Class 12: Adjoint and Inverse of a Matrix

$\displaystyle \text{Find the inverse of each of the following matrices by using elementary row transformations:}$

$\displaystyle \textbf{Question 1: }\begin{bmatrix}7&1\\4&-3\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}7&1\\4&-3\end{bmatrix}$
$\displaystyle \text{We know that }A=I_2A$
$\displaystyle \therefore\ \begin{bmatrix}7&1\\4&-3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&\frac17\\4&-3\end{bmatrix}=\begin{bmatrix}\frac17&0\\0&1\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow\frac17R_1\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&\frac17\\0&-\frac{25}{7}\end{bmatrix}=\begin{bmatrix}\frac17&0\\-\frac47&1\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow R_2-4R_1\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&\frac17\\0&1\end{bmatrix}=\begin{bmatrix}\frac17&0\\\frac4{25}&-\frac7{25}\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow-\frac7{25}R_2\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}\frac3{25}&\frac1{25}\\\frac4{25}&-\frac7{25}\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow R_1-\frac17R_2\right]$
$\displaystyle \therefore\ I_2=\frac1{25}\begin{bmatrix}3&1\\4&-7\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\frac1{25}\begin{bmatrix}3&1\\4&-7\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\begin{bmatrix}5&2\\2&1\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}5&2\\2&1\end{bmatrix}$
$\displaystyle \text{We know that }A=I_2A$
$\displaystyle \therefore\ \begin{bmatrix}5&2\\2&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0\\2&1\end{bmatrix}=\begin{bmatrix}1&-2\\0&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-2R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}1&-2\\-2&5\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-2R_1]$
$\displaystyle \therefore\ I_2=\begin{bmatrix}1&-2\\-2&5\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}1&-2\\-2&5\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\begin{bmatrix}1&2\\2&-1\end{bmatrix}\hspace{8.0cm}[\text{CBSE 2007}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}1&2\\2&-1\end{bmatrix}$
$\displaystyle \text{We know that }A=I_2A$
$\displaystyle \therefore\ \begin{bmatrix}1&2\\2&-1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&2\\0&-5\end{bmatrix}=\begin{bmatrix}1&0\\-2&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-2R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&2\\0&1\end{bmatrix}=\begin{bmatrix}1&0\\\frac25&-\frac15\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow-\frac15R_2\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}\frac15&\frac25\\\frac25&-\frac15\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-2R_2]$
$\displaystyle \therefore\ I_2=\frac15\begin{bmatrix}1&2\\2&-1\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\frac15\begin{bmatrix}1&2\\2&-1\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\begin{bmatrix}2&5\\1&3\end{bmatrix}\hspace{8.0cm}[\text{CBSE 2010}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}2&5\\1&3\end{bmatrix}$
$\displaystyle \text{We know that }A=I_2A$
$\displaystyle \therefore\ \begin{bmatrix}2&5\\1&3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&2\\1&3\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&2\\0&1\end{bmatrix}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-2R_2]$
$\displaystyle \therefore\ I_2=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\begin{bmatrix}3&10\\2&7\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}3&10\\2&7\end{bmatrix}$
$\displaystyle \text{We know that }A=I_2A$
$\displaystyle \therefore\ \begin{bmatrix}3&10\\2&7\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&3\\2&7\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&3\\0&1\end{bmatrix}=\begin{bmatrix}1&-1\\-2&3\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-2R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}7&-10\\-2&3\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-3R_2]$
$\displaystyle \therefore\ I_2=\begin{bmatrix}7&-10\\-2&3\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}7&-10\\-2&3\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\begin{bmatrix}0&1&2\\1&2&3\\3&1&1\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}0&1&2\\1&2&3\\3&1&1\end{bmatrix}$
$\displaystyle \text{We know that }A=I_3A$
$\displaystyle \therefore\ \begin{bmatrix}0&1&2\\1&2&3\\3&1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}0&1&2\\-2&1&2\\3&1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&-1\\0&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-R_3]$
