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Question
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}3 & 10 \\ 2 & 7\end{bmatrix}\]
Solution
\[A = \begin{bmatrix} 3 & 10\\2 7 \end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix} 3 & 10\\2 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix} 3 - 2 & 10 - 7\\ 2 & 7 \end{bmatrix} = \begin{bmatrix} 1 - 0 & 0 - 1\\ 0 & 1 \end{bmatrix} A [\text{ Applying }R_1 \to R_1 - R_2 ]\]
\[ \Rightarrow \begin{bmatrix} 1 & 3\\2 & 7 \end{bmatrix} = \begin{bmatrix} 1 & - 1\\0 & 1 \end{bmatrix}A\]
\[ \Rightarrow \begin{bmatrix} 1 & 3\\2 - 2 & 7 - 6 \end{bmatrix} = \begin{bmatrix} 1 & - 1\\0 - 2 &1 + 2 \end{bmatrix} [\text{ Applying }R_2 \to R_2 - 2 R_1 ]\]
\[ \Rightarrow \begin{bmatrix} 1 & 3\\0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & - 1\\ - 2 & 3 \end{bmatrix} A \]
\[ \Rightarrow \begin{bmatrix} 1 & 3 - 3\\0 & 1 \end{bmatrix} = \begin{bmatrix} 1 + 6 & - 1 - 9\\ - 2 & 3 \end{bmatrix}A [\text{ Applying }R_1 \to R_1 - 3 R_2 ]\]
\[ \Rightarrow \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & - 10\\ - 2 & 3 \end{bmatrix}A\]
\[ \Rightarrow A^{- 1} = \begin{bmatrix} 7 & - 10\\ - 2 & 3 \end{bmatrix}\]
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