\displaystyle \textbf{Question 1: }\text{Find the adjoint of each of the following matrices:}
\displaystyle \text{(i) }\begin{bmatrix}-3&5\\2&4\end{bmatrix}\qquad \text{(ii) }\begin{bmatrix}a&b\\c&d\end{bmatrix}
\displaystyle \text{(iii) }\begin{bmatrix}\cos\alpha&\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\qquad \text{(iv) }\begin{bmatrix}1&\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2}&1\end{bmatrix}
\displaystyle \text{Also, verify that }(\text{adj}\,A)A=|A|I=A(\text{adj}\,A)\text{ for each matrix.}
\displaystyle \text{Answer:}
\displaystyle \text{For a matrix }\begin{bmatrix}a&b\\c&d\end{bmatrix},\text{ we have }\text{adj}\,A=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.
\displaystyle \text{(i) Let }A=\begin{bmatrix}-3&5\\2&4\end{bmatrix}.
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}4&-5\\-2&-3\end{bmatrix}
\displaystyle (\text{adj}\,A)A=\begin{bmatrix}4&-5\\-2&-3\end{bmatrix}\begin{bmatrix}-3&5\\2&4\end{bmatrix}=\begin{bmatrix}-22&0\\0&-22\end{bmatrix}
\displaystyle |A|=(-3)(4)-(5)(2)=-22
\displaystyle |A|I=-22\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}-22&0\\0&-22\end{bmatrix}
\displaystyle A(\text{adj}\,A)=\begin{bmatrix}-3&5\\2&4\end{bmatrix}\begin{bmatrix}4&-5\\-2&-3\end{bmatrix}=\begin{bmatrix}-22&0\\0&-22\end{bmatrix}
\displaystyle \therefore\ (\text{adj}\,A)A=|A|I=A(\text{adj}\,A)
\displaystyle \text{Hence verified.}

\displaystyle \text{(ii) Let }B=\begin{bmatrix}a&b\\c&d\end{bmatrix}.
\displaystyle \therefore\ \text{adj}\,B=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}
\displaystyle (\text{adj}\,B)B=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}
\displaystyle |B|=ad-bc
\displaystyle |B|I=(ad-bc)\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}
\displaystyle B(\text{adj}\,B)=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}
\displaystyle \therefore\ (\text{adj}\,B)B=|B|I=B(\text{adj}\,B)
\displaystyle \text{Hence verified.}

\displaystyle \text{(iii) Let }C=\begin{bmatrix}\cos\alpha&\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}.
\displaystyle \therefore\ \text{adj}\,C=\begin{bmatrix}\cos\alpha&-\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}
\displaystyle (\text{adj}\,C)C=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha&0\\0&\cos^2\alpha-\sin^2\alpha\end{bmatrix}
\displaystyle |C|=\cos^2\alpha-\sin^2\alpha
\displaystyle |C|I=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha&0\\0&\cos^2\alpha-\sin^2\alpha\end{bmatrix}
\displaystyle C(\text{adj}\,C)=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha&0\\0&\cos^2\alpha-\sin^2\alpha\end{bmatrix}
\displaystyle \therefore\ (\text{adj}\,C)C=|C|I=C(\text{adj}\,C)
\displaystyle \text{Hence verified.}

\displaystyle \text{(iv) Let }D=\begin{bmatrix}1&\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2}&1\end{bmatrix}.
\displaystyle \therefore\ \text{adj}\,D=\begin{bmatrix}1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}
\displaystyle (\text{adj}\,D)D=\begin{bmatrix}1+\tan^2\frac{\alpha}{2}&0\\0&1+\tan^2\frac{\alpha}{2}\end{bmatrix}
\displaystyle |D|=1+\tan^2\frac{\alpha}{2}
\displaystyle |D|I=\begin{bmatrix}1+\tan^2\frac{\alpha}{2}&0\\0&1+\tan^2\frac{\alpha}{2}\end{bmatrix}
\displaystyle D(\text{adj}\,D)=\begin{bmatrix}1+\tan^2\frac{\alpha}{2}&0\\0&1+\tan^2\frac{\alpha}{2}\end{bmatrix}
\displaystyle \therefore\ (\text{adj}\,D)D=|D|I=D(\text{adj}\,D)
\displaystyle \text{Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the adjoint of each of the following matrices:}
\displaystyle \text{(i) }\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\qquad \text{(ii) }\begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}
\displaystyle \text{(iii) }\begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}\qquad \text{(iv) }\begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}
\displaystyle \text{Also, verify that }(\text{adj}\,A)A=|A|I=A(\text{adj}\,A)\text{ for each matrix.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }A=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}.
\displaystyle C_{11}=\begin{vmatrix}1&2\\2&1\end{vmatrix}=-3,\quad C_{12}=-\begin{vmatrix}2&2\\2&1\end{vmatrix}=2,\quad C_{13}=\begin{vmatrix}2&1\\2&2\end{vmatrix}=2
\displaystyle C_{21}=-\begin{vmatrix}2&2\\2&1\end{vmatrix}=2,\quad C_{22}=\begin{vmatrix}1&2\\2&1\end{vmatrix}=-3,\quad C_{23}=-\begin{vmatrix}1&2\\2&2\end{vmatrix}=2
\displaystyle C_{31}=\begin{vmatrix}2&2\\1&2\end{vmatrix}=2,\quad C_{32}=-\begin{vmatrix}1&2\\2&2\end{vmatrix}=2,\quad C_{33}=\begin{vmatrix}1&2\\2&1\end{vmatrix}=-3
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}^{T}=\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}
\displaystyle (\text{adj}\,A)A=\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}
\displaystyle |A|=1(-3)-2(-2)+2(2)=5
\displaystyle \therefore\ |A|I=\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}
\displaystyle A(\text{adj}\,A)=\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}
\displaystyle \therefore\ (\text{adj}\,A)A=|A|I=A(\text{adj}\,A)
\displaystyle \text{Hence verified.}

\displaystyle \text{(ii) Let }B=\begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}.
\displaystyle C_{11}=\begin{vmatrix}3&1\\1&1\end{vmatrix}=2,\quad C_{12}=-\begin{vmatrix}2&1\\-1&1\end{vmatrix}=-3,\quad C_{13}=\begin{vmatrix}2&3\\-1&1\end{vmatrix}=5
\displaystyle C_{21}=-\begin{vmatrix}2&5\\1&1\end{vmatrix}=3,\quad C_{22}=\begin{vmatrix}1&5\\-1&1\end{vmatrix}=6,\quad C_{23}=-\begin{vmatrix}1&2\\-1&1\end{vmatrix}=-3
\displaystyle C_{31}=\begin{vmatrix}2&5\\3&1\end{vmatrix}=-13,\quad C_{32}=-\begin{vmatrix}1&5\\2&1\end{vmatrix}=9,\quad C_{33}=\begin{vmatrix}1&2\\2&3\end{vmatrix}=-1
\displaystyle \therefore\ \text{adj}\,B=\begin{bmatrix}2&-3&5\\3&6&-3\\-13&9&-1\end{bmatrix}^{T}=\begin{bmatrix}2&3&-13\\-3&6&9\\5&-3&-1\end{bmatrix}
\displaystyle (\text{adj}\,B)B=\begin{bmatrix}21&0&0\\0&21&0\\0&0&21\end{bmatrix}
\displaystyle |B|=1(2)-2(3)+5(5)=21
\displaystyle \therefore\ |B|I=\begin{bmatrix}21&0&0\\0&21&0\\0&0&21\end{bmatrix}
\displaystyle B(\text{adj}\,B)=\begin{bmatrix}21&0&0\\0&21&0\\0&0&21\end{bmatrix}
\displaystyle \therefore\ (\text{adj}\,B)B=|B|I=B(\text{adj}\,B)
\displaystyle \text{Hence verified.}

\displaystyle \text{(iii) Let }C=\begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}.
\displaystyle C_{11}=\begin{vmatrix}2&5\\4&-1\end{vmatrix}=-22,\quad C_{12}=-\begin{vmatrix}4&5\\0&-1\end{vmatrix}=4,\quad C_{13}=\begin{vmatrix}4&2\\0&4\end{vmatrix}=16
\displaystyle C_{21}=-\begin{vmatrix}-1&3\\4&-1\end{vmatrix}=11,\quad C_{22}=\begin{vmatrix}2&3\\0&-1\end{vmatrix}=-2,\quad C_{23}=-\begin{vmatrix}2&-1\\0&4\end{vmatrix}=-8
\displaystyle C_{31}=\begin{vmatrix}-1&3\\2&5\end{vmatrix}=-11,\quad C_{32}=-\begin{vmatrix}2&3\\4&5\end{vmatrix}=2,\quad C_{33}=\begin{vmatrix}2&-1\\4&2\end{vmatrix}=8
\displaystyle \therefore\ \text{adj}\,C=\begin{bmatrix}-22&4&16\\11&-2&-8\\-11&2&8\end{bmatrix}^{T}=\begin{bmatrix}-22&11&-11\\4&-2&2\\16&-8&8\end{bmatrix}
\displaystyle (\text{adj}\,C)C=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
\displaystyle |C|=2(-22)-(-1)(-4)+3(16)=0
\displaystyle \therefore\ |C|I=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
\displaystyle C(\text{adj}\,C)=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
\displaystyle \therefore\ (\text{adj}\,C)C=|C|I=C(\text{adj}\,C)
\displaystyle \text{Hence verified.}