$\displaystyle \Rightarrow\ \begin{bmatrix}-3&0&1\\-2&1&2\\3&1&1\end{bmatrix}=\begin{bmatrix}1&0&-1\\0&1&-1\\0&0&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-R_3]$
$\displaystyle \Rightarrow\ \begin{bmatrix}-3&0&1\\-2&1&2\\0&1&2\end{bmatrix}=\begin{bmatrix}1&0&-1\\0&1&-1\\1&0&0\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow R_3+R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}-3&0&1\\0&3&4\\0&1&2\end{bmatrix}=\begin{bmatrix}1&0&-1\\-2&3&-1\\1&0&0\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow3R_2-2R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}-3&0&1\\0&3&4\\0&-2&-2\end{bmatrix}=\begin{bmatrix}1&0&-1\\-2&3&-1\\3&-3&1\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow R_3-R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}-3&0&1\\0&3&4\\0&1&1\end{bmatrix}=\begin{bmatrix}1&0&-1\\-2&3&-1\\-\frac32&\frac32&-\frac12\end{bmatrix}A\qquad\left[\text{Applying }R_3\rightarrow-\frac12R_3\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}-3&0&1\\0&-1&0\\0&1&1\end{bmatrix}=\begin{bmatrix}1&0&-1\\4&-3&1\\-\frac32&\frac32&-\frac12\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-4R_3]$
$\displaystyle \Rightarrow\ \begin{bmatrix}-3&0&1\\0&-1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&-1\\4&-3&1\\\frac52&-\frac32&\frac12\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow R_3+R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}-3&0&0\\0&-1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}-\frac32&\frac32&-\frac32\\4&-3&1\\\frac52&-\frac32&\frac12\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-R_3]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&-1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}\frac12&-\frac12&\frac12\\4&-3&1\\\frac52&-\frac32&\frac12\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow-\frac13R_1\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}\frac12&-\frac12&\frac12\\-4&3&-1\\\frac52&-\frac32&\frac12\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow-R_2]$
$\displaystyle \therefore\ I_3=\begin{bmatrix}\frac12&-\frac12&\frac12\\-4&3&-1\\\frac52&-\frac32&\frac12\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}\frac12&-\frac12&\frac12\\-4&3&-1\\\frac52&-\frac32&\frac12\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}$
$\displaystyle \text{We know that }A=I_3A$
$\displaystyle \therefore\ \begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac12\\5&1&0\\0&1&3\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\0&1&0\\0&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow\frac12R_1\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac12\\0&1&\frac52\\0&1&3\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\-\frac52&1&0\\0&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-5R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac12\\0&1&\frac52\\0&0&\frac12\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\-\frac52&1&0\\\frac52&-1&1\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow R_3-R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac12\\0&1&\frac52\\0&0&1\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\-\frac52&1&0\\5&-2&2\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow2R_3]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1+\frac12R_3\text{ and }R_2\rightarrow R_2-\frac52R_3]$
$\displaystyle \therefore\ I_3=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\begin{bmatrix}2&3&1\\2&4&1\\3&7&2\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}2&3&1\\2&4&1\\3&7&2\end{bmatrix}$
$\displaystyle \text{We know that }A=I_3A$
$\displaystyle \therefore\ \begin{bmatrix}2&3&1\\2&4&1\\3&7&2\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&\frac32&\frac12\\2&4&1\\3&7&2\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\0&1&0\\0&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow\frac12R_1\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&\frac32&\frac12\\0&1&0\\0&\frac52&\frac12\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\-1&1&0\\-\frac32&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-2R_1\text{ and }R_3\rightarrow R_3-3R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&\frac12\\0&1&0\\0&0&\frac12\end{bmatrix}=\begin{bmatrix}2&-\frac32&0\\-1&1&0\\1&-\frac52&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-\frac32R_2\text{ and }R_3\rightarrow R_3-\frac52R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&\frac12\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}2&-\frac32&0\\-1&1&0\\2&-5&2\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow2R_3]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&1&-1\\-1&1&0\\2&-5&2\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-\frac12R_3]$
