefore |A|=10
\displaystyle \text{Again, }|\ {adj \ }A|=|A|^{n-1}
\displaystyle \therefore |\ {adj \ }A|=|A|^{3-1}=|A|^2=(10)^2=100
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\displaystyle \textbf{Question 10. }\text{If }A=\begin{bmatrix}a&0&0\\0&a&0\\0&0&a\end{bmatrix},\text{ then }\det({adj \ } A)\text{ equals}
\displaystyle \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }a^{27}\qquad \text{(b) }a^{9}\qquad \text{(c) }a^{6}\qquad \text{(d) }a^{2}
\displaystyle \text{Answer:}
\displaystyle (c)\ \text{We have }A=\begin{bmatrix}a&0&0\\0&a&0\\0&0&a\end{bmatrix}
\displaystyle \Rightarrow |A|=a^3
\displaystyle \therefore |\ {adj \ }A|=|A|^{n-1}=|A|^2=(a^3)^2=a^6
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\displaystyle \textbf{Question 11. }\text{Find }{adj \ }(A),\text{ if }A=\begin{bmatrix}2&-1\\4&3\end{bmatrix}.
\displaystyle \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}2&-1\\4&3\end{bmatrix}
\displaystyle \text{Cofactors of }|A|\text{ are}
\displaystyle A_{11}=3,\quad A_{12}=-4,\quad A_{21}=1\text{ and }A_{22}=2
\displaystyle \therefore \ {adj \ }(A)=\begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix}=\begin{bmatrix}3&1\\-4&2\end{bmatrix}
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\displaystyle \textbf{Question 12. }\text{For }A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix},\text{ write }A^{-1}.
\displaystyle \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A=\begin{bmatrix}3&4\\1&-1\end{bmatrix},\text{ then}
\displaystyle |A|=\begin{vmatrix}3&4\\1&-1\end{vmatrix}=-3-4=-7\neq0
\displaystyle \text{So, }A\text{ is a non-singular matrix and }A^{-1}\text{ exists.}
\displaystyle \text{Now, cofactors of each element of }|A|\text{ are}
\displaystyle A_{11}=-1,\quad A_{12}=-(1)=-1
\displaystyle A_{21}=-(4)=-4,\quad A_{22}=3
\displaystyle \text{Now, }\ {adj \ }A=\begin{bmatrix}-1&-4\\-1&3\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}(\ {adj \ }A)=\frac{1}{-7}\begin{bmatrix}-1&-4\\-1&3\end{bmatrix}=\frac{1}{7}\begin{bmatrix}1&4\\1&-3\end{bmatrix}
\\

