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