\displaystyle \text{(iv) Let }D=\begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}.
\displaystyle C_{11}=\begin{vmatrix}1&0\\1&3\end{vmatrix}=3,\quad C_{12}=-\begin{vmatrix}5&0\\1&3\end{vmatrix}=-15,\quad C_{13}=\begin{vmatrix}5&1\\1&1\end{vmatrix}=4
\displaystyle C_{21}=-\begin{vmatrix}0&-1\\1&3\end{vmatrix}=-1,\quad C_{22}=\begin{vmatrix}2&-1\\1&3\end{vmatrix}=7,\quad C_{23}=-\begin{vmatrix}2&0\\1&1\end{vmatrix}=-2
\displaystyle C_{31}=\begin{vmatrix}0&-1\\1&0\end{vmatrix}=1,\quad C_{32}=-\begin{vmatrix}2&-1\\5&0\end{vmatrix}=-5,\quad C_{33}=\begin{vmatrix}2&0\\5&1\end{vmatrix}=2
\displaystyle \therefore\ \text{adj}\,D=\begin{bmatrix}3&-15&4\\-1&7&-2\\1&-5&2\end{bmatrix}^{T}=\begin{bmatrix}3&-1&1\\-15&7&-5\\4&-2&2\end{bmatrix}
\displaystyle (\text{adj}\,D)D=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}
\displaystyle |D|=2(3)+(-1)(4)=2
\displaystyle \therefore\ |D|I=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}
\displaystyle D(\text{adj}\,D)=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}
\displaystyle \therefore\ (\text{adj}\,D)D=|D|I=D(\text{adj}\,D)
\displaystyle \text{Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{For the matrix }A=\begin{bmatrix}1&-1&1\\2&3&0\\18&2&10\end{bmatrix},
\displaystyle \text{show that }A(\text{adj}\,A)=O.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}1&-1&1\\2&3&0\\18&2&10\end{bmatrix}
\displaystyle \text{The cofactors of the elements of }A\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}3&0\\2&10\end{vmatrix}=30,\quad C_{12}=-\begin{vmatrix}2&0\\18&10\end{vmatrix}=-20,\quad C_{13}=\begin{vmatrix}2&3\\18&2\end{vmatrix}=-50
\displaystyle C_{21}=-\begin{vmatrix}-1&1\\2&10\end{vmatrix}=12,\quad C_{22}=\begin{vmatrix}1&1\\18&10\end{vmatrix}=-8,\quad C_{23}=-\begin{vmatrix}1&-1\\18&2\end{vmatrix}=-20
\displaystyle C_{31}=\begin{vmatrix}-1&1\\3&0\end{vmatrix}=-3,\quad C_{32}=-\begin{vmatrix}1&1\\2&0\end{vmatrix}=2,\quad C_{33}=\begin{vmatrix}1&-1\\2&3\end{vmatrix}=5
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}30&-20&-50\\12&-8&-20\\-3&2&5\end{bmatrix}^{T}=\begin{bmatrix}30&12&-3\\-20&-8&2\\-50&-20&5\end{bmatrix}
\displaystyle A(\text{adj}\,A)=\begin{bmatrix}1&-1&1\\2&3&0\\18&2&10\end{bmatrix}\begin{bmatrix}30&12&-3\\-20&-8&2\\-50&-20&5\end{bmatrix}
\displaystyle =\begin{bmatrix}30+20-50&12+8-20&-3-2+5\\60-60+0&24-24+0&-6+6+0\\540-40-500&216-16-200&-54+4+50\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O
\displaystyle \therefore\ A(\text{adj}\,A)=O.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }A=\begin{bmatrix}-4&-3&-3\\1&0&1\\4&4&3\end{bmatrix},\text{ show that }\text{adj}\,A=A.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}-4&-3&-3\\1&0&1\\4&4&3\end{bmatrix}
\displaystyle \text{The cofactors of the elements of }A\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}0&1\\4&3\end{vmatrix}=-4,\quad C_{12}=-\begin{vmatrix}1&1\\4&3\end{vmatrix}=1,\quad C_{13}=\begin{vmatrix}1&0\\4&4\end{vmatrix}=4
\displaystyle C_{21}=-\begin{vmatrix}-3&-3\\4&3\end{vmatrix}=-3,\quad C_{22}=\begin{vmatrix}-4&-3\\4&3\end{vmatrix}=0,\quad C_{23}=-\begin{vmatrix}-4&-3\\4&4\end{vmatrix}=4
\displaystyle C_{31}=\begin{vmatrix}-3&-3\\0&1\end{vmatrix}=-3,\quad C_{32}=-\begin{vmatrix}-4&-3\\1&1\end{vmatrix}=1,\quad C_{33}=\begin{vmatrix}-4&-3\\1&0\end{vmatrix}=3
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}-4&1&4\\-3&0&4\\-3&1&3\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-4&-3&-3\\1&0&1\\4&4&3\end{bmatrix}=A
\displaystyle \therefore\ \text{adj}\,A=A.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }A=\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix},\text{ show that }\text{adj}\,A=3A^T.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}
\displaystyle \text{The cofactors of the elements of }A\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}1&-2\\-2&1\end{vmatrix}=-3,\quad C_{12}=-\begin{vmatrix}2&-2\\2&1\end{vmatrix}=-6,\quad C_{13}=\begin{vmatrix}2&1\\2&-2\end{vmatrix}=-6
\displaystyle C_{21}=-\begin{vmatrix}-2&-2\\-2&1\end{vmatrix}=6,\quad C_{22}=\begin{vmatrix}-1&-2\\2&1\end{vmatrix}=3,\quad C_{23}=-\begin{vmatrix}-1&-2\\2&-2\end{vmatrix}=-6
\displaystyle C_{31}=\begin{vmatrix}-2&-2\\1&-2\end{vmatrix}=6,\quad C_{32}=-\begin{vmatrix}-1&-2\\2&-2\end{vmatrix}=-6,\quad C_{33}=\begin{vmatrix}-1&-2\\2&1\end{vmatrix}=3
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}-3&-6&-6\\6&3&-6\\6&-6&3\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-3&6&6\\-6&3&-6\\-6&-6&3\end{bmatrix}
\displaystyle A^T=\begin{bmatrix}-1&2&2\\-2&1&-2\\-2&-2&1\end{bmatrix}
\displaystyle \therefore\ 3A^T=3\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 \therefore\ \text{adj}\,A=3A^T.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find }A(\text{adj}\,A)\text{ for the matrix }A=\begin{bmatrix}1&-2&3\\0&2&-1\\-4&5&2\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}1&-2&3\\0&2&-1\\-4&5&2\end{bmatrix}
\displaystyle \text{The cofactors of the elements of }A\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}2&-1\\5&2\end{vmatrix}=9,\quad C_{12}=-\begin{vmatrix}0&-1\\-4&2\end{vmatrix}=4,\quad C_{13}=\begin{vmatrix}0&2\\-4&5\end{vmatrix}=8
\displaystyle C_{21}=-\begin{vmatrix}-2&3\\5&2\end{vmatrix}=19,\quad C_{22}=\begin{vmatrix}1&3\\-4&2\end{vmatrix}=14,\quad C_{23}=-\begin{vmatrix}1&-2\\-4&5\end{vmatrix}=3
\displaystyle C_{31}=\begin{vmatrix}-2&3\\2&-1\end{vmatrix}=-4,\quad C_{32}=-\begin{vmatrix}1&3\\0&-1\end{vmatrix}=1,\quad C_{33}=\begin{vmatrix}1&-2\\0&2\end{vmatrix}=2
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}9&4&8\\19&14&3\\-4&1&2\end{bmatrix}^{T}=\begin{bmatrix}9&19&-4\\4&14&1\\8&3&2\end{bmatrix}
\displaystyle |A|=1\begin{vmatrix}2&-1\\5&2\end{vmatrix}-(-2)\begin{vmatrix}0&-1\\-4&2\end{vmatrix}+3\begin{vmatrix}0&2\\-4&5\end{vmatrix}
\displaystyle =1(9)+2(-4)+3(8)=9-8+24=25
\displaystyle \text{Since }A(\text{adj}\,A)=|A|I,
\displaystyle \therefore\ A(\text{adj}\,A)=25\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}25&0&0\\0&25&0\\0&0&25\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the inverse of each of the following matrices:}
\displaystyle \text{(i) }\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\qquad \text{(ii) }\begin{bmatrix}0&1\\1&0\end{bmatrix}
\displaystyle \text{(iii) }\begin{bmatrix}a&b\\c&\dfrac{1+bc}{a}\end{bmatrix},\ a\neq0\qquad \text{(iv) }\begin{bmatrix}2&5\\-3&1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{For }X=\begin{bmatrix}a&b\\c&d\end{bmatrix},\quad X^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix},\quad ad-bc\neq0.
\displaystyle \text{(i) Let }A=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}.
\displaystyle |A|=\cos^2\theta+\sin^2\theta=1\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle \text{adj}\,A=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}
\displaystyle \therefore\ A^{-1}=\frac{1}{|A|}\text{adj}\,A=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}

\displaystyle \text{(ii) Let }B=\begin{bmatrix}0&1\\1&0\end{bmatrix}.
\displaystyle |B|=(0)(0)-(1)(1)=-1\neq0
\displaystyle \therefore\ B\text{ is non-singular and hence invertible.}
\displaystyle \text{adj}\,B=\begin{bmatrix}0&-1\\-1&0\end{bmatrix}
\displaystyle \therefore\ B^{-1}=\frac{1}{-1}\begin{bmatrix}0&-1\\-1&0\end{bmatrix}=\begin{bmatrix}0&1\\1&0\end{bmatrix}

\displaystyle \text{(iii) Let }C=\begin{bmatrix}a&b\\c&\dfrac{1+bc}{a}\end{bmatrix},\quad a\neq0.
\displaystyle |C|=a\left(\frac{1+bc}{a}\right)-bc=1+bc-bc=1\neq0
\displaystyle \therefore\ C\text{ is non-singular and hence invertible.}
\displaystyle \text{adj}\,C=\begin{bmatrix}\dfrac{1+bc}{a}&-b\\-c&a\end{bmatrix}
\displaystyle \therefore\ C^{-1}=\frac{1}{|C|}\text{adj}\,C=\begin{bmatrix}\dfrac{1+bc}{a}&-b\\-c&a\end{bmatrix}

\displaystyle \text{(iv) Let }D=\begin{bmatrix}2&5\\-3&1\end{bmatrix}.
\displaystyle |D|=(2)(1)-(5)(-3)=2+15=17\neq0
\displaystyle \therefore\ D\text{ is non-singular and hence invertible.}
\displaystyle \text{adj}\,D=\begin{bmatrix}1&-5\\3&2\end{bmatrix}
\displaystyle \therefore\ D^{-1}=\frac{1}{|D|}\text{adj}\,D=\frac{1}{17}\begin{bmatrix}1&-5\\3&2\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the inverse of each of the following matrices:}
\displaystyle \text{(i) }\begin{bmatrix}1&2&3\\2&3&1\\3&1&2\end{bmatrix}\qquad \text{(ii) }\begin{bmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }A=\begin{bmatrix}1&2&3\\2&3&1\\3&1&2\end{bmatrix}.
\displaystyle \text{The cofactors of the elements of }A\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}3&1\\1&2\end{vmatrix}=5,\quad C_{12}=-\begin{vmatrix}2&1\\3&2\end{vmatrix}=-1,\quad C_{13}=\begin{vmatrix}2&3\\3&1\end{vmatrix}=-7
\displaystyle C_{21}=-\begin{vmatrix}2&3\\1&2\end{vmatrix}=-1,\quad C_{22}=\begin{vmatrix}1&3\\3&2\end{vmatrix}=-7,\quad C_{23}=-\begin{vmatrix}1&2\\3&1\end{vmatrix}=5
\displaystyle C_{31}=\begin{vmatrix}2&3\\3&1\end{vmatrix}=-7,\quad C_{32}=-\begin{vmatrix}1&3\\2&1\end{vmatrix}=5,\quad C_{33}=\begin{vmatrix}1&2\\2&3\end{vmatrix}=-1
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}5&-1&-7\\-1&-7&5\\-7&5&-1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}5&-1&-7\\-1&-7&5\\-7&5&-1\end{bmatrix}
\displaystyle |A|=1(5)+2(-1)+3(-7)=5-2-21=-18\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj}\,A
\displaystyle \therefore\ A^{-1}=-\frac{1}{18}\begin{bmatrix}5&-1&-7\\-1&-7&5\\-7&5&-1\end{bmatrix}

