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Question
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 6 \\ - 3 & 5\end{bmatrix}\]
Solution
\[A = \begin{bmatrix} 1 & 6\\ - 3 & 5 \end{bmatrix}\]
We know
\[A = IA\]
\[ \Rightarrow \begin{bmatrix} 1 & 6\\ - 3 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}A\]
\[ \Rightarrow \begin{bmatrix} 1 & 6\\ - 3 + 3 & 5 + 18 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 + 3 & 1 + 0 \end{bmatrix}A [\text{ Applying }R_2 \to R_2 + 3 R_1 ]\]
\[ \Rightarrow \begin{bmatrix} 1 & 6\\ 0 & 23 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 3 & 1 \end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix} 1 & 6 - 6\\ 0 & 23 \end{bmatrix} = \begin{bmatrix} 1 - \frac{18}{23} & 0 - \frac{6}{23}\\ 3 & 1 \end{bmatrix}A [\text{ Applying }R_1 \to R_1 - \frac{6}{23} R_2 ]\]
\[ \Rightarrow \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{5}{23} & \frac{- 6}{23}\\ \frac{3}{23} & \frac{1}{23} \end{bmatrix}A [\text{ Applying }R_2 \to \frac{1}{23} R_2 ]\]
\[ \Rightarrow A^{- 1} = \begin{bmatrix} \frac{5}{23} & \frac{- 6}{23}\\ \frac{3}{23} & \frac{1}{23} \end{bmatrix} = \frac{1}{23}\begin{bmatrix} 5 & - 6\\ 3 & 1 \end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{23}\begin{bmatrix} 5 & - 6\\ 3 & 1 \end{bmatrix}\]
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