Find the inverse of the following matrices by using elementry row…

Question

Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$

✓

Answer

$\text{A}=\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$We know A = IA
$\Rightarrow\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\Rightarrow\begin{bmatrix} 0 & 1 & 2 \\ -2 & 1 & 2 \\ 3 & 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
[Applying $R_2 → R_2 - R_3$]
$\Rightarrow\begin{bmatrix} -3 & 0 & 1 \\ -2 & 1 & 2 \\ 3 & 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
[Applying $R_1 → R_1 - R_3$]
$\Rightarrow\begin{bmatrix} -3 & 0 & 1 \\ -2 & 1 & 2 \\ 0 & 1 & 2 \end{bmatrix}=\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & -1 \\ 1 & 0 & 0 \end{bmatrix}\text{A}$
[Applying $R_3 → R_3 + R_1$]
$\Rightarrow\begin{bmatrix} -3 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}=\begin{bmatrix} 1 & 0 &-1 \\ -2 & 3 & -1 \\ 1 & 0 & 0 \end{bmatrix}\text{A}$
[Applying $R_2 → 3R_2 - 2R_1$]
$\Rightarrow\begin{bmatrix} -3 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & -2 & -2 \end{bmatrix}=\begin{bmatrix} 1 & 0 & -1 \\ -2 & 3 & -1 \\ 3 & -3 & 1 \end{bmatrix}\text{A}$
[Applying $R_3 → R_3 - R_2$]
$\Rightarrow\begin{bmatrix} -3 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & -1 \\ -2 & 3 & -1 \\ \frac{-3}{2} & \frac{3}{2} & \frac{-1}{2} \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_3+\text{R}_3-\frac{1}{2}\text{R}_3\Big]$
$\Rightarrow\begin{bmatrix} -3 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & -1 \\ 4 & -3 & 1 \\ \frac{-3}{2} & \frac{3}{2} & \frac{-1}{2} \end{bmatrix}\text{A}$
[Appying $R_2 → R_2 - 4R_3$]
$\Rightarrow\begin{bmatrix} -3 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & -1 \\ 4 & -3 & 1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{bmatrix}\text{A}$
[Applying $R_3 → R_3 + R_2$]
$\Rightarrow\begin{bmatrix} -3 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{3}{2} & \frac{-3}{2} & \frac{3}{2} \\ 4 & -3 & 1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{bmatrix}\text{A}$
[Applying $R_1 → R_1 - R_3$]
$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ 4 & -3 & 1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_1\rightarrow\frac{-1}{3}\text{ R}_1\Big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ -4 & -3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{bmatrix}\text{A}$
[Applying $R_2 → R_2$]