\displaystyle \textbf{Question 13. }\text{If for any }2\times2\text{ square matrix }A,\ A({adj \ } A)=\begin{bmatrix}8&0\\0&8\end{bmatrix},
\displaystyle \text{then write the value of }|A|.\hspace{2.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A(\ {adj \ }A)=\begin{bmatrix}8&0\\0&8\end{bmatrix}
\displaystyle \Rightarrow |A(\ {adj \ }A)|=\left|\begin{matrix}8&0\\0&8\end{matrix}\right|
\displaystyle \Rightarrow |A||\ {adj \ }A|=64-0
\displaystyle \Rightarrow |A||A|^{2-1}=64\qquad[\because |\ {adj \ }A|=|A|^{n-1}]
\displaystyle \Rightarrow |A|^2=64
\displaystyle \Rightarrow |A|=\pm8
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\displaystyle \textbf{Question 14. }\text{For what values of }k,\text{ the system of linear equations}
\displaystyle x+y+z=2
\displaystyle 2x+y-z=3
\displaystyle 3x+2y+kz=4
\displaystyle \text{has a unique solution?}\hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, system of linear equations is}
\displaystyle x+y+z=2
\displaystyle 2x+y-z=3
\displaystyle 3x+2y+kz=4
\displaystyle \text{It can be written in matrix form as}
\displaystyle AX=B
\displaystyle \text{where, }A=\begin{bmatrix}1&1&1\\2&1&-1\\3&2&k\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\3\\4\end{bmatrix}
\displaystyle \text{The condition for the system of linear equations to have a unique solution is,}
\displaystyle |A|\neq0
\displaystyle \therefore \begin{vmatrix}1&1&1\\2&1&-1\\3&2&k\end{vmatrix}\neq0
\displaystyle \Rightarrow 1(k+2)-1(2k+3)+1(4-3)\neq0
\displaystyle \Rightarrow k+2-2k-3+1\neq0
\displaystyle \Rightarrow -k\neq0
\displaystyle \Rightarrow k\neq0
\displaystyle \text{Hence, for a unique solution }k\text{ should be a non-zero real number.}
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\displaystyle \textbf{Question 15. }\text{Find }|{adj \ } A|,\text{ if }A=\begin{bmatrix}5&2\\7&3\end{bmatrix}.
\displaystyle \hspace{2.2cm}\text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}5&2\\7&3\end{bmatrix}
\displaystyle \text{Clearly, }|A|=\begin{vmatrix}5&2\\7&3\end{vmatrix}=15-14=1
\displaystyle \text{We know that, if }A\text{ is a non-singular matrix of order }n,
\displaystyle \text{then }|\ {adj \ }(A)|=|A|^{n-1}
\displaystyle \therefore |\ {adj \ }(A)|=(1)^{2-1}=1
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\displaystyle \textbf{Question 16. }\text{If }A\text{ is a square matrix of order }3\text{ such that }|{adj \ } A|=64,
\displaystyle \text{then find }|A|.\hspace{2.2cm}\text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle \text{We know that, for a square matrix of order }n,
\displaystyle |\ {adj \ }(A)|=|A|^{n-1}
\displaystyle \text{Here, the order of }A\text{ is }3,\text{ therefore }n=3
\displaystyle \text{Now, }|\ {adj \ }(A)|=|A|^{3-1}=|A|^2
\displaystyle \text{Given, }|\ {adj \ }(A)|=64\Rightarrow 64=|A|^2=(8)^2=4|A|^2
\displaystyle \Rightarrow |A|=\pm8\qquad[\text{taking square root}]
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\displaystyle \textbf{Question 27. }\text{Write }A^{-1}\text{ for }A=\begin{bmatrix}2&5\\1&3\end{bmatrix}.
\displaystyle \hspace{2.2cm}\text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}2&5\\1&3\end{bmatrix}
\displaystyle \text{Clearly, }\ {adj \ }A=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}
\displaystyle \text{and }\ |A|=\begin{vmatrix}2&5\\1&3\end{vmatrix}=6-5=1
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\ {adj \ }(A)=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}
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\displaystyle \textbf{Question 18. }\text{Given }A=\begin{bmatrix}2&-3\\-4&7\end{bmatrix},\text{ compute }A^{-1}\text{ and show that}
\displaystyle 2A^{-1}=9I-A.\hspace{2.2cm}\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}2&-3\\-4&7\end{bmatrix}
\displaystyle \text{Here, }|A|=\begin{vmatrix}2&-3\\-4&7\end{vmatrix}=14-12=2\neq0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Clearly, }\ {adj \ }(A)=\begin{bmatrix}7&3\\4&2\end{bmatrix}
\displaystyle \text{If }A=\begin{bmatrix}a&b\\c&d\end{bmatrix},\text{ then }\ {adj \ }(A)=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\ {adj \ }(A)=\frac{1}{2}\begin{bmatrix}7&3\\4&2\end{bmatrix}\qquad ...(i)
\displaystyle \text{Now, consider RHS }=9I-A
\displaystyle =9\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}2&-3\\-4&7\end{bmatrix}
\displaystyle =\begin{bmatrix}9&0\\0&9\end{bmatrix}-\begin{bmatrix}2&-3\\-4&7\end{bmatrix}
\displaystyle =\begin{bmatrix}7&3\\4&2\end{bmatrix}
\displaystyle =2A^{-1}\qquad[\text{Using Eq. }(i)]
\displaystyle \therefore 9I-A=2A^{-1}=\text{LHS}
\displaystyle \text{Hence proved.}
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\displaystyle \textbf{Question 19. }\text{If }A=\begin{bmatrix}2&3\\5&-2\end{bmatrix}\text{ be such that }A^{-1}=kA,\text{ then find the}
\displaystyle \text{value of }k.\hspace{2.2cm}\text{[CBSE 2018C, CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}2&3\\5&-2\end{bmatrix}
\displaystyle \text{Clearly, }\ {adj \ }A=\begin{bmatrix}-2&-3\\-5&2\end{bmatrix}
\displaystyle \text{and }\ |A|=\begin{vmatrix}2&3\\5&-2\end{vmatrix}=-4-15=-19
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\ {adj \ }A=\frac{1}{-19}\begin{bmatrix}-2&-3\\-5&2\end{bmatrix}
\displaystyle =\frac{1}{19}\begin{bmatrix}2&3\\5&-2\end{bmatrix}=\frac{1}{19}A
\displaystyle \text{On comparing with }A^{-1}=kA\text{ (given), we get}
\displaystyle k=\frac{1}{19}
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\displaystyle \textbf{Question 20. }\text{If }A=\begin{bmatrix}1&-2&3\\0&-1&4\\-2&2&1\end{bmatrix},\text{ then find }(A')^{-1}.
\displaystyle \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}1&-2&3\\0&-1&4\\-2&2&1\end{bmatrix}
\displaystyle \text{Now, }|A|=\begin{vmatrix}1&-2&3\\0&-1&4\\-2&2&1\end{vmatrix}
\displaystyle =1(-1-8)+2(0+8)+3(0-2)\qquad[\text{expanding along }R_1]
\displaystyle =-9+16-6=1\neq0
\displaystyle \text{So, }A\text{ is non-singular matrix and its inverse exists.}
\displaystyle \text{Cofactors of elements of }|A|\text{ are}
\displaystyle A_{11}=(-1)^{1+1}\begin{vmatrix}-1&4\\2&1\end{vmatrix}=(-1-8)=-9
\displaystyle A_{12}=(-1)^{1+2}\begin{vmatrix}0&4\\-2&1\end{vmatrix}=-(0+8)=-8
\displaystyle A_{13}=(-1)^{1+3}\begin{vmatrix}0&-1\\-2&2\end{vmatrix}=(0-2)=-2
\displaystyle A_{21}=(-1)^{2+1}\begin{vmatrix}-2&3\\2&1\end{vmatrix}=-(-2-6)=8
\displaystyle A_{22}=(-1)^{2+2}\begin{vmatrix}1&3\\-2&1\end{vmatrix}=(1+6)=7
\displaystyle A_{23}=(-1)^{2+3}\begin{vmatrix}1&-2\\-2&2\end{vmatrix}=-(2-4)=2
\displaystyle A_{31}=(-1)^{3+1}\begin{vmatrix}-2&3\\-1&4\end{vmatrix}=(-8+3)=-5
\displaystyle A_{32}=(-1)^{3+2}\begin{vmatrix}1&3\\0&4\end{vmatrix}=-(4-0)=-4
\displaystyle A_{33}=(-1)^{3+3}\begin{vmatrix}1&-2\\0&-1\end{vmatrix}=(-1+0)=-1
\displaystyle \text{Thus, }\ {adj \ }A=\begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{bmatrix}=\begin{bmatrix}-9&8&-5\\-8&7&-4\\-2&2&-1\end{bmatrix}
\displaystyle \text{Hence, }A^{-1}=\frac{1}{|A|}\ {adj \ }A=\begin{bmatrix}-9&8&-5\\-8&7&-4\\-2&2&-1\end{bmatrix}
\displaystyle \text{Now, }(A')^{-1}=(A^{-1})'=\begin{bmatrix}-9&8&-5\\-8&7&-4\\-2&2&-1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}
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\displaystyle \textbf{Question 21. }\text{Find the adjoint of the matrix }A=\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}
\displaystyle \text{and hence show that }A({adj \ } A)=|A|I_{3}. \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}
\displaystyle \text{Let }A_{ij}\text{ be the cofactor of an element }a_{ij}\text{ of }|A|.
\displaystyle \text{Then, cofactors of elements of }|A|\text{ are}
\displaystyle A_{11}=(-1)^{1+1}\begin{vmatrix}1&-2\\-2&1\end{vmatrix}=(1-4)=-3
\displaystyle A_{12}=(-1)^{1+2}\begin{vmatrix}2&-2\\2&1\end{vmatrix}=-(2+4)=-6
\displaystyle A_{13}=(-1)^{1+3}\begin{vmatrix}2&1\\2&-2\end{vmatrix}=(-4-2)=-6
\displaystyle A_{21}=(-1)^{2+1}\begin{vmatrix}-2&-2\\-2&1\end{vmatrix}=-(-2-4)=6
\displaystyle A_{22}=(-1)^{2+2}\begin{vmatrix}-1&-2\\2&1\end{vmatrix}=(-1+4)=3
\displaystyle A_{23}=(-1)^{2+3}\begin{vmatrix}-1&-2\\2&-2\end{vmatrix}=-(2+4)=-6
\displaystyle A_{31}=(-1)^{3+1}\begin{vmatrix}-2&-2\\1&-2\end{vmatrix}=(4+2)=6
\displaystyle A_{32}=(-1)^{3+2}\begin{vmatrix}-1&-2\\2&-2\end{vmatrix}=-(2+4)=-6
\displaystyle A_{33}=(-1)^{3+3}\begin{vmatrix}-1&-2\\2&1\end{vmatrix}=(-1+4)=3
\displaystyle \text{Clearly, the adjoint of the matrix }A\text{ is given by}
\displaystyle \ {adj \ }A=\begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{bmatrix}=\begin{bmatrix}-3&6&6\\-6&3&-6\\-6&-6&3\end{bmatrix}
\displaystyle \text{Now, }|A|=\begin{vmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{vmatrix}
\displaystyle =-1(1-4)+2(2+4)-2(-4-2)
\displaystyle =3+12+12=27
\displaystyle \text{and }A(\ {adj \ }A)=\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}\begin{bmatrix}-3&6&6\\-6&3&-6\\-6&-6&3\end{bmatrix}
\displaystyle =\begin{bmatrix}3+12+12&-6-6+12&-6+12-6\\-6-3+12&12+3+12&12-6-6\\-6+12-3&12-6-6&12+12+3\end{bmatrix}
\displaystyle =\begin{bmatrix}27&0&0\\0&27&0\\0&0&27\end{bmatrix}=27\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle =27I_3=|A|I_3
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 22. }\text{If }A=\begin{bmatrix}2&-1\\-1&2\end{bmatrix}\text{ and }I\text{ is the identity matrix of order }2,
\displaystyle \text{then show that }A^{2}=4A-3I.\text{ Hence, find }A^{-1}. \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}2&-1\\-1&2\end{bmatrix}
\displaystyle \text{Now, }A^2=A\cdot A=\begin{bmatrix}2&-1\\-1&2\end{bmatrix}\begin{bmatrix}2&-1\\-1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}4+1&-2-2\\-2-2&1+4\end{bmatrix}=\begin{bmatrix}5&-4\\-4&5\end{bmatrix}\qquad ...(i)
\displaystyle \text{and, }4A-3I=4\begin{bmatrix}2&-1\\-1&2\end{bmatrix}-3\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}8&-4\\-4&8\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}=\begin{bmatrix}5&-4\\-4&5\end{bmatrix}\qquad ...(ii)
\displaystyle \text{From Eqs. }(i)\text{ and }(ii),\text{ we get}
\displaystyle A^2=4A-3I\qquad ...(iii)
\displaystyle \text{Here, }|A|=\begin{vmatrix}2&-1\\-1&2\end{vmatrix}=4-1=3\neq0
\displaystyle \text{So, }A\text{ is non-singular matrix and its inverse exists.}
\displaystyle \text{Now, pre-multiplying both sides of Eq. }(iii)\text{ by }A^{-1},\text{ we get}
\displaystyle A^{-1}A^2=A^{-1}(4A-3I)
\displaystyle \Rightarrow (A^{-1}A)A=4A^{-1}A-3A^{-1}I
\displaystyle \Rightarrow IA=4I-3A^{-1}
\displaystyle \Rightarrow A=4I-3A^{-1}
\displaystyle \Rightarrow 3A^{-1}=4I-A
\displaystyle \Rightarrow A^{-1}=\frac{1}{3}(4I-A)
\displaystyle =\frac{1}{3}\left(4\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}2&-1\\-1&2\end{bmatrix}\right)
\displaystyle =\frac{1}{3}\left(\begin{bmatrix}4&0\\0&4\end{bmatrix}-\begin{bmatrix}2&-1\\-1&2\end{bmatrix}\right)
\displaystyle =\frac{1}{3}\begin{bmatrix}2&1\\1&2\end{bmatrix}=\begin{bmatrix}\frac{2}{3}&\frac{1}{3}\\\frac{1}{3}&\frac{2}{3}\end{bmatrix}
\\

\displaystyle \textbf{Question 23. }\text{If }A=\begin{bmatrix}2&3\\1&-4\end{bmatrix},\ B=\begin{bmatrix}1&-2\\-1&3\end{bmatrix},\text{ verify that}
\displaystyle (AB)^{-1}=B^{-1}A^{-1}. \hspace{2.2cm}\text{[CBSE 2015C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}2&3\\1&-4\end{bmatrix}\text{ and }B=\begin{bmatrix}1&-2\\-1&3\end{bmatrix}
\displaystyle \therefore AB=\begin{bmatrix}2&3\\1&-4\end{bmatrix}\begin{bmatrix}1&-2\\-1&3\end{bmatrix}
\displaystyle =\begin{bmatrix}2-3&-4+9\\1+4&-2-12\end{bmatrix}=\begin{bmatrix}-1&5\\5&-14\end{bmatrix}
\displaystyle \text{Now, }|A|=\begin{vmatrix}2&3\\1&-4\end{vmatrix}=-8-3=-11\neq0
\displaystyle |B|=\begin{vmatrix}1&-2\\-1&3\end{vmatrix}=3-2=1\neq0
\displaystyle \text{and }|AB|=\begin{vmatrix}-1&5\\5&-14\end{vmatrix}=14-25=-11\neq0
\displaystyle \text{Thus, }A,\ B\text{ and }AB\text{ are non-singular matrices, so their inverse exists.}
\displaystyle \text{Now, }\ {adj \ }(A)=\begin{bmatrix}-4&-3\\-1&2\end{bmatrix}
\displaystyle \qquad[\because \ {adj \ }\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}]
\displaystyle \ {adj \ }(B)=\begin{bmatrix}3&2\\1&1\end{bmatrix}
\displaystyle \text{and }\ {adj \ }(AB)=\begin{bmatrix}-14&-5\\-5&-1\end{bmatrix}
\displaystyle \therefore (AB)^{-1}=\frac{1}{|AB|}\ {adj \ }(AB)=\frac{1}{-11}\begin{bmatrix}-14&-5\\-5&-1\end{bmatrix}
\displaystyle =\frac{1}{11}\begin{bmatrix}14&5\\5&1\end{bmatrix}\qquad ...(i)
\displaystyle A^{-1}=\frac{1}{|A|}\ {adj \ }(A)=\frac{1}{-11}\begin{bmatrix}-4&-3\\-1&2\end{bmatrix}
\displaystyle =\frac{1}{11}\begin{bmatrix}4&3\\1&-2\end{bmatrix}
\displaystyle \text{and }\ B^{-1}=\frac{1}{|B|}\ {adj \ }(B)=\begin{bmatrix}3&2\\1&1\end{bmatrix}
\displaystyle \text{Now, }B^{-1}A^{-1}=\begin{bmatrix}3&2\\1&1\end{bmatrix}\frac{1}{11}\begin{bmatrix}4&3\\1&-2\end{bmatrix}
\displaystyle =\frac{1}{11}\begin{bmatrix}3&2\\1&1\end{bmatrix}\begin{bmatrix}4&3\\1&-2\end{bmatrix}
\displaystyle =\frac{1}{11}\begin{bmatrix}12+2&9-4\\4+1&3-2\end{bmatrix}
\displaystyle =\frac{1}{11}\begin{bmatrix}14&5\\5&1\end{bmatrix}=(AB)^{-1}\qquad[\text{from Eq. }(i)]
\displaystyle \text{Hence, }(AB)^{-1}=B^{-1}A^{-1}
\\