\displaystyle \text{(ii) Let }B=\begin{bmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{bmatrix}.
\displaystyle \text{The cofactors of the elements of }B\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}-1&-1\\3&-1\end{vmatrix}=4,\quad C_{12}=-\begin{vmatrix}1&-1\\2&-1\end{vmatrix}=-1,\quad C_{13}=\begin{vmatrix}1&-1\\2&3\end{vmatrix}=5
\displaystyle C_{21}=-\begin{vmatrix}2&5\\3&-1\end{vmatrix}=17,\quad C_{22}=\begin{vmatrix}1&5\\2&-1\end{vmatrix}=-11,\quad C_{23}=-\begin{vmatrix}1&2\\2&3\end{vmatrix}=1
\displaystyle C_{31}=\begin{vmatrix}2&5\\-1&-1\end{vmatrix}=3,\quad C_{32}=-\begin{vmatrix}1&5\\1&-1\end{vmatrix}=6,\quad C_{33}=\begin{vmatrix}1&2\\1&-1\end{vmatrix}=-3
\displaystyle \therefore\ \text{adj}\,B=\begin{bmatrix}4&-1&5\\17&-11&1\\3&6&-3\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}4&17&3\\-1&-11&6\\5&1&-3\end{bmatrix}
\displaystyle |B|=1(4)+2(-1)+5(5)=4-2+25=27\neq0
\displaystyle \therefore\ B\text{ is non-singular and hence invertible.}
\displaystyle B^{-1}=\frac{1}{|B|}\text{adj}\,B
\displaystyle \therefore\ B^{-1}=\frac{1}{27}\begin{bmatrix}4&17&3\\-1&-11&6\\5&1&-3\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the inverse of each of the following matrices:}
\displaystyle \text{(iii) }\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\qquad \text{(iv) }\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(iii) Let }C=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}.
\displaystyle \text{The cofactors of the elements of }C\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}2&-1\\-1&2\end{vmatrix}=3,\quad C_{12}=-\begin{vmatrix}-1&-1\\1&2\end{vmatrix}=1,\quad C_{13}=\begin{vmatrix}-1&2\\1&-1\end{vmatrix}=-1
\displaystyle C_{21}=-\begin{vmatrix}-1&1\\-1&2\end{vmatrix}=1,\quad C_{22}=\begin{vmatrix}2&1\\1&2\end{vmatrix}=3,\quad C_{23}=-\begin{vmatrix}2&-1\\1&-1\end{vmatrix}=1
\displaystyle C_{31}=\begin{vmatrix}-1&1\\2&-1\end{vmatrix}=-1,\quad C_{32}=-\begin{vmatrix}2&1\\-1&-1\end{vmatrix}=1,\quad C_{33}=\begin{vmatrix}2&-1\\-1&2\end{vmatrix}=3
\displaystyle \therefore\ \text{adj}\,C=\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}
\displaystyle |C|=2(3)+(-1)(1)+1(-1)=6-1-1=4\neq0
\displaystyle \therefore\ C\text{ is non-singular and hence invertible.}
\displaystyle C^{-1}=\frac{1}{|C|}\text{adj}\,C
\displaystyle \therefore\ C^{-1}=\frac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}

\displaystyle \text{(iv) Let }D=\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}.
\displaystyle \text{The cofactors of the elements of }D\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}1&0\\1&3\end{vmatrix}=3,\quad C_{12}=-\begin{vmatrix}5&0\\0&3\end{vmatrix}=-15,\quad C_{13}=\begin{vmatrix}5&1\\0&1\end{vmatrix}=5
\displaystyle C_{21}=-\begin{vmatrix}0&-1\\1&3\end{vmatrix}=-1,\quad C_{22}=\begin{vmatrix}2&-1\\0&3\end{vmatrix}=6,\quad C_{23}=-\begin{vmatrix}2&0\\0&1\end{vmatrix}=-2
\displaystyle C_{31}=\begin{vmatrix}0&-1\\1&0\end{vmatrix}=1,\quad C_{32}=-\begin{vmatrix}2&-1\\5&0\end{vmatrix}=-5,\quad C_{33}=\begin{vmatrix}2&0\\5&1\end{vmatrix}=2
\displaystyle \therefore\ \text{adj}\,D=\begin{bmatrix}3&-15&5\\-1&6&-2\\1&-5&2\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}
\displaystyle |D|=2(3)+(-1)(5)=6-5=1\neq0
\displaystyle \therefore\ D\text{ is non-singular and hence invertible.}
\displaystyle D^{-1}=\frac{1}{|D|}\text{adj}\,D
\displaystyle \therefore\ D^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the inverse of each of the following matrices:}
\displaystyle \text{(v) }\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}\qquad \text{(vi) }\begin{bmatrix}0&0&-1\\3&4&5\\-2&-4&-7\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(v) Let }E=\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}.
\displaystyle \text{The cofactors of the elements of }E\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}-3&4\\-3&4\end{vmatrix}=0,\quad C_{12}=-\begin{vmatrix}4&4\\3&4\end{vmatrix}=-4,\quad C_{13}=\begin{vmatrix}4&-3\\3&-3\end{vmatrix}=-3
\displaystyle C_{21}=-\begin{vmatrix}1&-1\\-3&4\end{vmatrix}=-1,\quad C_{22}=\begin{vmatrix}0&-1\\3&4\end{vmatrix}=3,\quad C_{23}=-\begin{vmatrix}0&1\\3&-3\end{vmatrix}=3
\displaystyle C_{31}=\begin{vmatrix}1&-1\\-3&4\end{vmatrix}=1,\quad C_{32}=-\begin{vmatrix}0&-1\\4&4\end{vmatrix}=-4,\quad C_{33}=\begin{vmatrix}0&1\\4&-3\end{vmatrix}=-4
\displaystyle \therefore\ \text{adj}\,E=\begin{bmatrix}0&-4&-3\\-1&3&3\\1&-4&-4\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}0&-1&1\\-4&3&-4\\-3&3&-4\end{bmatrix}
\displaystyle |E|=0(0)+1(-4)+(-1)(-3)=-4+3=-1\neq0
\displaystyle \therefore\ E\text{ is non-singular and hence invertible.}
\displaystyle E^{-1}=\frac{1}{|E|}\text{adj}\,E
\displaystyle \therefore\ E^{-1}=-\begin{bmatrix}0&-1&1\\-4&3&-4\\-3&3&-4\end{bmatrix}=\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}

\displaystyle \text{(vi) Let }F=\begin{bmatrix}0&0&-1\\3&4&5\\-2&-4&-7\end{bmatrix}.
\displaystyle \text{The cofactors of the elements of }F\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}4&5\\-4&-7\end{vmatrix}=-8,\quad C_{12}=-\begin{vmatrix}3&5\\-2&-7\end{vmatrix}=11,\quad C_{13}=\begin{vmatrix}3&4\\-2&-4\end{vmatrix}=-4
\displaystyle C_{21}=-\begin{vmatrix}0&-1\\-4&-7\end{vmatrix}=4,\quad C_{22}=\begin{vmatrix}0&-1\\-2&-7\end{vmatrix}=-2,\quad C_{23}=-\begin{vmatrix}0&0\\-2&-4\end{vmatrix}=0
\displaystyle C_{31}=\begin{vmatrix}0&-1\\4&5\end{vmatrix}=4,\quad C_{32}=-\begin{vmatrix}0&-1\\3&5\end{vmatrix}=-3,\quad C_{33}=\begin{vmatrix}0&0\\3&4\end{vmatrix}=0
\displaystyle \therefore\ \text{adj}\,F=\begin{bmatrix}-8&11&-4\\4&-2&0\\4&-3&0\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-8&4&4\\11&-2&-3\\-4&0&0\end{bmatrix}
\displaystyle |F|=0(-8)+0(11)+(-1)(-4)=4\neq0
\displaystyle \therefore\ F\text{ is non-singular and hence invertible.}
\displaystyle F^{-1}=\frac{1}{|F|}\text{adj}\,F
\displaystyle \therefore\ F^{-1}=\frac{1}{4}\begin{bmatrix}-8&4&4\\11&-2&-3\\-4&0&0\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the inverse of each of the following matrices:}
\displaystyle \text{(vii) }\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(vii) Let }G=\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}.
\displaystyle \text{The cofactors of the elements of }G\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}\cos\alpha&\sin\alpha\\\sin\alpha&-\cos\alpha\end{vmatrix}=-\cos^2\alpha-\sin^2\alpha=-1
\displaystyle C_{12}=-\begin{vmatrix}0&\sin\alpha\\0&-\cos\alpha\end{vmatrix}=0,\quad C_{13}=\begin{vmatrix}0&\cos\alpha\\0&\sin\alpha\end{vmatrix}=0
\displaystyle C_{21}=-\begin{vmatrix}0&0\\\sin\alpha&-\cos\alpha\end{vmatrix}=0,\quad C_{22}=\begin{vmatrix}1&0\\0&-\cos\alpha\end{vmatrix}=-\cos\alpha
\displaystyle C_{23}=-\begin{vmatrix}1&0\\0&\sin\alpha\end{vmatrix}=-\sin\alpha
\displaystyle C_{31}=\begin{vmatrix}0&0\\\cos\alpha&\sin\alpha\end{vmatrix}=0,\quad C_{32}=-\begin{vmatrix}1&0\\0&\sin\alpha\end{vmatrix}=-\sin\alpha
\displaystyle C_{33}=\begin{vmatrix}1&0\\0&\cos\alpha\end{vmatrix}=\cos\alpha
\displaystyle \therefore\ \text{adj}\,G=\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}
\displaystyle |G|=1(-1)+0+0=-1\neq0
\displaystyle \therefore\ G\text{ is non-singular and hence invertible.}
\displaystyle G^{-1}=\frac{1}{|G|}\text{adj}\,G
\displaystyle \therefore\ G^{-1}=-\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}=G
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the inverse of each of the following matrices and verify that } \\ A^{-1}A=I_3.
\displaystyle \text{(i) }\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}\qquad \text{(ii) }\begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }A=\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}.
\displaystyle \text{The cofactors of the elements of }A\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}4&3\\3&4\end{vmatrix}=7,\quad C_{12}=-\begin{vmatrix}1&3\\1&4\end{vmatrix}=-1,\quad C_{13}=\begin{vmatrix}1&4\\1&3\end{vmatrix}=-1
\displaystyle C_{21}=-\begin{vmatrix}3&3\\3&4\end{vmatrix}=-3,\quad C_{22}=\begin{vmatrix}1&3\\1&4\end{vmatrix}=1,\quad C_{23}=-\begin{vmatrix}1&3\\1&3\end{vmatrix}=0
\displaystyle C_{31}=\begin{vmatrix}3&3\\4&3\end{vmatrix}=-3,\quad C_{32}=-\begin{vmatrix}1&3\\1&3\end{vmatrix}=0,\quad C_{33}=\begin{vmatrix}1&3\\1&4\end{vmatrix}=1
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}7&-1&-1\\-3&1&0\\-3&0&1\end{bmatrix}^{T}=\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}
\displaystyle |A|=1(7)+3(-1)+3(-1)=7-3-3=1\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj}\,A=\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}
\displaystyle \text{Verification:}
\displaystyle A^{-1}A=\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}
\displaystyle =\begin{bmatrix}7-3-3&21-12-9&21-9-12\\-1+1+0&-3+4+0&-3+3+0\\-1+0+1&-3+0+3&-3+0+4\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3

