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Question
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}7 & 1 \\ 4 & - 3\end{bmatrix}\]
Solution
\[A = \begin{bmatrix} 7 & 1\\4 & - 3 \end{bmatrix}\]
We know
\[A = IA\]
\[ \Rightarrow \begin{bmatrix} 7 & 1\\4 & - 3 \end{bmatrix} = \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix} 1 & \frac{1}{7}\\ 4 & - 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{7} & 0\\ 0 & 1 \end{bmatrix}A \left(\text{ Applying }R_1 \to \frac{1}{7} R_1 \right)\]
\[ \Rightarrow \begin{bmatrix} 1 & \frac{1}{7}\\ 0 & - \frac{25}{7} \end{bmatrix} = \begin{bmatrix} \frac{1}{7} & 0\\ - \frac{4}{7} & 1 \end {bmatrix} A \left(\text{ Applying }R_2 \to R_2 - 4 R_1 \right)\]
\[ \Rightarrow \begin{bmatrix} 1 & \frac{1}{7}\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{7} & 0\\\frac{4}{25} & - \frac{7}{25} \end{bmatrix} A \left(\text{ Applying }R_2 \to - \frac{7}{25} R_2 \right)\]
\[ \Rightarrow \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{3}{25} & \frac{1}{25}\\\frac{4}{25} & - \frac{7}{25} \end{bmatrix} A \left(\text{ Applying }R_1 \to R_1 - \frac{1}{7} R_2 \right)\]
\[ \therefore A^{- 1} = \frac{1}{25}\begin{bmatrix} 3 & 1\\4 & - 7 \end{bmatrix}\]
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