\displaystyle \textbf{Question 24. }\text{If }A=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix}\text{ and }B^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix},\text{ find}
\displaystyle (AB)^{-1}. \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{We know that, }(AB)^{-1}=B^{-1}A^{-1}
\displaystyle \text{Here, }A=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix}
\displaystyle \Rightarrow |A|=1(3+0)-2(-1-0)-2(2-0)
\displaystyle =3+2-4=1\neq0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Cofactors of element of }|A|\text{ are}
\displaystyle A_{11}=(-1)^{1+1}\begin{vmatrix}3&0\\-2&1\end{vmatrix}=(3-0)=3
\displaystyle A_{12}=(-1)^{1+2}\begin{vmatrix}-1&0\\0&1\end{vmatrix}=-(-1-0)=1
\displaystyle A_{13}=(-1)^{1+3}\begin{vmatrix}-1&3\\0&-2\end{vmatrix}=(2-0)=2
\displaystyle A_{21}=(-1)^{2+1}\begin{vmatrix}2&-2\\-2&1\end{vmatrix}=-(2-4)=2
\displaystyle A_{22}=(-1)^{2+2}\begin{vmatrix}1&-2\\0&1\end{vmatrix}=1-0=1
\displaystyle A_{23}=(-1)^{2+3}\begin{vmatrix}1&2\\0&-2\end{vmatrix}=-(-2-0)=2
\displaystyle A_{31}=(-1)^{3+1}\begin{vmatrix}2&-2\\3&0\end{vmatrix}=(0+6)=6
\displaystyle A_{32}=(-1)^{3+2}\begin{vmatrix}1&-2\\-1&0\end{vmatrix}=-(0-2)=2
\displaystyle A_{33}=(-1)^{3+3}\begin{vmatrix}1&2\\-1&3\end{vmatrix}=(3+2)=5
\displaystyle \ {adj \ }A=\begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\ {adj \ }(A)=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}
\displaystyle \text{Now, }(AB)^{-1}=B^{-1}A^{-1}
\displaystyle =\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}
\displaystyle =\begin{bmatrix}9-1+2&6-1+2&18-2+5\\-45+6-10&-30+6-10&-90+12-25\\15-2+4&10-2+4&30-4+10\end{bmatrix}
\displaystyle =\begin{bmatrix}10&7&21\\-49&-34&-103\\17&12&36\end{bmatrix}
\\

\displaystyle \textbf{Question 25. }\text{If }A=\begin{bmatrix}3&2\\5&-7\end{bmatrix},\text{ then find }A^{-1}\text{ and use it to solve the}
\displaystyle \text{following system of equations:}
\displaystyle 3x+5y=11,\ 2x-7y=-3. \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}3&2\\5&-7\end{bmatrix}
\displaystyle |A|=-21-10=-31\neq0
\displaystyle \text{Since, }A\text{ is non-singular, so }A^{-1}\text{ exists.}
\displaystyle \text{Now, cofactors of elements of }|A|\text{ are}
\displaystyle A_{11}=(-1)^{1+1}(-7)=-7
\displaystyle A_{12}=(-1)^{1+2}(5)=-5
\displaystyle A_{21}=(-1)^{2+1}(2)=-2
\displaystyle A_{22}=(-1)^{2+2}(3)=3
\displaystyle \therefore \ {adj \ }A=\begin{bmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{bmatrix}^{T}
\displaystyle \Rightarrow \ {adj \ }A=\begin{bmatrix}-7&-5\\-2&3\end{bmatrix}^{T}=\begin{bmatrix}-7&-2\\-5&3\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{\ {adj \ }A}{|A|}=\frac{1}{(-31)}\begin{bmatrix}-7&-2\\-5&3\end{bmatrix}
\displaystyle =\begin{bmatrix}\frac{7}{31}&\frac{2}{31}\\\frac{5}{31}&-\frac{3}{31}\end{bmatrix}
\displaystyle \text{Now, the given system of equations is expressible as}
\displaystyle \begin{bmatrix}3&5\\2&-7\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}11\\-3\end{bmatrix}
\displaystyle \Rightarrow A^{T}X=B
\displaystyle \Rightarrow X=(A^{T})^{-1}B
\displaystyle =(A^{-1})^{T}B\qquad[\because (A^{T})^{-1}=(A^{-1})^{T}]
\displaystyle \Rightarrow \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\frac{7}{31}&\frac{5}{31}\\\frac{2}{31}&-\frac{3}{31}\end{bmatrix}\begin{bmatrix}11\\-3\end{bmatrix}
\displaystyle =\begin{bmatrix}\frac{77}{31}-\frac{15}{31}\\\frac{22}{31}+\frac{9}{31}\end{bmatrix}=\begin{bmatrix}\frac{62}{31}\\\frac{31}{31}\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}
\displaystyle \Rightarrow x=2\text{ and }y=1
\\

\displaystyle \textbf{Question 26. }\text{Solve the following system of equations by matrix}
\displaystyle \text{method:}\hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle x+2y+3z=6
\displaystyle 2x-y+z=2
\displaystyle 3x+2y-2z=3
\displaystyle \text{Answer:}
\displaystyle \text{The given system of equation can be written as}
\displaystyle AX=B,\text{ where}
\displaystyle A=\begin{bmatrix}1&2&3\\2&-1&1\\3&2&-2\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}6\\2\\3\end{bmatrix}
\displaystyle \text{Here, }|A|=1(2-2)-2(-4-3)+3(4+3)
\displaystyle =0+14+21=35\neq0
\displaystyle \text{So, }A^{-1}\text{ exists.}
\displaystyle \text{Cofactors of elements of }|A|\text{ are}
\displaystyle A_{11}=(-1)^{1+1}\begin{vmatrix}-1&1\\2&-2\end{vmatrix}=(2-2)=0
\displaystyle A_{12}=(-1)^{1+2}\begin{vmatrix}2&1\\3&-2\end{vmatrix}=-(-4-3)=7
\displaystyle A_{13}=(-1)^{1+3}\begin{vmatrix}2&-1\\3&2\end{vmatrix}=(4+3)=7
\displaystyle A_{21}=(-1)^{2+1}\begin{vmatrix}2&3\\2&-2\end{vmatrix}=-(-4-6)=10
\displaystyle A_{22}=(-1)^{2+2}\begin{vmatrix}1&3\\3&-2\end{vmatrix}=(-2-9)=-11
\displaystyle A_{23}=(-1)^{2+3}\begin{vmatrix}1&2\\3&2\end{vmatrix}=-(2-6)=4
\displaystyle A_{31}=(-1)^{3+1}\begin{vmatrix}2&3\\-1&1\end{vmatrix}=2+3=5
\displaystyle A_{32}=(-1)^{3+2}\begin{vmatrix}1&3\\2&1\end{vmatrix}=-(1-6)=5
\displaystyle A_{33}=(-1)^{3+3}\begin{vmatrix}1&2\\2&-1\end{vmatrix}=(-1-4)=-5
\displaystyle \therefore \ {adj \ }A=\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}0&10&5\\7&-11&5\\7&4&-5\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{|A|}(\ {adj \ }A)=\frac{1}{35}\begin{bmatrix}0&10&5\\7&-11&5\\7&4&-5\end{bmatrix}
\displaystyle \text{Now, }X=A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{35}\begin{bmatrix}0&10&5\\7&-11&5\\7&4&-5\end{bmatrix}\begin{bmatrix}6\\2\\3\end{bmatrix}
\displaystyle =\frac{1}{35}\begin{bmatrix}0+20+15\\42-22+15\\42+8-15\end{bmatrix}=\frac{1}{35}\begin{bmatrix}35\\35\\35\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}
\displaystyle \therefore x=1,\ y=1\text{ and }z=1
\\