$\displaystyle \therefore\ I_3=\begin{bmatrix}1&1&-1\\-1&1&0\\2&-5&2\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}1&1&-1\\-1&1&0\\2&-5&2\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}$
$\displaystyle \text{We know that }A=I_3A$
$\displaystyle \therefore\ \begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-1&\frac43\\2&-3&4\\0&-1&1\end{bmatrix}=\begin{bmatrix}\frac13&0&0\\0&1&0\\0&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow\frac13R_1\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-1&\frac43\\0&-1&\frac43\\0&-1&1\end{bmatrix}=\begin{bmatrix}\frac13&0&0\\-\frac23&1&0\\0&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-2R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-1&\frac43\\0&1&-\frac43\\0&-1&1\end{bmatrix}=\begin{bmatrix}\frac13&0&0\\\frac23&-1&0\\0&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow-R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&-\frac43\\0&0&-\frac13\end{bmatrix}=\begin{bmatrix}1&-1&0\\\frac23&-1&0\\\frac23&-1&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1+R_2\text{ and }R_3\rightarrow R_3+R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&-\frac43\\0&0&1\end{bmatrix}=\begin{bmatrix}1&-1&0\\\frac23&-1&0\\-2&3&-3\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow-3R_3]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&-1&0\\-2&3&-4\\-2&3&-3\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow R_2+\frac43R_3\right]$
$\displaystyle \therefore\ I_3=\begin{bmatrix}1&-1&0\\-2&3&-4\\-2&3&-3\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}1&-1&0\\-2&3&-4\\-2&3&-3\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\begin{bmatrix}1&2&0\\2&3&-1\\1&-1&3\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}1&2&0\\2&3&-1\\1&-1&3\end{bmatrix}$
$\displaystyle \text{We know that }A=I_3A$
$\displaystyle \therefore\ \begin{bmatrix}1&2&0\\2&3&-1\\1&-1&3\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&2&0\\0&-1&-1\\0&-3&3\end{bmatrix}=\begin{bmatrix}1&0&0\\-2&1&0\\-1&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-2R_1\text{ and }R_3\rightarrow R_3-R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&2&0\\0&1&1\\0&-3&3\end{bmatrix}=\begin{bmatrix}1&0&0\\2&-1&0\\-1&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow-R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-2\\0&1&1\\0&0&6\end{bmatrix}=\begin{bmatrix}-3&2&0\\2&-1&0\\5&-3&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-2R_2\text{ and }R_3\rightarrow R_3+3R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-2\\0&1&1\\0&0&1\end{bmatrix}=\begin{bmatrix}-3&2&0\\2&-1&0\\\frac56&-\frac12&\frac16\end{bmatrix}A\qquad\left[\text{Applying }R_3\rightarrow\frac16R_3\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}-\frac43&1&\frac13\\\frac76&-\frac12&-\frac16\\\frac56&-\frac12&\frac16\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1+2R_3\text{ and }R_2\rightarrow R_2-R_3]$
$\displaystyle \therefore\ I_3=\begin{bmatrix}-\frac43&1&\frac13\\\frac76&-\frac12&-\frac16\\\frac56&-\frac12&\frac16\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}-\frac43&1&\frac13\\\frac76&-\frac12&-\frac16\\\frac56&-\frac12&\frac16\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\begin{bmatrix}2&-1&3\\1&2&4\\3&1&1\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}2&-1&3\\1&2&4\\3&1&1\end{bmatrix}$
$\displaystyle \text{We know that }A=I_3A$
$\displaystyle \therefore\ \begin{bmatrix}2&-1&3\\1&2&4\\3&1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-\frac12&\frac32\\1&2&4\\3&1&1\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\0&1&0\\0&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow\frac12R_1\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-\frac12&\frac32\\0&\frac52&\frac52\\0&\frac52&-\frac72\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\-\frac12&1&0\\-\frac32&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-R_1\text{ and }R_3\rightarrow R_3-3R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-\frac12&\frac32\\0&1&1\\0&\frac52&-\frac72\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\-\frac15&\frac25&0\\-\frac32&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow\frac25R_2\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&2\\0&1&1\\0&0&-6\end{bmatrix}=\begin{bmatrix}\frac25&\frac15&0\\-\frac15&\frac25&0\\-1&-1&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1+\frac12R_2\text{ and }R_3\rightarrow R_3-\frac52R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&2\\0&1&1\\0&0&1\end{bmatrix}=\begin{bmatrix}\frac25&\frac15&0\\-\frac15&\frac25&0\\\frac16&\frac16&-\frac16\end{bmatrix}A\qquad\left[\text{Applying }R_3\rightarrow-\frac16R_3\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}\frac1{15}&-\frac2{15}&\frac13\\-\frac{11}{30}&\frac7{30}&\frac16\\\frac16&\frac16&-\frac16\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-R_3\text{ and }R_1\rightarrow R_1-2R_3]$
