Question
Find the inverse of the following matrices by using elementry row transformation:
$\begin{bmatrix} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{bmatrix}$
$\begin{bmatrix} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{bmatrix}$
Here, A = IA
$\begin{bmatrix} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\frac{1}{2}\text{ R}_1$
$\begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2}\\ 2 & 4 & 1 \\ 3 & -7 & 2 \end{bmatrix}=\begin{bmatrix}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_2\rightarrow\text{R}_2-2\text{R}_1,\text{R}_3\rightarrow\text{R}_3-3\text{R}_1$
$\begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2}\\ 0 & 1 & 0 \\ 0 & \frac{5}{2} & \frac{1}{2} \end{bmatrix}=\begin{bmatrix}\frac{1}{2} & 0 & 0\\-1 & 1 & 0 \\ \frac{-3}{2} & 0 & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\text{R}_1-\frac{3}{2}\text{R}_2,\text{R}_3\rightarrow\text{R}_3-\frac{5}{2}\text{R}_2$
$\begin{bmatrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{bmatrix}=\begin{bmatrix} 2 & \frac{-3}{2} & 0 \\ -1 & 1 & 0 \\ 1 & \frac{-5}{2} & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_3\rightarrow 2\text{R}_3$
$\begin{bmatrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 2 & \frac{-3}{2} & 0 \\ -1 & 1 & 0 \\ 2 & -5 & 2 \end{bmatrix}\text{A}$
$\text{Applying R}_3\rightarrow\text{R}_1-\frac{1}{2}\text{R}_3$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 & -1 \\ -1 & 1 & 0 \\ 2 & -5 & 2 \end{bmatrix}\text{A}$
Hence, $\text{A}^{-1}=\begin{bmatrix} 1 & 1 & -1 \\ -1 & 1 & 0 \\ 2 & -5 & 2 \end{bmatrix}$
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