\displaystyle \textbf{Question 27. }\text{Find the inverse of the matrix}
\displaystyle A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}. \text{ Using the inverse,}
\displaystyle A^{-1},\text{ solve the system of linear equations}
\displaystyle x-y+2z=1,\ 2y-3z=1,\ 3x-2y+4z=3. \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}
\displaystyle \therefore |A|=1(8-6)+1(0+9)+2(0-6)
\displaystyle =2+9-12
\displaystyle =-1\neq0
\displaystyle \text{So, }A\text{ is invertible.}
\displaystyle \text{Cofactors of elements of }|A|\text{ are}
\displaystyle A_{11}=(-1)^{1+1}\begin{vmatrix}2&-3\\-2&4\end{vmatrix}=8-6=2
\displaystyle A_{12}=(-1)^{1+2}\begin{vmatrix}0&-3\\3&4\end{vmatrix}=-(0+9)=-9
\displaystyle A_{13}=(-1)^{1+3}\begin{vmatrix}0&2\\3&-2\end{vmatrix}=(0-6)=-6
\displaystyle A_{21}=(-1)^{2+1}\begin{vmatrix}-1&2\\-2&4\end{vmatrix}=-(-4+4)=0
\displaystyle A_{22}=(-1)^{2+2}\begin{vmatrix}1&2\\3&4\end{vmatrix}=(4-6)=-2
\displaystyle A_{23}=(-1)^{2+3}\begin{vmatrix}1&-1\\3&-2\end{vmatrix}=-(-2+3)=-1
\displaystyle A_{31}=(-1)^{3+1}\begin{vmatrix}-1&2\\2&-3\end{vmatrix}=3-4=-1
\displaystyle A_{32}=(-1)^{3+2}\begin{vmatrix}1&2\\0&-3\end{vmatrix}=-(-3-0)=3
\displaystyle A_{33}=(-1)^{3+3}\begin{vmatrix}1&-1\\0&2\end{vmatrix}=(2-0)=2
\displaystyle \therefore \ {adj \ }A=\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}2&-9&-6\\0&-2&-1\\-1&3&2\end{bmatrix}^{T}=\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{\ {adj \ }A}{|A|}=\frac{1}{-1}\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}
\displaystyle \text{Now, the given system of equation can be written in the form of }AX=B.
\displaystyle \text{Where, }A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}1\\1\\3\end{bmatrix}
\displaystyle \text{As, }|A|=-1\neq0,\text{ so given system of equation has a unique solution given by }X=A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}\begin{bmatrix}1\\1\\3\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-2+0+3\\9+2-9\\6+1-6\end{bmatrix}\Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\1\end{bmatrix}
\displaystyle \text{Hence, }x=1,\ y=2\text{ and }z=1.
\\

\displaystyle \textbf{Question 28. }\text{If }A=\begin{bmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{bmatrix},\ B=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix},\text{ then find }AB
\displaystyle \text{and use it to solve the following system of equations}
\displaystyle x-2y=3
\displaystyle 2x-y-z=2
\displaystyle -2y+z=3 \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{bmatrix}\text{ and }B=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix}
\displaystyle \therefore AB=\begin{bmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{bmatrix}\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-3+4+0&-6+2+4&0+4-4\\2-2+0&4-1-2&0-2+2\\2-2+0&4-1-3&0-2+3\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle \therefore B^{-1}=A
\displaystyle \text{Now, the given system of equation can be written as}
\displaystyle B^{T}X=C
\displaystyle \text{Where, }X=\begin{bmatrix}x\\y\\z\end{bmatrix},\ C=\begin{bmatrix}3\\2\\3\end{bmatrix}
\displaystyle \therefore X=(B^{T})^{-1}C=(B^{-1})^{T}C\qquad[\because (B^{T})^{-1}=(B^{-1})^{T}]
\displaystyle =A^{T}C
\displaystyle =\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}\begin{bmatrix}3\\2\\3\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-9+4+6\\-6+2+3\\-12+4+9\end{bmatrix}\Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\-1\\1\end{bmatrix}
\displaystyle \therefore x=1,\ y=-1\text{ and }z=1
\\

\displaystyle \textbf{Question 29. }\text{If }A=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},\text{ then find }A^{-1}.
\displaystyle \text{Using }A^{-1},\text{ solve the following system of equations:}
\displaystyle 2x-3y+5z=11
\displaystyle 3x+2y-4z=-5
\displaystyle x+y-2z=-3\hspace{2.2cm}\text{[CBSE 2020, CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}
\displaystyle \therefore |A|=2(-4+4)+3(-6+4)+5(3-2)
\displaystyle =0-6+5=-1\neq0
\displaystyle \text{So, }A\text{ is invertible.}
\displaystyle \text{Cofactors of }A\text{ are}
\displaystyle C_{11}=(-1)^{1+1}\begin{vmatrix}2&-4\\1&-2\end{vmatrix}=(-4+4)=0
\displaystyle C_{12}=(-1)^{1+2}\begin{vmatrix}3&-4\\1&-2\end{vmatrix}=-(-6+4)=2
\displaystyle C_{13}=(-1)^{1+3}\begin{vmatrix}3&2\\1&1\end{vmatrix}=(3-2)=1
\displaystyle C_{21}=(-1)^{2+1}\begin{vmatrix}-3&5\\1&-2\end{vmatrix}=-(6-5)=-1
\displaystyle C_{22}=(-1)^{2+2}\begin{vmatrix}2&5\\1&-2\end{vmatrix}=(-4-5)=-9
\displaystyle C_{23}=(-1)^{2+3}\begin{vmatrix}2&-3\\1&1\end{vmatrix}=-(2+3)=-5
\displaystyle C_{31}=(-1)^{3+1}\begin{vmatrix}-3&5\\2&-4\end{vmatrix}=(12-10)=2
\displaystyle C_{32}=(-1)^{3+2}\begin{vmatrix}2&5\\3&-4\end{vmatrix}=-(-8-15)=23
\displaystyle C_{33}=(-1)^{3+3}\begin{vmatrix}2&-3\\3&2\end{vmatrix}=(4+9)=13
\displaystyle \therefore \ {adj \ }A=\begin{bmatrix}0&2&1\\-1&-9&-5\\2&23&13\end{bmatrix}^{T}=\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}
\displaystyle \text{Hence, }A^{-1}=\frac{1}{|A|}(\ {adj \ }A)=\frac{1}{-1}\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}
\displaystyle =\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}
\displaystyle \text{Now, the given system of equation can be written in the form }AX=B,\text{ where}
\displaystyle A=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}11\\-5\\-3\end{bmatrix}
\displaystyle \text{As, }|A|=-1\neq0,\text{ so given system of equation has a unique solution given by }X=A^{-1}B.
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\begin{bmatrix}11\\-5\\-3\end{bmatrix}
\displaystyle =\begin{bmatrix}0-5+6\\-22-45+69\\-11-25+39\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}
\displaystyle \text{Hence, }x=1,\ y=2\text{ and }z=3
\\

\displaystyle \textbf{Question 30. }\text{If }A=\begin{bmatrix}1&3&2\\2&0&-1\\1&2&3\end{bmatrix},\text{ then show that}
\displaystyle A^{3}-4A^{2}-3A+11I=O.
\displaystyle \text{Hence find }A^{-1}.\hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}1&3&2\\2&0&-1\\1&2&3\end{bmatrix}
\displaystyle \text{Now, }A^2=A\cdot A=\begin{bmatrix}9&7&5\\1&4&1\\8&9&9\end{bmatrix}
\displaystyle \text{and }A^3=A^2\cdot A=\begin{bmatrix}28&37&26\\10&5&1\\35&42&34\end{bmatrix}
\displaystyle \text{Now, consider LHS }=A^3-4A^2-3A+11I
\displaystyle =\begin{bmatrix}28&37&26\\10&5&1\\35&42&34\end{bmatrix}-4\begin{bmatrix}9&7&5\\1&4&1\\8&9&9\end{bmatrix}
\displaystyle \qquad -3\begin{bmatrix}1&3&2\\2&0&-1\\1&2&3\end{bmatrix}+11\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O=\text{RHS}
\displaystyle \text{Now, }A^3-4A^2-3A+11I=O
\displaystyle \text{Pre-multiplying by }A^{-1},\text{ we get}
\displaystyle A^{-1}A^3-4A^{-1}A^2-3A^{-1}A+11A^{-1}=O
\displaystyle \Rightarrow A^2-4A-3I+11A^{-1}=O\qquad[\because A^{-1}A=I]
\displaystyle \Rightarrow 11A^{-1}=3I+4A-A^2
\displaystyle \Rightarrow 11A^{-1}=3\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+4\begin{bmatrix}1&3&2\\2&0&-1\\1&2&3\end{bmatrix}-\begin{bmatrix}9&7&5\\1&4&1\\8&9&9\end{bmatrix}
\displaystyle =\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}+\begin{bmatrix}4&12&8\\8&0&-4\\4&8&12\end{bmatrix}-\begin{bmatrix}9&7&5\\1&4&1\\8&9&9\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&5&3\\7&-1&-5\\-4&-1&6\end{bmatrix}
\displaystyle \Rightarrow A^{-1}=\frac{1}{11}\begin{bmatrix}-2&5&3\\7&-1&-5\\-4&-1&6\end{bmatrix}
\\