$\displaystyle \therefore\ I_3=\begin{bmatrix}\frac1{15}&-\frac2{15}&\frac13\\-\frac{11}{30}&\frac7{30}&\frac16\\\frac16&\frac16&-\frac16\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}\frac1{15}&-\frac2{15}&\frac13\\-\frac{11}{30}&\frac7{30}&\frac16\\\frac16&\frac16&-\frac16\end{bmatrix}=-\frac1{30}\begin{bmatrix}-2&4&-10\\11&-7&-5\\-5&-5&5\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\begin{bmatrix}1&1&2\\3&1&1\\2&3&1\end{bmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}1&1&2\\3&1&1\\2&3&1\end{bmatrix}$
$\displaystyle \text{We know that }A=I_3A$
$\displaystyle \therefore\ \begin{bmatrix}1&1&2\\3&1&1\\2&3&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&1&2\\0&-2&-5\\0&1&-3\end{bmatrix}=\begin{bmatrix}1&0&0\\-3&1&0\\-2&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-3R_1\text{ and }R_3\rightarrow R_3-2R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&1&2\\0&1&\frac52\\0&1&-3\end{bmatrix}=\begin{bmatrix}1&0&0\\\frac32&-\frac12&0\\-2&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow-\frac12R_2\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac12\\0&1&\frac52\\0&0&-\frac{11}{2}\end{bmatrix}=\begin{bmatrix}-\frac12&\frac12&0\\\frac32&-\frac12&0\\-\frac72&\frac12&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-R_2\text{ and }R_3\rightarrow R_3-R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac12\\0&1&\frac52\\0&0&1\end{bmatrix}=\begin{bmatrix}-\frac12&\frac12&0\\\frac32&-\frac12&0\\\frac7{11}&-\frac1{11}&-\frac2{11}\end{bmatrix}A\qquad\left[\text{Applying }R_3\rightarrow-\frac2{11}R_3\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}-\frac2{11}&\frac5{11}&-\frac1{11}\\-\frac1{11}&-\frac3{11}&\frac5{11}\\\frac7{11}&-\frac1{11}&-\frac2{11}\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-\frac52R_3\text{ and }R_1\rightarrow R_1+\frac12R_3]$
$\displaystyle \therefore\ I_3=\frac1{11}\begin{bmatrix}-2&5&-1\\-1&-3&5\\7&-1&-2\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\frac1{11}\begin{bmatrix}-2&5&-1\\-1&-3&5\\7&-1&-2\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\begin{bmatrix}2&-1&4\\4&0&2\\3&-2&7\end{bmatrix}\hspace{8.0cm}[\text{CBSE 2008}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}2&-1&4\\4&0&2\\3&-2&7\end{bmatrix}$
$\displaystyle \text{We know that }A=I_3A$
$\displaystyle \therefore\ \begin{bmatrix}2&-1&4\\4&0&2\\3&-2&7\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-\frac12&2\\4&0&2\\3&-2&7\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\0&1&0\\0&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow\frac12R_1\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-\frac12&2\\0&2&-6\\0&-\frac12&1\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\-2&1&0\\-\frac32&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-4R_1\text{ and }R_3\rightarrow R_3-3R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-\frac12&2\\0&1&-3\\0&-\frac12&1\end{bmatrix}=\begin{bmatrix}\frac12&0&0\\-1&\frac12&0\\-\frac32&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow\frac12R_2\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&\frac12\\0&1&-3\\0&0&-\frac12\end{bmatrix}=\begin{bmatrix}0&\frac14&0\\-1&\frac12&0\\-2&\frac14&1\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow R_1+\frac12R_2\text{ and }R_3\rightarrow R_3+\frac12R_2\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&\frac12\\0&1&-3\\0&0&1\end{bmatrix}=\begin{bmatrix}0&\frac14&0\\-1&\frac12&0\\4&-\frac12&-2\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow-2R_3]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}-2&\frac12&1\\11&-1&-6\\4&-\frac12&-2\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow R_1-\frac12R_3\text{ and }R_2\rightarrow R_2+3R_3\right]$
$\displaystyle \therefore\ I_3=\begin{bmatrix}-2&\frac12&1\\11&-1&-6\\4&-\frac12&-2\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}-2&\frac12&1\\11&-1&-6\\4&-\frac12&-2\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\begin{bmatrix}3&0&-1\\2&3&0\\0&4&1\end{bmatrix}\hspace{8.0cm}[\text{CBSE 2009}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}3&0&-1\\2&3&0\\0&4&1\end{bmatrix}$
$\displaystyle \text{We know that }A=I_3A$