\displaystyle \text{(ii) Let }B=\begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}.
\displaystyle \text{The cofactors of the elements of }B\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}4&1\\7&2\end{vmatrix}=1,\quad C_{12}=-\begin{vmatrix}3&1\\3&2\end{vmatrix}=-3,\quad C_{13}=\begin{vmatrix}3&4\\3&7\end{vmatrix}=9
\displaystyle C_{21}=-\begin{vmatrix}3&1\\7&2\end{vmatrix}=1,\quad C_{22}=\begin{vmatrix}2&1\\3&2\end{vmatrix}=1,\quad C_{23}=-\begin{vmatrix}2&3\\3&7\end{vmatrix}=-5
\displaystyle C_{31}=\begin{vmatrix}3&1\\4&1\end{vmatrix}=-1,\quad C_{32}=-\begin{vmatrix}2&1\\3&1\end{vmatrix}=1,\quad C_{33}=\begin{vmatrix}2&3\\3&4\end{vmatrix}=-1
\displaystyle \therefore\ \text{adj}\,B=\begin{bmatrix}1&-3&9\\1&1&-5\\-1&1&-1\end{bmatrix}^{T}=\begin{bmatrix}1&1&-1\\-3&1&1\\9&-5&-1\end{bmatrix}
\displaystyle |B|=2(1)+3(-3)+1(9)=2-9+9=2\neq0
\displaystyle \therefore\ B\text{ is non-singular and hence invertible.}
\displaystyle B^{-1}=\frac{1}{|B|}\text{adj}\,B=\frac{1}{2}\begin{bmatrix}1&1&-1\\-3&1&1\\9&-5&-1\end{bmatrix}
\displaystyle \text{Verification:}
\displaystyle B^{-1}B=\frac{1}{2}\begin{bmatrix}1&1&-1\\-3&1&1\\9&-5&-1\end{bmatrix}\begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}
\displaystyle =\frac{1}{2}\begin{bmatrix}2+3-3&3+4-7&1+1-2\\-6+3+3&-9+4+7&-3+1+2\\18-15-3&27-20-7&9-5-2\end{bmatrix}
\displaystyle =\frac{1}{2}\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{For each of the following pairs of matrices, verify that } \\ (AB)^{-1}=B^{-1}A^{-1}.
\displaystyle \text{(i) }A=\begin{bmatrix}3&2\\7&5\end{bmatrix},\quad B=\begin{bmatrix}4&6\\3&2\end{bmatrix}
\displaystyle \text{(ii) }A=\begin{bmatrix}2&1\\5&3\end{bmatrix},\quad B=\begin{bmatrix}4&5\\3&4\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }A=\begin{bmatrix}3&2\\7&5\end{bmatrix},\quad B=\begin{bmatrix}4&6\\3&2\end{bmatrix}
\displaystyle AB=\begin{bmatrix}3&2\\7&5\end{bmatrix}\begin{bmatrix}4&6\\3&2\end{bmatrix}=\begin{bmatrix}18&22\\43&52\end{bmatrix}
\displaystyle |AB|=(18)(52)-(22)(43)=936-946=-10\neq0
\displaystyle \therefore\ AB\text{ is non-singular and hence invertible.}
\displaystyle (AB)^{-1}=\frac{1}{-10}\begin{bmatrix}52&-22\\-43&18\end{bmatrix}\quad\ldots(1)
\displaystyle |B|=(4)(2)-(6)(3)=8-18=-10\neq0
\displaystyle \therefore\ B^{-1}=\frac{1}{-10}\begin{bmatrix}2&-6\\-3&4\end{bmatrix}
\displaystyle |A|=(3)(5)-(2)(7)=15-14=1\neq0
\displaystyle \therefore\ A^{-1}=\begin{bmatrix}5&-2\\-7&3\end{bmatrix}
\displaystyle B^{-1}A^{-1}=\frac{1}{-10}\begin{bmatrix}2&-6\\-3&4\end{bmatrix}\begin{bmatrix}5&-2\\-7&3\end{bmatrix}
\displaystyle =\frac{1}{-10}\begin{bmatrix}10+42&-4-18\\-15-28&6+12\end{bmatrix}
\displaystyle =\frac{1}{-10}\begin{bmatrix}52&-22\\-43&18\end{bmatrix}\quad\ldots(2)
\displaystyle \text{From equations (1) and (2), }(AB)^{-1}=B^{-1}A^{-1}.
\displaystyle \text{Hence verified.}
\displaystyle \\