\displaystyle \textbf{Question 31. }\text{Show that for the matrix }A=\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix},
\displaystyle A^{3}-6A^{2}+5A+11I=O.
\displaystyle \text{Hence, find }A^{-1}.\hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}
\displaystyle \therefore A^2=AA=\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}
\displaystyle =\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}
\displaystyle \text{and }A^3=A^2A=\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}
\displaystyle =\begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}
\displaystyle \therefore A^3-6A^2+5A+11I_3
\displaystyle =\begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}-6\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}
\displaystyle \qquad +5\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}+11\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}8-24+5+11&7-12+5+0&1-6+5+0\\-23+18+5+0&27-48+10+11&-69+84-15+0\\32-42+10+0&-13+18-5+0&58-84+15+11\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O
\displaystyle \Rightarrow A^3-6A^2+5A+11I_3=O
\displaystyle \text{Now, multiplying both sides by }A^{-1},\text{ we get}
\displaystyle A^{-1}(A^3-6A^2+5A+11I_3)=A^{-1}O
\displaystyle \Rightarrow A^2-6A+5I+11A^{-1}=O
\displaystyle \Rightarrow 11A^{-1}=-A^2+6A-5I
\displaystyle \Rightarrow 11A^{-1}=-\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}
\displaystyle \qquad +6\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}-5\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-4+6-5&-2+6+0&-1+6+0\\3+6+0&-8+12-5&14-18+0\\-7+12+0&3-6+0&-14+18-5\end{bmatrix}
\displaystyle =\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}
\displaystyle \Rightarrow A^{-1}=\frac{1}{11}\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}
\\

\displaystyle \textbf{Question 32. }\text{If }A=\begin{bmatrix}1&3&4\\2&1&2\\5&1&1\end{bmatrix},\text{ find }A^{-1}.
\displaystyle \text{Hence, solve the system of equations}
\displaystyle x+3y+4z=8
\displaystyle 2x+y+2z=5
\displaystyle \text{and }5x+y+z=7\hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }|A|=\begin{vmatrix}1&3&4\\2&1&2\\5&1&1\end{vmatrix}
\displaystyle =1(1-2)-3(2-10)+4(2-5)
\displaystyle =-1+24-12=11\neq0
\displaystyle \text{Thus, }A\text{ is invertible.}
\displaystyle \text{Clearly, the system has a unique solution given by}
\displaystyle X=A^{-1}B
\displaystyle \text{Now, the cofactors of }|A|\text{ are}
\displaystyle A_{11}=-1,\ A_{12}=8,\ A_{13}=-3,
\displaystyle A_{21}=-1,\ A_{22}=-19,\ A_{23}=14,
\displaystyle A_{31}=2,\ A_{32}=6\text{ and }A_{33}=-5
\displaystyle \therefore \ {adj \ }(A)=\begin{bmatrix}-1&8&-3\\-1&-19&14\\2&6&-5\end{bmatrix}^{T}=\begin{bmatrix}-1&-1&2\\8&-19&6\\-3&14&-5\end{bmatrix}
\displaystyle \text{So, }A^{-1}=\frac{1}{|A|}\ {adj \ }(A)
\displaystyle =\frac{1}{11}\begin{bmatrix}-1&-1&2\\8&-19&6\\-3&14&-5\end{bmatrix}
\displaystyle =\begin{bmatrix}-\frac{1}{11}&-\frac{1}{11}&\frac{2}{11}\\\frac{8}{11}&-\frac{19}{11}&\frac{6}{11}\\-\frac{3}{11}&\frac{14}{11}&-\frac{5}{11}\end{bmatrix}
\displaystyle \text{The given equations are}
\displaystyle x+3y+4z=8\qquad ...(i)
\displaystyle 2x+y+2z=5\qquad ...(ii)
\displaystyle 5x+y+z=7\qquad ...(iii)
\displaystyle \text{which can be written in matrix form as }AX=B,
\displaystyle \text{where }A=\begin{bmatrix}1&3&4\\2&1&2\\5&1&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}8\\5\\7\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-\frac{1}{11}&-\frac{1}{11}&\frac{2}{11}\\\frac{8}{11}&-\frac{19}{11}&\frac{6}{11}\\-\frac{3}{11}&\frac{14}{11}&-\frac{5}{11}\end{bmatrix}\begin{bmatrix}8\\5\\7\end{bmatrix}
\displaystyle =\begin{bmatrix}-\frac{8}{11}-\frac{5}{11}+\frac{14}{11}\\\frac{64}{11}-\frac{95}{11}+\frac{42}{11}\\-\frac{24}{11}+\frac{70}{11}-\frac{35}{11}\end{bmatrix}
\displaystyle =\begin{bmatrix}\frac{1}{11}\\\frac{11}{11}\\\frac{11}{11}\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}
\displaystyle \therefore x=1,\ y=1\text{ and }z=1
\\

\displaystyle \textbf{Question 33. }\text{If }A=\begin{bmatrix}1&1&1\\1&0&2\\3&1&1\end{bmatrix},\text{ find }A^{-1}.\text{ Hence, solve the system of}
\displaystyle \text{equations }x+y+z=6,\ x+2z=7,\ 3x+y+z=12.
\displaystyle \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}1&1&1\\1&0&2\\3&1&1\end{bmatrix}
\displaystyle \therefore |A|=1(0-2)-1(1-6)+1(1-0)
\displaystyle =-2+5+1
\displaystyle =4\neq0
\displaystyle \text{So, }A\text{ is invertible.}
\displaystyle \text{Clearly, }\ {adj \ }(A)=\begin{bmatrix}-2&0&2\\5&-2&-1\\1&2&-1\end{bmatrix}
\displaystyle \Rightarrow A^{-1}=\frac{1}{|A|}\ {adj \ }A=\frac{1}{4}\begin{bmatrix}-2&0&2\\5&-2&-1\\1&2&-1\end{bmatrix}
\displaystyle \text{Given, system of the equations is}
\displaystyle x+y+z=6,\ x+2z=7,\ 3x+y+z=12,
\displaystyle \text{which can be written in matrix form as }AX=B,\text{ where}
\displaystyle A=\begin{bmatrix}1&1&1\\1&0&2\\3&1&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}6\\7\\12\end{bmatrix}
\displaystyle \text{Now, }AX=B\Rightarrow X=A^{-1}B
\displaystyle \Rightarrow X=\frac{1}{4}\begin{bmatrix}-2&0&2\\5&-2&-1\\1&2&-1\end{bmatrix}\begin{bmatrix}6\\7\\12\end{bmatrix}
\displaystyle \Rightarrow X=\frac{1}{4}\begin{bmatrix}-12+0+24\\30-14-12\\6+14-12\end{bmatrix}=\frac{1}{4}\begin{bmatrix}12\\4\\8\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\1\\2\end{bmatrix}
\displaystyle \text{Hence, }x=3,\ y=1\text{ and }z=2
\\

\displaystyle \textbf{Question 34. }\text{Given }A=\begin{bmatrix}5&0&4\\2&3&2\\1&2&1\end{bmatrix},\ B^{-1}=\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix},\text{ compute}
\displaystyle (AB)^{-1}.\hspace{2.2cm}\text{[CBSE 2018C]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}5&0&4\\2&3&2\\1&2&1\end{bmatrix}
\displaystyle \text{Now, }|A|=\begin{vmatrix}5&0&4\\2&3&2\\1&2&1\end{vmatrix}=5(3-4)-0+4(4-3)
\displaystyle =-5+4=-1\neq0
\displaystyle \text{Thus, }A\text{ is a non-singular matrix, so }A^{-1}\text{ exists.}
\displaystyle \text{Now, cofactors of elements of }|A|\text{ are}
\displaystyle A_{11}=(-1)^2\begin{vmatrix}3&2\\2&1\end{vmatrix}=(3-4)=-1
\displaystyle A_{12}=(-1)^3\begin{vmatrix}2&2\\1&1\end{vmatrix}=-(2-2)=0
\displaystyle A_{13}=(-1)^4\begin{vmatrix}2&3\\1&2\end{vmatrix}=(4-3)=1
\displaystyle A_{21}=(-1)^3\begin{vmatrix}0&4\\2&1\end{vmatrix}=-(0-8)=8
\displaystyle A_{22}=(-1)^4\begin{vmatrix}5&4\\1&1\end{vmatrix}=(5-4)=1
\displaystyle A_{23}=(-1)^5\begin{vmatrix}5&0\\1&2\end{vmatrix}=-(10-0)=-10
\displaystyle A_{31}=(-1)^4\begin{vmatrix}0&4\\3&2\end{vmatrix}=(0-12)=-12
\displaystyle A_{32}=(-1)^5\begin{vmatrix}5&4\\2&2\end{vmatrix}=-(10-8)=-2
\displaystyle A_{33}=(-1)^6\begin{vmatrix}5&0\\2&3\end{vmatrix}=(15-0)=15
\displaystyle \therefore \ {adj \ }A=\begin{bmatrix}-1&8&-12\\0&1&-2\\1&-10&15\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{\ {adj \ }A}{|A|}=\frac{1}{-1}\begin{bmatrix}-1&8&-12\\0&1&-2\\1&-10&15\end{bmatrix}
\displaystyle =\begin{bmatrix}1&-8&12\\0&-1&2\\-1&10&-15\end{bmatrix}
\displaystyle \text{Now, }(AB)^{-1}=B^{-1}A^{-1}
\displaystyle =\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}\begin{bmatrix}1&-8&12\\0&-1&2\\-1&10&-15\end{bmatrix}
\displaystyle =\begin{bmatrix}1+0-3&-8-3+30&12+6-45\\1+0-3&-8-4+30&12+8-45\\1+0-4&-8-3+40&12+6-60\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&19&-27\\-2&18&-25\\-3&29&-42\end{bmatrix}
\\

