\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}.\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.}

 

\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}.\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}\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.}\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}

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