\displaystyle \text{(ii) }A=\begin{bmatrix}2&1\\5&3\end{bmatrix},\quad B=\begin{bmatrix}4&5\\3&4\end{bmatrix}
\displaystyle AB=\begin{bmatrix}2&1\\5&3\end{bmatrix}\begin{bmatrix}4&5\\3&4\end{bmatrix}=\begin{bmatrix}11&14\\29&37\end{bmatrix}
\displaystyle |AB|=(11)(37)-(14)(29)=407-406=1\neq0
\displaystyle \therefore\ AB\text{ is non-singular and hence invertible.}
\displaystyle (AB)^{-1}=\begin{bmatrix}37&-14\\-29&11\end{bmatrix}\quad\ldots(1)
\displaystyle |B|=(4)(4)-(5)(3)=16-15=1\neq0
\displaystyle \therefore\ B^{-1}=\begin{bmatrix}4&-5\\-3&4\end{bmatrix}
\displaystyle |A|=(2)(3)-(1)(5)=6-5=1\neq0
\displaystyle \therefore\ A^{-1}=\begin{bmatrix}3&-1\\-5&2\end{bmatrix}
\displaystyle B^{-1}A^{-1}=\begin{bmatrix}4&-5\\-3&4\end{bmatrix}\begin{bmatrix}3&-1\\-5&2\end{bmatrix}
\displaystyle =\begin{bmatrix}12+25&-4-10\\-9-20&3+8\end{bmatrix}=\begin{bmatrix}37&-14\\-29&11\end{bmatrix}\quad\ldots(2)
\displaystyle \text{From equations (1) and (2), }(AB)^{-1}=B^{-1}A^{-1}.
\displaystyle \text{Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Let }A=\begin{bmatrix}3&2\\7&5\end{bmatrix}\text{ and }B=\begin{bmatrix}6&7\\8&9\end{bmatrix}.\text{ Find }(AB)^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3&2\\7&5\end{bmatrix},\qquad B=\begin{bmatrix}6&7\\8&9\end{bmatrix}
\displaystyle AB=\begin{bmatrix}3&2\\7&5\end{bmatrix}\begin{bmatrix}6&7\\8&9\end{bmatrix}=\begin{bmatrix}34&39\\82&94\end{bmatrix}
\displaystyle |AB|=(34)(94)-(39)(82)=3196-3198=-2\neq0
\displaystyle \therefore\ AB\text{ is non-singular and hence invertible.}
\displaystyle \text{adj}(AB)=\begin{bmatrix}94&-39\\-82&34\end{bmatrix}
\displaystyle \therefore\ (AB)^{-1}=\frac{1}{|AB|}\text{adj}(AB)=-\frac{1}{2}\begin{bmatrix}94&-39\\-82&34\end{bmatrix}
\displaystyle =\begin{bmatrix}-47&\dfrac{39}{2}\\41&-17\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Given }A=\begin{bmatrix}2&-3\\-4&7\end{bmatrix},\text{ compute }A^{-1}\text{ and show that } \\ 2A^{-1}=9I-A.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}2&-3\\-4&7\end{bmatrix}
\displaystyle |A|=(2)(7)-(-3)(-4)=14-12=2\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle \text{adj}\,A=\begin{bmatrix}7&3\\4&2\end{bmatrix}
\displaystyle \therefore\ A^{-1}=\frac{1}{|A|}\text{adj}\,A=\frac{1}{2}\begin{bmatrix}7&3\\4&2\end{bmatrix}
\displaystyle \text{Now, LHS}=2A^{-1}=2\left(\frac{1}{2}\begin{bmatrix}7&3\\4&2\end{bmatrix}\right)=\begin{bmatrix}7&3\\4&2\end{bmatrix}
\displaystyle \text{RHS}=9I-A=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}=\begin{bmatrix}7&3\\4&2\end{bmatrix}
\displaystyle \therefore\ \text{LHS}=\text{RHS}.
\displaystyle \therefore\ 2A^{-1}=9I-A.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }A=\begin{bmatrix}4&5\\2&1\end{bmatrix},\text{ show that }A-3I=2\left(I+3A^{-1}\right).
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}4&5\\2&1\end{bmatrix}
\displaystyle |A|=(4)(1)-(5)(2)=4-10=-6\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle \text{adj}\,A=\begin{bmatrix}1&-5\\-2&4\end{bmatrix}
\displaystyle \therefore\ A^{-1}=\frac{1}{|A|}\text{adj}\,A=-\frac{1}{6}\begin{bmatrix}1&-5\\-2&4\end{bmatrix}
\displaystyle \text{LHS}=A-3I=\begin{bmatrix}4&5\\2&1\end{bmatrix}-3\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}4&5\\2&1\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}=\begin{bmatrix}1&5\\2&-2\end{bmatrix}
\displaystyle \text{RHS}=2\left(I+3A^{-1}\right)
\displaystyle =2\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix}-\frac{1}{2}\begin{bmatrix}1&-5\\-2&4\end{bmatrix}\right\}
\displaystyle =2\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}\dfrac{1}{2}&-\dfrac{5}{2}\\-1&2\end{bmatrix}\right\}
\displaystyle =2\begin{bmatrix}\dfrac{1}{2}&\dfrac{5}{2}\\1&-1\end{bmatrix}=\begin{bmatrix}1&5\\2&-2\end{bmatrix}
\displaystyle \therefore\ \text{LHS}=\text{RHS}.
\displaystyle \therefore\ A-3I=2\left(I+3A^{-1}\right).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Let }A=\begin{bmatrix}a&b\\c&\dfrac{1+bc}{a}\end{bmatrix},\ a\neq0.\text{ Find }A^{-1}\text{ and show that } \\ aA^{-1}=(a^2+bc+1)I-aA.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}a&b\\c&\dfrac{1+bc}{a}\end{bmatrix},\quad a\neq0
\displaystyle |A|=a\left(\frac{1+bc}{a}\right)-bc=1+bc-bc=1\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle \text{adj}\,A=\begin{bmatrix}\dfrac{1+bc}{a}&-b\\-c&a\end{bmatrix}
\displaystyle \therefore\ A^{-1}=\frac{1}{|A|}\text{adj}\,A=\begin{bmatrix}\dfrac{1+bc}{a}&-b\\-c&a\end{bmatrix}
\displaystyle \text{LHS}=aA^{-1}=a\begin{bmatrix}\dfrac{1+bc}{a}&-b\\-c&a\end{bmatrix}=\begin{bmatrix}1+bc&-ab\\-ac&a^2\end{bmatrix}
\displaystyle \text{RHS}=(a^2+bc+1)I-aA
\displaystyle =(a^2+bc+1)\begin{bmatrix}1&0\\0&1\end{bmatrix}-a\begin{bmatrix}a&b\\c&\dfrac{1+bc}{a}\end{bmatrix}
\displaystyle =\begin{bmatrix}a^2+bc+1&0\\0&a^2+bc+1\end{bmatrix}-\begin{bmatrix}a^2&ab\\ac&1+bc\end{bmatrix}
\displaystyle =\begin{bmatrix}1+bc&-ab\\-ac&a^2\end{bmatrix}
\displaystyle \therefore\ \text{LHS}=\text{RHS}.
\displaystyle \therefore\ aA^{-1}=(a^2+bc+1)I-aA.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Given }A=\begin{bmatrix}5&0&4\\2&3&2\\1&2&1\end{bmatrix},\quad B^{-1}=\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}.\text{ Compute }(AB)^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}5&0&4\\2&3&2\\1&2&1\end{bmatrix},\qquad B^{-1}=\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}
\displaystyle \text{We know that }(AB)^{-1}=B^{-1}A^{-1}.
\displaystyle \text{For the matrix }A,\text{ the cofactors are:}
\displaystyle C_{11}=\begin{vmatrix}3&2\\2&1\end{vmatrix}=-1,\quad C_{12}=-\begin{vmatrix}2&2\\1&1\end{vmatrix}=0,\quad C_{13}=\begin{vmatrix}2&3\\1&2\end{vmatrix}=1
\displaystyle C_{21}=-\begin{vmatrix}0&4\\2&1\end{vmatrix}=8,\quad C_{22}=\begin{vmatrix}5&4\\1&1\end{vmatrix}=1,\quad C_{23}=-\begin{vmatrix}5&0\\1&2\end{vmatrix}=-10
\displaystyle C_{31}=\begin{vmatrix}0&4\\3&2\end{vmatrix}=-12,\quad C_{32}=-\begin{vmatrix}5&4\\2&2\end{vmatrix}=-2,\quad C_{33}=\begin{vmatrix}5&0\\2&3\end{vmatrix}=15
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}-1&0&1\\8&1&-10\\-12&-2&15\end{bmatrix}^{T}=\begin{bmatrix}-1&8&-12\\0&1&-2\\1&-10&15\end{bmatrix}
\displaystyle |A|=5(-1)+0+4(1)=-5+4=-1\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj}\,A=-\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 \therefore\ (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 \therefore\ (AB)^{-1}=\begin{bmatrix}-2&19&-27\\-2&18&-25\\-3&29&-42\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Let }F(\alpha)=\begin{bmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}\text{ and }G(\beta)=\begin{bmatrix}\cos\beta&0&\sin\beta\\0&1&0\\-\sin\beta&0&\cos\beta\end{bmatrix}. \\ \text{ Show that:}
\displaystyle \text{(i) }\left[F(\alpha)\right]^{-1}=F(-\alpha)\qquad \text{(ii) }\left[G(\beta)\right]^{-1}=G(-\beta)
\displaystyle \text{(iii) }\left[F(\alpha)G(\beta)\right]^{-1}=G(-\beta)F(-\alpha)
\displaystyle \text{Answer:}
\displaystyle \text{(i) }F(\alpha)=\begin{bmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}
\displaystyle F(-\alpha)=\begin{bmatrix}\cos(-\alpha)&-\sin(-\alpha)&0\\\sin(-\alpha)&\cos(-\alpha)&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos\alpha&\sin\alpha&0\\-\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}
\displaystyle \text{The cofactors of the elements of }F(\alpha)\text{ are:}
\displaystyle C_{11}=\cos\alpha,\quad C_{12}=-\sin\alpha,\quad C_{13}=0
\displaystyle C_{21}=\sin\alpha,\quad C_{22}=\cos\alpha,\quad C_{23}=0
\displaystyle C_{31}=0,\quad C_{32}=0,\quad C_{33}=\cos^2\alpha+\sin^2\alpha=1
\displaystyle \therefore\ \text{adj}\,F(\alpha)=\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 \left|F(\alpha)\right|=\cos^2\alpha+\sin^2\alpha=1\neq0
\displaystyle \therefore\ F(\alpha)\text{ is non-singular and hence invertible.}
\displaystyle \left[F(\alpha)\right]^{-1}=\frac{1}{\left|F(\alpha)\right|}\text{adj}\,F(\alpha)
\displaystyle =\begin{bmatrix}\cos\alpha&\sin\alpha&0\\-\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}=F(-\alpha)
\displaystyle \therefore\ \left[F(\alpha)\right]^{-1}=F(-\alpha).
\displaystyle \\

\displaystyle \text{(ii) }G(\beta)=\begin{bmatrix}\cos\beta&0&\sin\beta\\0&1&0\\-\sin\beta&0&\cos\beta\end{bmatrix}
\displaystyle G(-\beta)=\begin{bmatrix}\cos(-\beta)&0&\sin(-\beta)\\0&1&0\\-\sin(-\beta)&0&\cos(-\beta)\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos\beta&0&-\sin\beta\\0&1&0\\\sin\beta&0&\cos\beta\end{bmatrix}
\displaystyle \text{The cofactors of the elements of }G(\beta)\text{ are:}
\displaystyle C_{11}=\cos\beta,\quad C_{12}=0,\quad C_{13}=\sin\beta
\displaystyle C_{21}=0,\quad C_{22}=\cos^2\beta+\sin^2\beta=1,\quad C_{23}=0
\displaystyle C_{31}=-\sin\beta,\quad C_{32}=0,\quad C_{33}=\cos\beta
\displaystyle \therefore\ \text{adj}\,G(\beta)=\begin{bmatrix}\cos\beta&0&\sin\beta\\0&1&0\\-\sin\beta&0&\cos\beta\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}\cos\beta&0&-\sin\beta\\0&1&0\\\sin\beta&0&\cos\beta\end{bmatrix}
\displaystyle \left|G(\beta)\right|=\cos^2\beta+\sin^2\beta=1\neq0
\displaystyle \therefore\ G(\beta)\text{ is non-singular and hence invertible.}
\displaystyle \left[G(\beta)\right]^{-1}=\frac{1}{\left|G(\beta)\right|}\text{adj}\,G(\beta)
\displaystyle =\begin{bmatrix}\cos\beta&0&-\sin\beta\\0&1&0\\\sin\beta&0&\cos\beta\end{bmatrix}=G(-\beta)
\displaystyle \therefore\ \left[G(\beta)\right]^{-1}=G(-\beta).
\displaystyle \\