\displaystyle \textbf{Question 35. }\text{If }A=\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix},\text{ find }A^{-1}.\text{ Using }A^{-1}\text{ solve the}
\displaystyle \text{system of equations}
\displaystyle \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=2,
\displaystyle \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=5,
\displaystyle \text{and }\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=-4\hspace{2.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{x}=p,\ \frac{1}{y}=q\text{ and }\frac{1}{z}=r.
\displaystyle \text{Then, the given equations become}
\displaystyle 2p+3q+10r=2
\displaystyle 4p-6q+5r=5
\displaystyle 6p+9q-20r=-4
\displaystyle \text{This system can be written as }AX=B,\text{ where}
\displaystyle A=\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix},\ X=\begin{bmatrix}p\\q\\r\end{bmatrix},\ B=\begin{bmatrix}2\\5\\-4\end{bmatrix}
\displaystyle \text{Here, }|A|=\begin{vmatrix}2&3&10\\4&-6&5\\6&9&-20\end{vmatrix}
\displaystyle =2(120-45)-3(-80-30)+10(36+36)
\displaystyle =150+330+720=1200\neq0
\displaystyle \text{Thus, }A\text{ is non-singular, therefore its inverse exists.}
\displaystyle \text{Therefore, the above system has a unique solution given by}
\displaystyle X=A^{-1}B
\displaystyle \text{Cofactors of }A\text{ are}
\displaystyle C_{11}=75,\ C_{21}=150,\ C_{31}=75
\displaystyle C_{12}=110,\ C_{22}=-100,\ C_{32}=30
\displaystyle C_{13}=72,\ C_{23}=0\text{ and }C_{33}=-24
\displaystyle \therefore \ {adj \ }(A)=\begin{bmatrix}75&110&72\\150&-100&0\\75&30&-24\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{|A|}(\ {adj \ }A)=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}
\displaystyle \text{Now, }X=A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}p\\q\\r\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}\begin{bmatrix}2\\5\\-4\end{bmatrix}
\displaystyle =\frac{1}{1200}\begin{bmatrix}150+750-300\\220-500-120\\144+0+96\end{bmatrix}
\displaystyle =\frac{1}{1200}\begin{bmatrix}600\\-400\\240\end{bmatrix}=\begin{bmatrix}\frac{1}{2}\\-\frac{1}{3}\\\frac{1}{5}\end{bmatrix}
\displaystyle \Rightarrow p=\frac{1}{2},\ q=-\frac{1}{3},\ r=\frac{1}{5}
\displaystyle \therefore x=2,\ y=-3\text{ and }z=5
\\

\displaystyle \textbf{Question 36. }\text{Determine the product of }\begin{bmatrix}-4&4&4\\-7&1&3\\5&-3&-1\end{bmatrix}\begin{bmatrix}1&-1&1\\1&-2&-2\\2&1&3\end{bmatrix}
\displaystyle \text{and then use to solve the system of equations}
\displaystyle x-y+z=4,
\displaystyle x-2y-2z=9
\displaystyle \text{and }2x+y+3z=1\hspace{2.2cm}\text{[CBSE 2017; CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }B=\begin{bmatrix}-4&4&4\\-7&1&3\\5&-3&-1\end{bmatrix}\text{ and }A=\begin{bmatrix}1&-1&1\\1&-2&-2\\2&1&3\end{bmatrix}
\displaystyle \text{Now, }BA=\begin{bmatrix}-4&4&4\\-7&1&3\\5&-3&-1\end{bmatrix}\begin{bmatrix}1&-1&1\\1&-2&-2\\2&1&3\end{bmatrix}
\displaystyle =\begin{bmatrix}-4+4+8&4-8+4&-4-8+12\\-7+1+6&7-2+3&-7-2+9\\5-3-2&-5+6-1&5+6-3\end{bmatrix}
\displaystyle =\begin{bmatrix}8&0&0\\0&8&0\\0&0&8\end{bmatrix}=8\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=8I
\displaystyle \Rightarrow BA=8I
\displaystyle \Rightarrow BA(A^{-1})=8I\cdot A^{-1}\qquad[\text{post-multiplying both sides by }A^{-1}]
\displaystyle \Rightarrow B(AA^{-1})=8IA^{-1}
\displaystyle \Rightarrow B=8A^{-1}\qquad[\because AA^{-1}=I]
\displaystyle \therefore A^{-1}=\frac{1}{8}\begin{bmatrix}-4&4&4\\-7&1&3\\5&-3&-1\end{bmatrix}
\displaystyle \text{Hence, }X=\frac{1}{8}\begin{bmatrix}-4&4&4\\-7&1&3\\5&-3&-1\end{bmatrix}\begin{bmatrix}4\\9\\-1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{8}\begin{bmatrix}-16+36-4\\-28+9-3\\20-27+1\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}16\\-22\\-6\end{bmatrix}=\begin{bmatrix}2\\-\frac{11}{4}\\-\frac{3}{4}\end{bmatrix}
\displaystyle \text{On comparing corresponding elements, we get}
\displaystyle x=2,\ y=-\frac{11}{4}\text{ and }z=-\frac{3}{4}
\\

\displaystyle \textbf{Question 37. }\text{Use product }\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}\text{ to solve the system of equations}
\displaystyle x-y+2z=1,\quad 2y-3z=1,\quad 3x-2y+4z=2. \hspace{2.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\text{ and }B=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}
\displaystyle AB=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I
\displaystyle \therefore B=A^{-1}
\displaystyle \text{The given system can be written as}
\displaystyle \begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\1\\2\end{bmatrix}
\displaystyle \therefore \begin{bmatrix}x\\y\\z\end{bmatrix}=A^{-1}\begin{bmatrix}1\\1\\2\end{bmatrix}=B\begin{bmatrix}1\\1\\2\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}\begin{bmatrix}1\\1\\2\end{bmatrix}
\displaystyle =\begin{bmatrix}-2+0+2\\9+2-6\\6+1-4\end{bmatrix}=\begin{bmatrix}0\\5\\3\end{bmatrix}
\displaystyle \therefore x=0,\ y=5,\ z=3
\displaystyle \\

\displaystyle \textbf{Question 38. }\text{If }A=\begin{bmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix},\text{ find adj }A
\displaystyle \text{and verify that }A({adj \ } A)=({adj \ } A)A=|A|I_{3}. \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}
\displaystyle \text{Clearly, the cofactors of elements of }|A|\text{ are given by}
\displaystyle A_{11}=\cos\alpha,\ A_{12}=-\sin\alpha,\ A_{13}=0;
\displaystyle A_{21}=\sin\alpha,\ A_{22}=\cos\alpha,\ A_{23}=0
\displaystyle A_{31}=0,\ A_{32}=0\text{ and }A_{33}=1
\displaystyle \therefore \ {adj \ }(A)=\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}\cos\alpha&\sin\alpha&0\\-\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}
\displaystyle \text{Now, }A(\ {adj \ }A)=\begin{bmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}\begin{bmatrix}\cos\alpha&\sin\alpha&0\\-\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos^2\alpha+\sin^2\alpha&0&0\\0&\sin^2\alpha+\cos^2\alpha&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\qquad ...(i)
\displaystyle (\ {adj \ }A)A=\begin{bmatrix}\cos\alpha&\sin\alpha&0\\-\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}\begin{bmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos^2\alpha+\sin^2\alpha&0&0\\0&\sin^2\alpha+\cos^2\alpha&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\qquad ...(ii)
\displaystyle \text{and }|A|=\begin{vmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{vmatrix}
\displaystyle =1\cdot(\cos^2\alpha+\sin^2\alpha)=1\qquad[\text{expanding along }R_3]
\displaystyle \therefore |A|I_3=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\qquad ...(iii)
\displaystyle \text{From Eqs. }(i),\ (ii)\text{ and }(iii),\text{ we get}
\displaystyle A(\ {adj \ }A)=(\ {adj \ }A)A=|A|I_3
\\

