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Question
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & 5 \\ 1 & 3\end{bmatrix}\]
Solution
\[A = \begin{bmatrix} 2 & 5\\1 & 3 \end{bmatrix}\]
We know
\[A = I A\]
\[ \Rightarrow \begin{bmatrix} 2 & 5\\1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix}A\]
\[ \Rightarrow \begin{bmatrix} 2 - 1 & 5 - 3\\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 1 - 0 & 0 - 1 \\ 0 & 1 \end{bmatrix}A [\text{ Applying }R_1 \to R_1 - R_2 ]\]
\[ \Rightarrow \begin{bmatrix} 1 & 2\\1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & - 1\\0 & 1 \end{bmatrix}A\]
\[ \Rightarrow \begin{bmatrix} 1 & 2\\1 - 1 & 3 - 2 \end{bmatrix} = \begin{bmatrix} 1 & - 1\\0 - 1 & 1 + 1 \end{bmatrix}A [\text{ Applying }R_2 \to R_2 - R_1 ]\]
\[ \Rightarrow \begin{bmatrix} 1 & 2\\0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & - 1 \\ - 1 & 2 \end{bmatrix}A\]
\[ \Rightarrow \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} = \begin{bmatrix} 1 + 2 & - 1 - 4 \\ - 1 & 2 \end{bmatrix}A [\text{ Applying }R_1 \to R_1 - 2 R_2 ]\]
\[ \Rightarrow \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & - 5\\ - 1 & 2 \end{bmatrix}A\]
\[ \Rightarrow A^{- 1} = \begin{bmatrix} 3 & - 5\\ - 1 & 2 \end{bmatrix}\]
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