\displaystyle \text{(iii) We know that }(AB)^{-1}=B^{-1}A^{-1}.
\displaystyle \therefore\ \left[F(\alpha)G(\beta)\right]^{-1}=\left[G(\beta)\right]^{-1}\left[F(\alpha)\right]^{-1}
\displaystyle =G(-\beta)F(-\alpha)\qquad\text{[Using parts (i) and (ii)]}
\displaystyle \therefore\ \left[F(\alpha)G(\beta)\right]^{-1}=G(-\beta)F(-\alpha).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }A=\begin{bmatrix}2&3\\1&2\end{bmatrix},\text{ verify that }A^2-4A+I=O,\text{ where } \\ I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\text{ and }O=\begin{bmatrix}0&0\\0&0\end{bmatrix}.\text{ Hence, find }A^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}2&3\\1&2\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}=\begin{bmatrix}7&12\\4&7\end{bmatrix}
\displaystyle A^2-4A+I=\begin{bmatrix}7&12\\4&7\end{bmatrix}-4\begin{bmatrix}2&3\\1&2\end{bmatrix}+\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}7&12\\4&7\end{bmatrix}-\begin{bmatrix}8&12\\4&8\end{bmatrix}+\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}7-8+1&12-12+0\\4-4+0&7-8+1\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O
\displaystyle \therefore\ A^2-4A+I=O.
\displaystyle \Rightarrow\ 4A-A^2=I
\displaystyle \Rightarrow\ A(4I-A)=I
\displaystyle \text{Also, }(4I-A)A=4A-A^2=I.
\displaystyle \therefore\ A^{-1}=4I-A
\displaystyle =4\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}2&3\\1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}4&0\\0&4\end{bmatrix}-\begin{bmatrix}2&3\\1&2\end{bmatrix}=\begin{bmatrix}2&-3\\-1&2\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Show that }A=\begin{bmatrix}-8&5\\2&4\end{bmatrix}\text{ satisfies the equation } \\ A^2+4A-42I=O.\text{ Hence, find }A^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}-8&5\\2&4\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}-8&5\\2&4\end{bmatrix}\begin{bmatrix}-8&5\\2&4\end{bmatrix}=\begin{bmatrix}74&-20\\-8&26\end{bmatrix}
\displaystyle A^2+4A-42I=\begin{bmatrix}74&-20\\-8&26\end{bmatrix}+4\begin{bmatrix}-8&5\\2&4\end{bmatrix}-42\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}74&-20\\-8&26\end{bmatrix}+\begin{bmatrix}-32&20\\8&16\end{bmatrix}-\begin{bmatrix}42&0\\0&42\end{bmatrix}
\displaystyle =\begin{bmatrix}74-32-42&-20+20-0\\-8+8-0&26+16-42\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O
\displaystyle \therefore\ A^2+4A-42I=O.
\displaystyle \Rightarrow\ A^2+4A=42I
\displaystyle \Rightarrow\ A(A+4I)=42I
\displaystyle \Rightarrow\ A\left[\frac{1}{42}(A+4I)\right]=I
\displaystyle \text{Also, }\left[\frac{1}{42}(A+4I)\right]A=I.
\displaystyle \therefore\ A^{-1}=\frac{1}{42}(A+4I)
\displaystyle =\frac{1}{42}\left\{\begin{bmatrix}-8&5\\2&4\end{bmatrix}+\begin{bmatrix}4&0\\0&4\end{bmatrix}\right\}
\displaystyle =\frac{1}{42}\begin{bmatrix}-4&5\\2&8\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }A=\begin{bmatrix}3&1\\-1&2\end{bmatrix},\text{ show that }A^2-5A+7I=O.\text{ Hence, find }A^{-1}.\hspace{1.0cm}\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}=\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}
\displaystyle A^2-5A+7I=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}
\displaystyle =\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O
\displaystyle \therefore\ A^2-5A+7I=O.
\displaystyle \Rightarrow\ 5A-A^2=7I
\displaystyle \Rightarrow\ A(5I-A)=7I
\displaystyle \Rightarrow\ A\left[\frac{1}{7}(5I-A)\right]=I
\displaystyle \text{Also, }\left[\frac{1}{7}(5I-A)\right]A=I.
\displaystyle \therefore\ A^{-1}=\frac{1}{7}(5I-A)
\displaystyle =\frac{1}{7}\left\{5\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}3&1\\-1&2\end{bmatrix}\right\}
\displaystyle =\frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }A=\begin{bmatrix}4&3\\2&5\end{bmatrix},\text{ find }x\text{ and }y\text{ such that }A^2=xA-yI.\text{ Hence, evaluate }A^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}4&3\\2&5\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}4&3\\2&5\end{bmatrix}\begin{bmatrix}4&3\\2&5\end{bmatrix}=\begin{bmatrix}22&27\\18&31\end{bmatrix}
\displaystyle \text{Since }A^2=xA-yI,\text{ we have }A^2-xA+yI=O.
\displaystyle \therefore\ \begin{bmatrix}22&27\\18&31\end{bmatrix}-\begin{bmatrix}4x&3x\\2x&5x\end{bmatrix}+\begin{bmatrix}y&0\\0&y\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow\ \begin{bmatrix}22-4x+y&27-3x\\18-2x&31-5x+y\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \therefore\ 22-4x+y=0,\quad 27-3x=0,\quad 18-2x=0,\quad 31-5x+y=0
\displaystyle 27-3x=0\Rightarrow x=9
\displaystyle \text{Substituting }x=9\text{ in }22-4x+y=0,\text{ we get}
\displaystyle 22-36+y=0\Rightarrow y=14
\displaystyle \therefore\ x=9\text{ and }y=14.
\displaystyle \text{Hence, }A^2-9A+14I=O
\displaystyle \Rightarrow\ 9A-A^2=14I
\displaystyle \Rightarrow\ A(9I-A)=14I
\displaystyle \Rightarrow\ A\left[\frac1{14}(9I-A)\right]=I
\displaystyle \text{Also, }\left[\frac1{14}(9I-A)\right]A=I.
\displaystyle \therefore\ A^{-1}=\frac1{14}(9I-A)
\displaystyle =\frac1{14}\left\{\begin{bmatrix}9&0\\0&9\end{bmatrix}-\begin{bmatrix}4&3\\2&5\end{bmatrix}\right\}
\displaystyle =\frac1{14}\begin{bmatrix}5&-3\\-2&4\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix},\text{ find the value of }\lambda\text{ so that }A^2=\lambda A-2I. \\ \text{ Hence, find }A^{-1}.\hspace{1.0cm}\text{[CBSE 2007]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}
\displaystyle \text{Given that }A^2=\lambda A-2I.
\displaystyle \therefore\ \begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\lambda\begin{bmatrix}3&-2\\4&-2\end{bmatrix}-2\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}3\lambda-2&-2\lambda\\4\lambda&-2\lambda-2\end{bmatrix}
\displaystyle \text{Equating the corresponding elements,}
\displaystyle -2\lambda=-2\Rightarrow\lambda=1
\displaystyle \therefore\ \lambda=1.
\displaystyle \text{Substituting }\lambda=1\text{ in }A^2=\lambda A-2I,\text{ we get}
\displaystyle A^2=A-2I
\displaystyle \Rightarrow\ A-A^2=2I
\displaystyle \Rightarrow\ A(I-A)=2I
\displaystyle \Rightarrow\ A\left[\frac12(I-A)\right]=I
\displaystyle \text{Also, }\left[\frac12(I-A)\right]A=I.
\displaystyle \therefore\ A^{-1}=\frac12(I-A)
\displaystyle =\frac12\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\right\}
\displaystyle =\frac12\begin{bmatrix}-2&2\\-4&3\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Show that }A=\begin{bmatrix}5&3\\-1&-2\end{bmatrix}\text{ satisfies the equation } \\ x^2-3x-7=0.\text{ Thus, find }A^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}5&3\\-1&-2\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}5&3\\-1&-2\end{bmatrix}\begin{bmatrix}5&3\\-1&-2\end{bmatrix}=\begin{bmatrix}22&9\\-3&1\end{bmatrix}
\displaystyle \text{If }I_2\text{ is the identity matrix of order }2,\text{ then}
\displaystyle A^2-3A-7I_2=\begin{bmatrix}22&9\\-3&1\end{bmatrix}-3\begin{bmatrix}5&3\\-1&-2\end{bmatrix}-7\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}22-15-7&9-9-0\\-3+3+0&1+6-7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O
\displaystyle \therefore\ A^2-3A-7I_2=O.
\displaystyle \therefore\ A\text{ satisfies the equation }x^2-3x-7=0.
\displaystyle \Rightarrow\ A^2-3A=7I_2
\displaystyle \Rightarrow\ A(A-3I_2)=7I_2
\displaystyle \Rightarrow\ A\left[\frac{1}{7}(A-3I_2)\right]=I_2
\displaystyle \text{Also, }\left[\frac{1}{7}(A-3I_2)\right]A=I_2.
\displaystyle \therefore\ A^{-1}=\frac{1}{7}(A-3I_2)
\displaystyle =\frac{1}{7}\left\{\begin{bmatrix}5&3\\-1&-2\end{bmatrix}-3\begin{bmatrix}1&0\\0&1\end{bmatrix}\right\}
\displaystyle =\frac{1}{7}\begin{bmatrix}2&3\\-1&-5\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Show that }A=\begin{bmatrix}6&5\\7&6\end{bmatrix}\text{ satisfies the equation } \\ x^2-12x+1=0.\text{ Thus, find }A^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}6&5\\7&6\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}6&5\\7&6\end{bmatrix}\begin{bmatrix}6&5\\7&6\end{bmatrix}=\begin{bmatrix}71&60\\84&71\end{bmatrix}
\displaystyle \text{If }I_2\text{ is the identity matrix of order }2,\text{ then}
\displaystyle A^2-12A+I_2=\begin{bmatrix}71&60\\84&71\end{bmatrix}-12\begin{bmatrix}6&5\\7&6\end{bmatrix}+\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}71-72+1&60-60+0\\84-84+0&71-72+1\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O
\displaystyle \therefore\ A^2-12A+I_2=O.
\displaystyle \therefore\ A\text{ satisfies the equation }x^2-12x+1=0.
\displaystyle \Rightarrow\ A^2-12A=-I_2
\displaystyle \Rightarrow\ 12A-A^2=I_2
\displaystyle \Rightarrow\ A(12I_2-A)=I_2
\displaystyle \Rightarrow\ A(12I_2-A)=I_2
\displaystyle \text{Also, }(12I_2-A)A=I_2.
\displaystyle \therefore\ A^{-1}=12I_2-A
\displaystyle =12\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}6&5\\7&6\end{bmatrix}
\displaystyle =\begin{bmatrix}12-6&0-5\\0-7&12-6\end{bmatrix}=\begin{bmatrix}6&-5\\-7&6\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{For the matrix }A=\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix},\text{ show that } \\ A^3-6A^2+5A+11I_3=O.\text{ Hence, find }A^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}
\displaystyle A^2=\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}1+1+2&1+2-1&1-3+3\\1+2-6&1+4+3&1-6-9\\2-1+6&2-2-3&2+3+9\end{bmatrix}=\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}
\displaystyle 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}4+2+2&4+4-1&4-6+3\\-3+8-28&-3+16+14&-3-24-42\\7-3+28&7-6-14&7+9+42\end{bmatrix}
\displaystyle =\begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}