\displaystyle \textbf{Question 39. }\text{A total amount of }\text{Rs.}\ 7000\text{ is deposited in three different}
\displaystyle \text{savings bank accounts with annual interest rates of }5\%,\ 8\%\text{ and }8\frac{1}{2}\%\text{ respectively.}
\displaystyle \text{The total annual interest from these three accounts is }\text{Rs.}\ 550.\text{ Equal amounts}
\displaystyle \text{have been deposited in the }5\%\text{ and }8\%\text{ savings accounts. Find the amount}
\displaystyle \text{deposited in each of the three accounts, with the help of matrices.}
\displaystyle \hspace{2.2cm}\text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let Rs. }x,\ \text{Rs. }y\text{ and Rs. }z\text{ be invested in saving bank accounts at}
\displaystyle \text{the rate of }5\%,\ 8\%\text{ and }8\frac{1}{2}\%,\text{ respectively.}
\displaystyle \text{Then, according to given condition we have the following system of equations}
\displaystyle x+y+z=7000\qquad ...(i)
\displaystyle \text{and }\frac{5x}{100}+\frac{8y}{100}+\frac{17z}{200}=550
\displaystyle \Rightarrow 10x+16y+17z=110000\qquad ...(ii)
\displaystyle \text{and }x-y=0\qquad ...(iii)
\displaystyle \text{This system of equations can be written in matrix form as }AX=B
\displaystyle \text{where, }A=\begin{bmatrix}1&1&1\\10&16&17\\1&-1&0\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}7000\\110000\\0\end{bmatrix}
\displaystyle \text{Here, }|A|=\begin{vmatrix}1&1&1\\10&16&17\\1&-1&0\end{vmatrix}
\displaystyle \Rightarrow |A|=1(0+17)-1(0-17)+1(-10-16)
\displaystyle =17+17-26=8\neq0
\displaystyle \text{So, }A\text{ is non-singular matrix and its inverse exists.}
\displaystyle \text{Now, cofactors of elements of }|A|\text{ are}
\displaystyle A_{11}=(-1)^2\begin{vmatrix}16&17\\-1&0\end{vmatrix}=1(0+17)=17
\displaystyle A_{12}=(-1)^3\begin{vmatrix}10&17\\1&0\end{vmatrix}=-1(0-17)=17
\displaystyle A_{13}=(-1)^4\begin{vmatrix}10&16\\1&-1\end{vmatrix}=1(-10-16)=-26
\displaystyle A_{21}=(-1)^3\begin{vmatrix}1&1\\-1&0\end{vmatrix}=-1(0+1)=-1
\displaystyle A_{22}=(-1)^4\begin{vmatrix}1&1\\1&0\end{vmatrix}=1(0-1)=-1
\displaystyle A_{23}=(-1)^5\begin{vmatrix}1&1\\1&-1\end{vmatrix}=-1(-1-1)=2
\displaystyle A_{31}=(-1)^4\begin{vmatrix}1&1\\16&17\end{vmatrix}=1(17-16)=1
\displaystyle A_{32}=(-1)^5\begin{vmatrix}1&1\\10&17\end{vmatrix}=-1(17-10)=-7
\displaystyle A_{33}=(-1)^6\begin{vmatrix}1&1\\10&16\end{vmatrix}=1(16-10)=6
\displaystyle \therefore \ {adj \ }(A)=\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}17&17&-26\\-1&-1&2\\1&-7&6\end{bmatrix}^{T}=\begin{bmatrix}17&-1&1\\17&-1&-7\\-26&2&6\end{bmatrix}
\displaystyle \text{Now, }A^{-1}=\frac{\ {adj \ }(A)}{|A|}=\frac{1}{8}\begin{bmatrix}17&-1&1\\17&-1&-7\\-26&2&6\end{bmatrix}
\displaystyle \text{and the solution of given system is given by}
\displaystyle X=A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{8}\begin{bmatrix}17&-1&1\\17&-1&-7\\-26&2&6\end{bmatrix}\begin{bmatrix}7000\\110000\\0\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}119000-110000+0\\119000-110000+0\\-182000+220000+0\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}9000\\9000\\38000\end{bmatrix}=\begin{bmatrix}1125\\1125\\4750\end{bmatrix}
\displaystyle \text{On comparing the corresponding elements, we get }x=1125,\ y=1125,\ z=4750
\displaystyle \text{Hence, the amount deposited in each type of account is }₹1125,\ ₹1125\text{ and }₹4750,\text{ respectively.}
\\

\displaystyle \textbf{Question 40. }\text{Using matrices, solve the following system of}
\displaystyle \text{equations.}
\displaystyle x-y+2z=7
\displaystyle 3x+4y-5z=-5
\displaystyle \text{and }2x-y+3z=12.\hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, system of equations is}
\displaystyle x-y+2z=7
\displaystyle 3x+4y-5z=-5
\displaystyle \text{and }2x-y+3z=12
\displaystyle \text{In matrix form, it can be written as}
\displaystyle AX=B\qquad ...(i)
\displaystyle \text{where, }A=\begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}7\\-5\\12\end{bmatrix}
\displaystyle \text{Here, }|A|=1(12-5)+1(9+10)+2(-3-8)
\displaystyle =1(7)+1(19)+2(-11)
\displaystyle =7+19-22=4
\displaystyle \Rightarrow |A|\neq0
\displaystyle \text{So, }A\text{ is non-singular and its inverse exists.}
\displaystyle \text{Now, cofactors of elements of }|A|\text{ are}
\displaystyle A_{11}=(-1)^2\begin{vmatrix}4&-5\\-1&3\end{vmatrix}=1(12-5)=7
\displaystyle A_{12}=(-1)^3\begin{vmatrix}3&-5\\2&3\end{vmatrix}=-1(9+10)=-19
\displaystyle A_{13}=(-1)^4\begin{vmatrix}3&4\\2&-1\end{vmatrix}=1(-3-8)=-11
\displaystyle A_{21}=(-1)^3\begin{vmatrix}-1&2\\-1&3\end{vmatrix}=-1(-3+2)=1
\displaystyle A_{22}=(-1)^4\begin{vmatrix}1&2\\2&3\end{vmatrix}=1(3-4)=-1
\displaystyle A_{23}=(-1)^5\begin{vmatrix}1&-1\\2&-1\end{vmatrix}=-1(-1+2)=-1
\displaystyle A_{31}=(-1)^4\begin{vmatrix}-1&2\\4&-5\end{vmatrix}=1(5-8)=-3
\displaystyle A_{32}=(-1)^5\begin{vmatrix}1&2\\3&-5\end{vmatrix}=-1(-5-6)=11
\displaystyle A_{33}=(-1)^6\begin{vmatrix}1&-1\\3&4\end{vmatrix}=1(4+3)=7
\displaystyle \therefore \ {adj \ }(A)=\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}7&-19&-11\\1&-1&-1\\-3&11&7\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}
\displaystyle \text{and }A^{-1}=\frac{\ {adj \ }(A)}{|A|}
\displaystyle \Rightarrow A^{-1}=\frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}
\displaystyle \text{Now, from Eq. }(i),\text{ we get}
\displaystyle X=A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}\begin{bmatrix}7\\-5\\12\end{bmatrix}
\displaystyle =\frac{1}{4}\begin{bmatrix}49-5-36\\-133+5+132\\-77+5+84\end{bmatrix}
\displaystyle =\frac{1}{4}\begin{bmatrix}8\\4\\12\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}
\displaystyle \text{On comparing corresponding elements, we get}
\displaystyle x=2,\ y=1\text{ and }z=3
\\

\displaystyle \textbf{Question 41. }\text{Using matrices, solve the following system of linear equations.}
\displaystyle x+y-z=3,\quad 2x+3y+z=10,\quad 3x-y-7z=1 \hspace{2.2cm}\text{[CBSE 2012; CBSE 2009C]}
\displaystyle \text{Answer:}
\displaystyle \text{The given equations can be written as }AX=B,\text{ where}
\displaystyle A=\begin{bmatrix}1&1&-1\\2&3&1\\3&-1&-7\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}3\\10\\1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&-1\\2&3&1\\3&-1&-7\end{vmatrix}
\displaystyle =1(-21+1)-1(-14-3)+(-1)(-2-9)
\displaystyle =-20+17+11=8
\displaystyle \therefore |A|\neq0,\text{ so }A^{-1}\text{ exists.}
\displaystyle A^{-1}=\frac{1}{8}\begin{bmatrix}-20&8&4\\17&-4&-3\\-11&4&1\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle =\frac{1}{8}\begin{bmatrix}-20&8&4\\17&-4&-3\\-11&4&1\end{bmatrix}\begin{bmatrix}3\\10\\1\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}-60+80+4\\51-40-3\\-33+40+1\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}24\\8\\8\end{bmatrix}=\begin{bmatrix}3\\1\\1\end{bmatrix}
\displaystyle \therefore x=3,\ y=1,\ z=1
\displaystyle \\

\displaystyle \textbf{Question 42. }\text{Using matrices, solve the following system of equations.}
\displaystyle 3x+4y+7z=4,\quad 2x-y+3z=-3,\quad x+2y-3z=8 \hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{The given equations can be written as }AX=B,\text{ where}
\displaystyle A=\begin{bmatrix}3&4&7\\2&-1&3\\1&2&-3\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}4\\-3\\8\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&4&7\\2&-1&3\\1&2&-3\end{vmatrix}
\displaystyle =3(3-6)-4(-6-3)+7(4+1)
\displaystyle =-9+36+35=62
\displaystyle \therefore |A|\neq0,\text{ so }A^{-1}\text{ exists.}
\displaystyle A^{-1}=\frac{1}{62}\begin{bmatrix}-3&26&19\\9&-16&5\\5&-2&-11\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle =\frac{1}{62}\begin{bmatrix}-3&26&19\\9&-16&5\\5&-2&-11\end{bmatrix}\begin{bmatrix}4\\-3\\8\end{bmatrix}
\displaystyle =\frac{1}{62}\begin{bmatrix}-12-78+152\\36+48+40\\20+6-88\end{bmatrix}
\displaystyle =\frac{1}{62}\begin{bmatrix}62\\124\\-62\end{bmatrix}=\begin{bmatrix}1\\2\\-1\end{bmatrix}
\displaystyle \therefore x=1,\ y=2,\ z=-1
\displaystyle \\

