Question
Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix}3 & 10 \\ 2 & 7 \end{bmatrix}$
✓
Answer
$\text{A}=\begin{bmatrix}3 & 10 \\ 2 & 7 \end{bmatrix}$Now, A = IA
$\begin{bmatrix}3 & 10 \\ 2 & 7 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\frac{1}{3}\text{ R}_1$
$\begin{bmatrix} 1 & \frac{10}{7} \\ 2 & 7 \end{bmatrix}=\begin{bmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
Applying $R_2 \rightarrow R_2 - 2R_1$
$\begin{bmatrix} 1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix}\frac{1}{3} & 0 \\ \frac{-2}{3} & 1 \end{bmatrix}\text{A}$
Applying $R_2 \rightarrow 3R_2$
$\begin{bmatrix}1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix}\frac{1}{3} & 0 \\ -2 & 3 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\text{R}_1-\frac{10}{3}\text{ R}_3$
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\text{A}$
$I = BA$
Hence, B is the inverse of A.
$\begin{bmatrix}3 & 10 \\ 2 & 7 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\frac{1}{3}\text{ R}_1$
$\begin{bmatrix} 1 & \frac{10}{7} \\ 2 & 7 \end{bmatrix}=\begin{bmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
Applying $R_2 \rightarrow R_2 - 2R_1$
$\begin{bmatrix} 1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix}\frac{1}{3} & 0 \\ \frac{-2}{3} & 1 \end{bmatrix}\text{A}$
Applying $R_2 \rightarrow 3R_2$
$\begin{bmatrix}1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix}\frac{1}{3} & 0 \\ -2 & 3 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\text{R}_1-\frac{10}{3}\text{ R}_3$
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\text{A}$
$I = BA$
Hence, B is the inverse of A.
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