\displaystyle 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}+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&1-6+5\\-23+18+5&27-48+10+11&-69+84-15\\32-42+10&-13+18-5&58-84+15+11\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O
\displaystyle \therefore\ A^3-6A^2+5A+11I_3=O.
\displaystyle \Rightarrow\ A^3-6A^2+5A=-11I_3
\displaystyle \Rightarrow\ A(A^2-6A+5I_3)=-11I_3
\displaystyle \Rightarrow\ A\left[-\frac{1}{11}(A^2-6A+5I_3)\right]=I_3
\displaystyle \text{Also, }\left[-\frac{1}{11}(A^2-6A+5I_3)\right]A=I_3.
\displaystyle \therefore\ A^{-1}=-\frac{1}{11}(A^2-6A+5I_3)
\displaystyle =-\frac{1}{11}\left\{\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}-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}\right\}
\displaystyle =-\frac{1}{11}\begin{bmatrix}3&-4&-5\\-9&1&4\\-5&3&1\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Show that the matrix }A=\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}\text{ satisfies the equation } \\ A^3-A^2-3A-I_3=O.\text{ Hence, find }A^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}
\displaystyle =\begin{bmatrix}1+0-6&0+0-8&-2+0-2\\-2+2+6&0+1+8&4-2+2\\3-8+3&0-4+4&-6+8+1\end{bmatrix}
\displaystyle =\begin{bmatrix}-5&-8&-4\\6&9&4\\-2&0&3\end{bmatrix}
\displaystyle A^3=A^2A=\begin{bmatrix}-5&-8&-4\\6&9&4\\-2&0&3\end{bmatrix}\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-5+16-12&0+8-16&10-16-4\\6-18+12&0-9+16&-12+18+4\\-2+0+9&0+0+12&4+0+3\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&-8&-10\\0&7&10\\7&12&7\end{bmatrix}
\displaystyle A^3-A^2-3A-I_3
\displaystyle =\begin{bmatrix}-1&-8&-10\\0&7&10\\7&12&7\end{bmatrix}-\begin{bmatrix}-5&-8&-4\\6&9&4\\-2&0&3\end{bmatrix}-3\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}-\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-1+5-3-1&-8+8&-10+4+6\\0-6+6&7-9+3-1&10-4-6\\7+2-9&12-12&7-3-3-1\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O
\displaystyle \therefore\ A^3-A^2-3A-I_3=O.
\displaystyle \text{Hence proved.}
\displaystyle A^3-A^2-3A=I_3
\displaystyle \Rightarrow\ A(A^2-A-3I_3)=I_3
\displaystyle \text{Also, }(A^2-A-3I_3)A=I_3.
\displaystyle \therefore\ A^{-1}=A^2-A-3I_3
\displaystyle =\begin{bmatrix}-5&-8&-4\\6&9&4\\-2&0&3\end{bmatrix}-\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}-3\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-5-1-3&-8&-4+2\\6+2&9+1-3&4-2\\-2-3&-4&3-1-3\end{bmatrix}
\displaystyle =\begin{bmatrix}-9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{If }A=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix},\text{ verify that } \\ A^3-6A^2+9A-4I=O\text{ and hence find }A^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}4+1+1&-2-2-1&2+1+2\\-2-2-1&1+4+1&-1-2-2\\2+1+2&-1-2-2&1+1+4\end{bmatrix}
\displaystyle =\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}
\displaystyle A^3=A^2A=\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}12+5+5&-6-10-5&6+5+10\\-10-6-5&5+12+5&-5-6-10\\10+5+6&-5-10-6&5+5+12\end{bmatrix}
\displaystyle =\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}
\displaystyle A^3-6A^2+9A-4I
\displaystyle =\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}-6\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}+9\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}-4\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}22-36+18-4&-21+30-9&21-30+9\\-21+30-9&22-36+18-4&-21+30-9\\21-30+9&-21+30-9&22-36+18-4\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O
\displaystyle \therefore\ A^3-6A^2+9A-4I=O.
\displaystyle \text{Hence proved.}
\displaystyle \Rightarrow\ A^3-6A^2+9A=4I
\displaystyle \Rightarrow\ A(A^2-6A+9I)=4I
\displaystyle \Rightarrow\ A\left[\frac14(A^2-6A+9I)\right]=I
\displaystyle \text{Also, }\left[\frac14(A^2-6A+9I)\right]A=I.
\displaystyle \therefore\ A^{-1}=\frac14(A^2-6A+9I)
\displaystyle =\frac14\left\{\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}-6\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}+9\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\right\}
\displaystyle =\frac14\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If }A=\frac{1}{9}\begin{bmatrix}-8&1&4\\4&4&7\\1&-8&4\end{bmatrix},\text{ prove that }A^{-1}=A^T.
\displaystyle \text{Answer:}
\displaystyle A=\frac{1}{9}\begin{bmatrix}-8&1&4\\4&4&7\\1&-8&4\end{bmatrix}
\displaystyle \therefore\ A^T=\frac{1}{9}\begin{bmatrix}-8&4&1\\1&4&-8\\4&7&4\end{bmatrix}
\displaystyle AA^T=\frac{1}{81}\begin{bmatrix}-8&1&4\\4&4&7\\1&-8&4\end{bmatrix}\begin{bmatrix}-8&4&1\\1&4&-8\\4&7&4\end{bmatrix}
\displaystyle =\frac{1}{81}\begin{bmatrix}64+1+16&-32+4+28&-8-8+16\\-32+4+28&16+16+49&4-32+28\\-8-8+16&4-32+28&1+64+16\end{bmatrix}
\displaystyle =\frac{1}{81}\begin{bmatrix}81&0&0\\0&81&0\\0&0&81\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3
\displaystyle \text{Similarly, }A^TA=I_3.
\displaystyle \therefore\ A^T\text{ is the inverse of }A.
\displaystyle \therefore\ A^{-1}=A^T.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If }A=\begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix},\text{ show that }A^{-1}=A^3.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}\begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}
\displaystyle =\begin{bmatrix}9-6+0&-9+9-4&12-12+4\\6-6+0&-6+9-4&8-12+4\\0-2+0&0+3-1&0-4+1\end{bmatrix}
\displaystyle =\begin{bmatrix}3&-4&4\\0&-1&0\\-2&2&-3\end{bmatrix}
\displaystyle A^3=A^2A=\begin{bmatrix}3&-4&4\\0&-1&0\\-2&2&-3\end{bmatrix}\begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}
\displaystyle =\begin{bmatrix}9-8+0&-9+12-4&12-16+4\\0-2+0&0+3+0&0-4+0\\-6+4+0&6-6+3&-8+8-3\end{bmatrix}
\displaystyle =\begin{bmatrix}1&-1&0\\-2&3&-4\\-2&3&-3\end{bmatrix}
\displaystyle A^3A=\begin{bmatrix}1&-1&0\\-2&3&-4\\-2&3&-3\end{bmatrix}\begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}
\displaystyle =\begin{bmatrix}3-2+0&-3+3+0&4-4+0\\-6+6+0&6-9+4&-8+12-4\\-6+6+0&6-9+3&-8+12-3\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3
\displaystyle \therefore\ A^3A=I_3.
\displaystyle \text{Also, }AA^3=A^4=A^3A=I_3.
\displaystyle \therefore\ A^{-1}=A^3.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{If }A=\begin{bmatrix}-1&2&0\\-1&1&1\\0&1&0\end{bmatrix},\text{ show that }A^2=A^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}-1&2&0\\-1&1&1\\0&1&0\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}-1&2&0\\-1&1&1\\0&1&0\end{bmatrix}\begin{bmatrix}-1&2&0\\-1&1&1\\0&1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}1-2+0&-2+2+0&0+2+0\\1-1+0&-2+1+1&0+1+0\\0-1+0&0+1+0&0+1+0\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&0&2\\0&0&1\\-1&1&1\end{bmatrix}
\displaystyle A^2A=\begin{bmatrix}-1&0&2\\0&0&1\\-1&1&1\end{bmatrix}\begin{bmatrix}-1&2&0\\-1&1&1\\0&1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}1+0+0&-2+0+2&0\\0+0+0&0+0+1&0\\1-1+0&-2+1+1&1\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3
\displaystyle \therefore\ A^2A=I_3.
\displaystyle \text{Also, }AA^2=A^3=A^2A=I_3.
\displaystyle \therefore\ A^2=A^{-1}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Solve the matrix equation }\begin{bmatrix}5&4\\1&1\end{bmatrix}X=\begin{bmatrix}1&-2\\1&3\end{bmatrix},\text{ where }X\text{ is a } \\ 2\times2\text{ matrix.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\begin{bmatrix}5&4\\1&1\end{bmatrix}\text{ and }B=\begin{bmatrix}1&-2\\1&3\end{bmatrix}
\displaystyle \text{Then }AX=B
\displaystyle |A|=\begin{vmatrix}5&4\\1&1\end{vmatrix}=5-4=1\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle A^{-1}=\frac{1}{|A|}\begin{bmatrix}1&-4\\-1&5\end{bmatrix}=\begin{bmatrix}1&-4\\-1&5\end{bmatrix}
\displaystyle \Rightarrow\ A^{-1}(AX)=A^{-1}B
\displaystyle \Rightarrow\ (A^{-1}A)X=A^{-1}B
\displaystyle \Rightarrow\ IX=A^{-1}B
\displaystyle \Rightarrow\ X=A^{-1}B
\displaystyle =\begin{bmatrix}1&-4\\-1&5\end{bmatrix}\begin{bmatrix}1&-2\\1&3\end{bmatrix}
\displaystyle =\begin{bmatrix}1-4&-2-12\\-1+5&2+15\end{bmatrix}=\begin{bmatrix}-3&-14\\4&17\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{Find the matrix }X\text{ satisfying the matrix equation } \\ X\begin{bmatrix}5&3\\-1&-2\end{bmatrix}=\begin{bmatrix}14&7\\7&7\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\begin{bmatrix}5&3\\-1&-2\end{bmatrix}\text{ and }B=\begin{bmatrix}14&7\\7&7\end{bmatrix}
\displaystyle \text{Then }XA=B
\displaystyle |A|=\begin{vmatrix}5&3\\-1&-2\end{vmatrix}=-10+3=-7\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle A^{-1}=\frac{1}{|A|}\begin{bmatrix}-2&-3\\1&5\end{bmatrix}=-\frac17\begin{bmatrix}-2&-3\\1&5\end{bmatrix}
\displaystyle \Rightarrow\ (XA)A^{-1}=BA^{-1}
\displaystyle \Rightarrow\ X(AA^{-1})=BA^{-1}
\displaystyle \Rightarrow\ XI=BA^{-1}
\displaystyle \Rightarrow\ X=BA^{-1}
\displaystyle =\begin{bmatrix}14&7\\7&7\end{bmatrix}\left(-\frac17\right)\begin{bmatrix}-2&-3\\1&5\end{bmatrix}
\displaystyle =-\frac17\begin{bmatrix}-28+7&-42+35\\-14+7&-21+35\end{bmatrix}
\displaystyle =-\frac17\begin{bmatrix}-21&-7\\-7&14\end{bmatrix}=\begin{bmatrix}3&1\\1&-2\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Find the matrix }X\text{ for which }\begin{bmatrix}3&2\\7&5\end{bmatrix}X\begin{bmatrix}-1&1\\-2&1\end{bmatrix}=\begin{bmatrix}2&-1\\0&4\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\begin{bmatrix}3&2\\7&5\end{bmatrix},\ B=\begin{bmatrix}-1&1\\-2&1\end{bmatrix}\text{ and }C=\begin{bmatrix}2&-1\\0&4\end{bmatrix}
\displaystyle \text{Then }AXB=C
\displaystyle |A|=\begin{vmatrix}3&2\\7&5\end{vmatrix}=15-14=1\neq0