$\displaystyle \therefore\ \begin{bmatrix}3&0&-1\\2&3&0\\0&4&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac13\\2&3&0\\0&4&1\end{bmatrix}=\begin{bmatrix}\frac13&0&0\\0&1&0\\0&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow\frac13R_1\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac13\\0&3&\frac23\\0&4&1\end{bmatrix}=\begin{bmatrix}\frac13&0&0\\-\frac23&1&0\\0&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-2R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac13\\0&1&\frac29\\0&4&1\end{bmatrix}=\begin{bmatrix}\frac13&0&0\\-\frac29&\frac13&0\\0&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow\frac13R_2\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac13\\0&1&\frac29\\0&0&\frac19\end{bmatrix}=\begin{bmatrix}\frac13&0&0\\-\frac29&\frac13&0\\\frac89&-\frac43&1\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow R_3-4R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac13\\0&1&\frac29\\0&0&1\end{bmatrix}=\begin{bmatrix}\frac13&0&0\\-\frac29&\frac13&0\\8&-12&9\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow9R_3]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}3&-4&3\\-2&3&-2\\8&-12&9\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow R_2-\frac29R_3\text{ and }R_1\rightarrow R_1+\frac13R_3\right]$
$\displaystyle \therefore\ I_3=\begin{bmatrix}3&-4&3\\-2&3&-2\\8&-12&9\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}3&-4&3\\-2&3&-2\\8&-12&9\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\begin{bmatrix}1&3&-2\\-3&0&-1\\2&1&0\end{bmatrix}\hspace{8.0cm}[\text{CBSE 2011}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}1&3&-2\\-3&0&-1\\2&1&0\end{bmatrix}$
$\displaystyle \text{We know that }A=I_3A$
$\displaystyle \therefore\ \begin{bmatrix}1&3&-2\\-3&0&-1\\2&1&0\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&3&-2\\0&9&-7\\0&-5&4\end{bmatrix}=\begin{bmatrix}1&0&0\\3&1&0\\-2&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2+3R_1\text{ and }R_3\rightarrow R_3-2R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&3&-2\\0&1&-\frac79\\0&-5&4\end{bmatrix}=\begin{bmatrix}1&0&0\\\frac13&\frac19&0\\-2&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow\frac19R_2\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&\frac13\\0&1&-\frac79\\0&0&\frac19\end{bmatrix}=\begin{bmatrix}0&-\frac13&0\\\frac13&\frac19&0\\-\frac13&\frac59&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1-3R_2\text{ and }R_3\rightarrow R_3+5R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&\frac13\\0&1&-\frac79\\0&0&1\end{bmatrix}=\begin{bmatrix}0&-\frac13&0\\\frac13&\frac19&0\\-3&5&9\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow9R_3]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&-2&-3\\-2&4&7\\-3&5&9\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow R_2+\frac79R_3\text{ and }R_1\rightarrow R_1-\frac13R_3\right]$
$\displaystyle \therefore\ I_3=\begin{bmatrix}1&-2&-3\\-2&4&7\\-3&5&9\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}1&-2&-3\\-2&4&7\\-3&5&9\end{bmatrix}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\begin{bmatrix}-1&1&2\\1&2&3\\3&1&1\end{bmatrix}\hspace{8.0cm}[\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=\begin{bmatrix}-1&1&2\\1&2&3\\3&1&1\end{bmatrix}$
$\displaystyle \text{We know that }A=I_3A$
$\displaystyle \therefore\ \begin{bmatrix}-1&1&2\\1&2&3\\3&1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-1&-2\\1&2&3\\3&1&1\end{bmatrix}=\begin{bmatrix}-1&0&0\\0&1&0\\0&0&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow-R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-1&-2\\0&3&5\\0&4&7\end{bmatrix}=\begin{bmatrix}-1&0&0\\1&1&0\\3&0&1\end{bmatrix}A\qquad[\text{Applying }R_2\rightarrow R_2-R_1\text{ and }R_3\rightarrow R_3-3R_1]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&-1&-2\\0&1&\frac53\\0&4&7\end{bmatrix}=\begin{bmatrix}-1&0&0\\\frac13&\frac13&0\\3&0&1\end{bmatrix}A\qquad\left[\text{Applying }R_2\rightarrow\frac13R_2\right]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac13\\0&1&\frac53\\0&0&\frac13\end{bmatrix}=\begin{bmatrix}-\frac23&\frac13&0\\\frac13&\frac13&0\\\frac53&-\frac43&1\end{bmatrix}A\qquad[\text{Applying }R_1\rightarrow R_1+R_2\text{ and }R_3\rightarrow R_3-4R_2]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&-\frac13\\0&1&\frac53\\0&0&1\end{bmatrix}=\begin{bmatrix}-\frac23&\frac13&0\\\frac13&\frac13&0\\5&-4&3\end{bmatrix}A\qquad[\text{Applying }R_3\rightarrow3R_3]$
$\displaystyle \Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&-1&1\\-8&7&-5\\5&-4&3\end{bmatrix}A\qquad\left[\text{Applying }R_1\rightarrow R_1+\frac13R_3\text{ and }R_2\rightarrow R_2-\frac53R_3\right]$
$\displaystyle \therefore\ I_3=\begin{bmatrix}1&-1&1\\-8&7&-5\\5&-4&3\end{bmatrix}A$
$\displaystyle \therefore\ A^{-1}=\begin{bmatrix}1&-1&1\\-8&7&-5\\5&-4&3\end{bmatrix}$
$\displaystyle \\$