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Question
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}\]
Solution
\[A = \begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{- 1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_1 \to \frac{1}{2} R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{- 1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ \frac{- 5}{2} & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 - 5 R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{- 1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2}\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ \frac{- 5}{2} & 1 & 0 \\ \frac{5}{2} & - 1 & 1\end{bmatrix} A \left[\text{ Applying }R_3 \to R_3 - R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{- 1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ - \frac{5}{2} & 1 & 0 \\ 5 & - 2 & 2\end{bmatrix} A \left[\text{ Applying }R_3 \to 2 R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix} A \left[\text{ Applying }R_1 \to R_1 + \frac{1}{2} R_3\text{ and }R_2 \to R_2 - \frac{5}{2} R_3 \right]\]
\[ \Rightarrow A^{- 1} = \begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix}\]
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