\displaystyle |B|=\begin{vmatrix}-1&1\\-2&1\end{vmatrix}=-1+2=1\neq0
\displaystyle \therefore\ A\text{ and }B\text{ are non-singular and hence invertible.}
\displaystyle A^{-1}=\frac{1}{|A|}\begin{bmatrix}5&-2\\-7&3\end{bmatrix}=\begin{bmatrix}5&-2\\-7&3\end{bmatrix}
\displaystyle B^{-1}=\frac{1}{|B|}\begin{bmatrix}1&-1\\2&-1\end{bmatrix}=\begin{bmatrix}1&-1\\2&-1\end{bmatrix}
\displaystyle \Rightarrow\ A^{-1}(AXB)B^{-1}=A^{-1}CB^{-1}
\displaystyle \Rightarrow\ (A^{-1}A)X(BB^{-1})=A^{-1}CB^{-1}
\displaystyle \Rightarrow\ I_2XI_2=A^{-1}CB^{-1}
\displaystyle \Rightarrow\ X=A^{-1}CB^{-1}
\displaystyle =\begin{bmatrix}5&-2\\-7&3\end{bmatrix}\begin{bmatrix}2&-1\\0&4\end{bmatrix}\begin{bmatrix}1&-1\\2&-1\end{bmatrix}
\displaystyle =\begin{bmatrix}5&-2\\-7&3\end{bmatrix}\begin{bmatrix}2-2&-2+1\\0+8&0-4\end{bmatrix}
\displaystyle =\begin{bmatrix}5&-2\\-7&3\end{bmatrix}\begin{bmatrix}0&-1\\8&-4\end{bmatrix}
\displaystyle =\begin{bmatrix}0-16&-5+8\\0+24&7-12\end{bmatrix}=\begin{bmatrix}-16&3\\24&-5\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Find the matrix }X\text{ satisfying the equation } \\ \begin{bmatrix}2&1\\5&3\end{bmatrix}X\begin{bmatrix}5&3\\3&2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\begin{bmatrix}2&1\\5&3\end{bmatrix}\text{ and }B=\begin{bmatrix}5&3\\3&2\end{bmatrix}
\displaystyle \text{Then }AXB=I_2
\displaystyle |A|=\begin{vmatrix}2&1\\5&3\end{vmatrix}=6-5=1\neq0
\displaystyle |B|=\begin{vmatrix}5&3\\3&2\end{vmatrix}=10-9=1\neq0
\displaystyle \therefore\ A\text{ and }B\text{ are non-singular and hence invertible.}
\displaystyle A^{-1}=\frac{1}{|A|}\begin{bmatrix}3&-1\\-5&2\end{bmatrix}=\begin{bmatrix}3&-1\\-5&2\end{bmatrix}
\displaystyle B^{-1}=\frac{1}{|B|}\begin{bmatrix}2&-3\\-3&5\end{bmatrix}=\begin{bmatrix}2&-3\\-3&5\end{bmatrix}
\displaystyle \Rightarrow\ A^{-1}(AXB)B^{-1}=A^{-1}I_2B^{-1}
\displaystyle \Rightarrow\ (A^{-1}A)X(BB^{-1})=A^{-1}B^{-1}
\displaystyle \Rightarrow\ I_2XI_2=A^{-1}B^{-1}
\displaystyle \Rightarrow\ X=A^{-1}B^{-1}
\displaystyle =\begin{bmatrix}3&-1\\-5&2\end{bmatrix}\begin{bmatrix}2&-3\\-3&5\end{bmatrix}
\displaystyle =\begin{bmatrix}6+3&-9-5\\-10-6&15+10\end{bmatrix}=\begin{bmatrix}9&-14\\-16&25\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{If }A=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix},\text{ find }A^{-1}\text{ and prove that } \\ A^2-4A-5I_3=O.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&2&2\\2&1&2\\2&2&1\end{vmatrix}=1(1-4)-2(2-4)+2(4-2)
\displaystyle =-3+4+4=5\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle A^2=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}
\displaystyle =\begin{bmatrix}1+4+4&2+2+4&2+4+2\\2+2+4&4+1+4&4+2+2\\2+4+2&4+2+2&4+4+1\end{bmatrix}
\displaystyle =\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}
\displaystyle A^2-4A-5I_3=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}-4\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}-5\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}9-4-5&8-8&8-8\\8-8&9-4-5&8-8\\8-8&8-8&9-4-5\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O
\displaystyle \therefore\ A^2-4A-5I_3=O.
\displaystyle \text{Also, }A^2-4A-5I_3=O
\displaystyle \Rightarrow\ A^2-4A=5I_3
\displaystyle \Rightarrow\ A(A-4I_3)=5I_3
\displaystyle \Rightarrow\ A\left(\frac{A-4I_3}{5}\right)=I_3
\displaystyle \text{Also, }\left(\frac{A-4I_3}{5}\right)A=I_3.
\displaystyle \therefore\ A^{-1}=\frac{A-4I_3}{5}
\displaystyle =\frac{1}{5}\left[\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}-4\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\right]
\displaystyle =\frac{1}{5}\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{If }A\text{ is a square matrix of order }n,\text{ prove that }\left|A\,\text{adj}\,A\right|=|A|^n.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=[a_{ij}]\text{ be a square matrix of order }n.
\displaystyle \text{If }C_{ij}\text{ is the cofactor of }a_{ij}\text{ in }A,\text{ then}
\displaystyle \text{adj}\,A=[C_{ij}]^T
\displaystyle \therefore\ (\text{adj}\,A)_{rj}=C_{jr}
\displaystyle \bigl(A\,\text{adj}\,A\bigr)_{ij}=\sum_{r=1}^{n}a_{ir}(\text{adj}\,A)_{rj}
\displaystyle =\sum_{r=1}^{n}a_{ir}C_{jr}
\displaystyle =\begin{cases}|A|,&i=j,\\0,&i\ne j.\end{cases}
\displaystyle \therefore\ A\,\text{adj}\,A=\begin{bmatrix}|A|&0&\cdots&0\\0&|A|&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&|A|\end{bmatrix}
\displaystyle =|A|I_n
\displaystyle \therefore\ \left|A\,\text{adj}\,A\right|=\left||A|I_n\right|
\displaystyle =|A|^n|I_n|
\displaystyle =|A|^n
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{If }A^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}\text{ and }B=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix},\text{ find }(AB)^{-1}.\hspace{1.0cm}[\text{CBSE 2006}]
\displaystyle \text{Answer:}
\displaystyle \text{We know that }(AB)^{-1}=B^{-1}A^{-1}.
\displaystyle |B|=\begin{vmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{vmatrix}
\displaystyle =1(3)-2(-1)+(-2)(2)=3+2-4=1\neq0
\displaystyle \therefore\ B\text{ is non-singular and hence invertible.}
\displaystyle \text{The cofactors of the elements of }B\text{ are}
\displaystyle C_{11}=3,\ C_{12}=1,\ C_{13}=2
\displaystyle C_{21}=2,\ C_{22}=1,\ C_{23}=2
\displaystyle C_{31}=6,\ C_{32}=2,\ C_{33}=5
\displaystyle \therefore\ \text{adj}\,B=\begin{bmatrix}3&1&2\\2&1&2\\6&2&5\end{bmatrix}^{T}=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}
\displaystyle B^{-1}=\frac{1}{|B|}\text{adj}\,B=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}
\displaystyle \therefore\ (AB)^{-1}=B^{-1}A^{-1}
\displaystyle =\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}
\displaystyle =\begin{bmatrix}9-30+30&-3+12-12&3-10+12\\3-15+10&-1+6-4&1-5+4\\6-30+25&-2+12-10&2-10+10\end{bmatrix}
\displaystyle =\begin{bmatrix}9&-3&5\\-2&1&0\\1&0&2\end{bmatrix}
\displaystyle \therefore\ (AB)^{-1}=\begin{bmatrix}9&-3&5\\-2&1&0\\1&0&2\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{If }A=\begin{bmatrix}1&-2&3\\0&-1&4\\-2&2&1\end{bmatrix},\text{ find }(A^T)^{-1}.\hspace{1.0cm}[\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle \text{We know that }(A^T)^{-1}=(A^{-1})^T.
\displaystyle |A|=\begin{vmatrix}1&-2&3\\0&-1&4\\-2&2&1\end{vmatrix}
\displaystyle =1(-1-8)-(-2)(0+8)+3(0-2)
\displaystyle =-9+16-6=1\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle \text{The cofactors of the elements of }A\text{ are}
\displaystyle C_{11}=-9,\ C_{12}=-8,\ C_{13}=-2
\displaystyle C_{21}=8,\ C_{22}=7,\ C_{23}=2
\displaystyle C_{31}=-5,\ C_{32}=-4,\ C_{33}=-1
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}-9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-9&8&-5\\-8&7&-4\\-2&2&-1\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj}\,A=\begin{bmatrix}-9&8&-5\\-8&7&-4\\-2&2&-1\end{bmatrix}
\displaystyle \therefore\ (A^T)^{-1}=(A^{-1})^T
\displaystyle =\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}
\displaystyle \therefore\ (A^T)^{-1}=\begin{bmatrix}-9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Find the adjoint of the matrix }A=\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix} \\ \text{ and hence show that }A(\text{adj}\,A)=|A|I_3.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}
\displaystyle \text{The cofactors of the elements of }A\text{ are}
\displaystyle C_{11}=-3,\ C_{12}=-6,\ C_{13}=-6
\displaystyle C_{21}=6,\ C_{22}=3,\ C_{23}=-6
\displaystyle C_{31}=6,\ C_{32}=-6,\ C_{33}=3
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}-3&-6&-6\\6&3&-6\\6&-6&3\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-3&6&6\\-6&3&-6\\-6&-6&3\end{bmatrix}
\displaystyle |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 A(\text{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\\-6-6+12&12+3+12&12-6-6\\-6+12-6&12-6-6&12+12+3\end{bmatrix}
\displaystyle =\begin{bmatrix}27&0&0\\0&27&0\\0&0&27\end{bmatrix}
\displaystyle =27\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=27I_3
\displaystyle =|A|I_3
\displaystyle \therefore\ A(\text{adj}\,A)=|A|I_3.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If }A=\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}, \\ \text{ find }A^{-1}\text{ and show that }A^{-1}=\frac{1}{2}\left(A^2-3I_3\right).\hspace{1.0cm}[\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}0&1&1\\1&0&1\\1&1&0\end{vmatrix}
\displaystyle =0-1(0-1)+1(1-0)=1+1=2\neq0
\displaystyle \therefore\ A\text{ is non-singular and hence invertible.}
\displaystyle \text{The cofactors of the elements of }A\text{ are}
\displaystyle C_{11}=-1,\ C_{12}=1,\ C_{13}=1
\displaystyle C_{21}=1,\ C_{22}=-1,\ C_{23}=1
\displaystyle C_{31}=1,\ C_{32}=1,\ C_{33}=-1
\displaystyle \therefore\ \text{adj}\,A=\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}
\displaystyle \therefore\ A^{-1}=\frac{1}{|A|}\text{adj}\,A
\displaystyle =\frac12\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}0+1+1&0+0+1&0+1+0\\0+0+1&1+0+1&1+0+0\\0+1+0&1+0+0&1+1+0\end{bmatrix}
\displaystyle =\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}
\displaystyle \frac12\left(A^2-3I_3\right)=\frac12\left[\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}-3\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\right]
\displaystyle =\frac12\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}
\displaystyle =A^{-1}
\displaystyle \therefore\ A^{-1}=\frac12\left(A^2-3I_3\right).
\displaystyle \text{Hence proved.}
\displaystyle \\

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