\displaystyle \textbf{Question 43. }\text{Using matrices, solve the following system of equations.}
\displaystyle 2x+3y+3z=5,\quad x-2y+z=-4,\quad 3x-y-2z=3 \hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{The given equations can be written as }AX=B,\text{ where}
\displaystyle A=\begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}5\\-4\\3\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{vmatrix}
\displaystyle =2(4+1)-3(-2-3)+3(-1+6)
\displaystyle =10+15+15=40
\displaystyle \therefore |A|\neq0,\text{ so }A^{-1}\text{ exists.}
\displaystyle A^{-1}=\frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle =\frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\begin{bmatrix}5\\-4\\3\end{bmatrix}
\displaystyle =\frac{1}{40}\begin{bmatrix}25-12+27\\25+52+3\\25-44-21\end{bmatrix}
\displaystyle =\frac{1}{40}\begin{bmatrix}40\\80\\-40\end{bmatrix}=\begin{bmatrix}1\\2\\-1\end{bmatrix}
\displaystyle \therefore x=1,\ y=2,\ z=-1
\displaystyle \\

\displaystyle \textbf{Question 44. }\text{If }A=\begin{bmatrix}1&2&1\\-1&1&1\\1&-3&1\end{bmatrix},\text{ then find }A^{-1}\text{ and hence solve the system of equations}
\displaystyle x+2y+z=4,\quad -x+y+z=0,\quad x-3y+z=4. \hspace{2.2cm}\text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle |A|=\begin{vmatrix}1&2&1\\-1&1&1\\1&-3&1\end{vmatrix}
\displaystyle =1(1+3)-2(-1-1)+1(3-1)
\displaystyle =4+4+2=10
\displaystyle A^{-1}=\frac{1}{10}\begin{bmatrix}4&-5&1\\2&0&-2\\2&5&3\end{bmatrix}
\displaystyle \text{The given equations can be written as }AX=B,\text{ where}
\displaystyle X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}4\\0\\4\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle =\frac{1}{10}\begin{bmatrix}4&-5&1\\2&0&-2\\2&5&3\end{bmatrix}\begin{bmatrix}4\\0\\4\end{bmatrix}
\displaystyle =\frac{1}{10}\begin{bmatrix}16+0+4\\8+0-8\\8+0+12\end{bmatrix}
\displaystyle =\frac{1}{10}\begin{bmatrix}20\\0\\20\end{bmatrix}=\begin{bmatrix}2\\0\\2\end{bmatrix}
\displaystyle \therefore x=2,\ y=0,\ z=2
\displaystyle \\

\displaystyle \textbf{Question 45. }\text{Find }A^{-1},\text{ where }A=\begin{bmatrix}1&2&-3\\2&3&2\\3&-3&-4\end{bmatrix}.\text{ Hence, solve the system of equations,}
\displaystyle x+2y-3z=-4,\quad 2x+3y+2z=2,\quad 3x-3y-4z=11. \hspace{2.2cm}\text{[CBSE 2012C, 2010, 2008]}
\displaystyle \text{Answer:}
\displaystyle |A|=\begin{vmatrix}1&2&-3\\2&3&2\\3&-3&-4\end{vmatrix}
\displaystyle =1(-12+6)-2(-8-6)-3(-6-9)
\displaystyle =-6+28+45=67
\displaystyle A^{-1}=\frac{1}{67}\begin{bmatrix}-6&17&13\\14&5&-8\\-15&9&-1\end{bmatrix}
\displaystyle \text{The given equations can be written as }AX=B,\text{ where}
\displaystyle X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}-4\\2\\11\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle =\frac{1}{67}\begin{bmatrix}-6&17&13\\14&5&-8\\-15&9&-1\end{bmatrix}\begin{bmatrix}-4\\2\\11\end{bmatrix}
\displaystyle =\frac{1}{67}\begin{bmatrix}24+34+143\\-56+10-88\\60+18-11\end{bmatrix}
\displaystyle =\frac{1}{67}\begin{bmatrix}201\\-134\\67\end{bmatrix}=\begin{bmatrix}3\\-2\\1\end{bmatrix}
\displaystyle \therefore x=3,\ y=-2,\ z=1
\displaystyle \\

\displaystyle \textbf{Question 46. }\text{Using matrices, solve the following system of equations.}
\displaystyle 4x+3y+2z=60,\quad x+2y+3z=45,\quad 6x+2y+3z=70 \hspace{2.2cm}\text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}4&3&2\\1&2&3\\6&2&3\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}60\\45\\70\end{bmatrix}
\displaystyle AX=B
\displaystyle |A|=\begin{vmatrix}4&3&2\\1&2&3\\6&2&3\end{vmatrix}=4(6-6)-3(3-18)+2(2-12)
\displaystyle =0+45-20=25
\displaystyle A^{-1}=\frac{1}{25}\begin{bmatrix}0&-5&5\\15&0&-10\\-10&10&5\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle =\frac{1}{25}\begin{bmatrix}0&-5&5\\15&0&-10\\-10&10&5\end{bmatrix}\begin{bmatrix}60\\45\\70\end{bmatrix}
\displaystyle =\frac{1}{25}\begin{bmatrix}-225+350\\900-700\\-600+450+350\end{bmatrix}
\displaystyle =\frac{1}{25}\begin{bmatrix}125\\200\\200\end{bmatrix}=\begin{bmatrix}5\\8\\8\end{bmatrix}
\displaystyle \therefore x=5,\ y=8,\ z=8
\displaystyle \\

\displaystyle \textbf{Question 47. }\text{Using matrices, solve the following system of equations.}
\displaystyle x+2y+z=7,\quad x+3z=11,\quad 2x-3y=1 \hspace{2.2cm}\text{[CBSE 2011; CBSE 2008C]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}1&2&1\\1&0&3\\2&-3&0\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}7\\11\\1\end{bmatrix}
\displaystyle AX=B
\displaystyle |A|=\begin{vmatrix}1&2&1\\1&0&3\\2&-3&0\end{vmatrix}=1(9)-2(0-6)+1(-3-0)
\displaystyle =9+12-3=18
\displaystyle A^{-1}=\frac{1}{18}\begin{bmatrix}9&-3&6\\6&-2&-2\\-3&7&-2\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle =\frac{1}{18}\begin{bmatrix}9&-3&6\\6&-2&-2\\-3&7&-2\end{bmatrix}\begin{bmatrix}7\\11\\1\end{bmatrix}
\displaystyle =\frac{1}{18}\begin{bmatrix}63-33+6\\42-22-2\\-21+77-2\end{bmatrix}
\displaystyle =\frac{1}{18}\begin{bmatrix}36\\18\\54\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}
\displaystyle \therefore x=2,\ y=1,\ z=3
\displaystyle \\

\displaystyle \textbf{Question 48. }\text{Using matrices, solve the following system of equations.}
\displaystyle x+2y-3z=-4,\quad 2x+3y+2z=2,\quad 3x-3y-4z=11 \hspace{2.2cm}\text{[CBSE 2011, 2008]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}1&2&-3\\2&3&2\\3&-3&-4\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}-4\\2\\11\end{bmatrix}
\displaystyle AX=B
\displaystyle |A|=\begin{vmatrix}1&2&-3\\2&3&2\\3&-3&-4\end{vmatrix}=-6+28+45=67
\displaystyle A^{-1}=\frac{1}{67}\begin{bmatrix}-6&17&13\\14&5&-8\\-15&9&-1\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle =\frac{1}{67}\begin{bmatrix}-6&17&13\\14&5&-8\\-15&9&-1\end{bmatrix}\begin{bmatrix}-4\\2\\11\end{bmatrix}
\displaystyle =\frac{1}{67}\begin{bmatrix}24+34+143\\-56+10-88\\60+18-11\end{bmatrix}
\displaystyle =\frac{1}{67}\begin{bmatrix}201\\-134\\67\end{bmatrix}=\begin{bmatrix}3\\-2\\1\end{bmatrix}
\displaystyle \therefore x=3,\ y=-2,\ z=1
\displaystyle \\

\displaystyle \textbf{Question 49. }\text{If }A=\begin{bmatrix}0&-2&1\\0&-1&1\\2&0&-3\end{bmatrix},\text{ then find }A^{-1}\text{ and hence solve the following system of equations}
\displaystyle x-2y+z=0,\quad -y+z=-2,\quad 2x-3z=10. \hspace{2.2cm}\text{[CBSE 2011 C]}
\displaystyle \text{Answer:}
\displaystyle |A|=\begin{vmatrix}0&-2&1\\0&-1&1\\2&0&-3\end{vmatrix}
\displaystyle =0-(-2)(-2)+1(2)=-2
\displaystyle A^{-1}=\begin{bmatrix}-\frac{3}{2}&3&\frac{1}{2}\\-1&1&0\\-1&2&0\end{bmatrix}
\displaystyle \text{The given equations can be written as }AX=B,\text{ where}
\displaystyle X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}0\\-2\\10\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle =\begin{bmatrix}-\frac{3}{2}&3&\frac{1}{2}\\-1&1&0\\-1&2&0\end{bmatrix}\begin{bmatrix}0\\-2\\10\end{bmatrix}
\displaystyle =\begin{bmatrix}0-6+5\\0-2+0\\0-4+0\end{bmatrix}=\begin{bmatrix}-1\\-2\\-4\end{bmatrix}
\displaystyle \therefore x=-1,\ y=-2,\ z=-4
\displaystyle \\

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