Question
Using elementary row transformations, find the inverse of the matrix $\text{A}=\begin{bmatrix}1 & 2&3 \\2 & 5&7\\-2&-4&-5 \end{bmatrix}.$
✓
Answer
$A^{-1} = IA ($Inverse of matrix$)$
$\text{A}^{-1}=\begin{bmatrix}1&0&0 \\0 &1&0\\0&0&1 \end{bmatrix}\begin{bmatrix}1&2&3\\2&5&7\\-2&-4&-5 \end{bmatrix}$
$\text{R}_2\rightarrow\text{R}_2-2\text{R}_1$
$\text{R}_3\rightarrow\text{R}_3-2\text{R}_1$
$\text{A}^{-1}=\begin{bmatrix}1&0&0 \\-2 &1&0\\2&0&1 \end{bmatrix}\begin{bmatrix}1&2&3\\0&1&1\\0&0&1 \end{bmatrix}$
$\text{R}_1\rightarrow\text{R}_1-3\text{R}_3$
$\text{R}_2\rightarrow\text{R}_2-\text{R}_3$
$\text{A}^{-1}=\begin{bmatrix}-5&0&-3 \\-4 &1&-1\\2&0&1 \end{bmatrix}\begin{bmatrix}1&2&0\\0&1&0\\0&0&1 \end{bmatrix}$
$\text{R}_1\rightarrow\text{R}_1-2\text{R}_2$
$\text{A}^{-1}=\begin{bmatrix}3&-2&-1 \\-4 &1&-1\\2&0&1 \end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$
$\text{A}^{-1}=\begin{bmatrix}3&-2&-1\\-4&1&-1\\2&0&1 \end{bmatrix}$
$\text{A}^{-1}=\begin{bmatrix}1&0&0 \\0 &1&0\\0&0&1 \end{bmatrix}\begin{bmatrix}1&2&3\\2&5&7\\-2&-4&-5 \end{bmatrix}$
$\text{R}_2\rightarrow\text{R}_2-2\text{R}_1$
$\text{R}_3\rightarrow\text{R}_3-2\text{R}_1$
$\text{A}^{-1}=\begin{bmatrix}1&0&0 \\-2 &1&0\\2&0&1 \end{bmatrix}\begin{bmatrix}1&2&3\\0&1&1\\0&0&1 \end{bmatrix}$
$\text{R}_1\rightarrow\text{R}_1-3\text{R}_3$
$\text{R}_2\rightarrow\text{R}_2-\text{R}_3$
$\text{A}^{-1}=\begin{bmatrix}-5&0&-3 \\-4 &1&-1\\2&0&1 \end{bmatrix}\begin{bmatrix}1&2&0\\0&1&0\\0&0&1 \end{bmatrix}$
$\text{R}_1\rightarrow\text{R}_1-2\text{R}_2$
$\text{A}^{-1}=\begin{bmatrix}3&-2&-1 \\-4 &1&-1\\2&0&1 \end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$
$\text{A}^{-1}=\begin{bmatrix}3&-2&-1\\-4&1&-1\\2&0&1 \end{bmatrix}$
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