If possible, using elementary row transformations, find the inverse of the following matrices.
$⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & - 1 5 & 1 & 0 0 & 1 & 3 \end{matrix} ⎤ ⎥ ⎦$

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Solution

For getting the inverse of the given matrix a by row elementary operations we may write the given matrix as
$∴ ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & - 1 5 & 1 & 0 0 & 1 & 3 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ A \Rightarrow ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & - 1 3 & 1 & 1 0 & 1 & 3 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 1 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ A \Rightarrow ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & - 1 1 & 1 & 2 2 & 1 & 2 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 - 2 & 1 & 0 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ A \Rightarrow ⎡ ⎢ ⎢ ⎣ \begin{matrix} 2 & 0 & - 1 0 & 1 & \frac{5}{2} 4 & 1 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 \frac{- 2}{2} & 1 & 0 2 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦ A \Rightarrow ⎡ ⎢ ⎢ ⎣ \begin{matrix} 2 & 0 & - 1 0 & 1 & \frac{5}{2} 0 & 1 & 3 \end{matrix} ⎤ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 \frac{- 5}{2} & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦ A \Rightarrow ⎡ ⎢ ⎢ ⎣ \begin{matrix} 2 & 0 & - 1 0 & 1 & \frac{5}{2} 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 \frac{- 5}{2} & 1 & 0 \frac{5}{2} & - 1 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ A \Rightarrow ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & \frac{- 1}{2} 0 & 1 & \frac{5}{2} 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{2} & 0 & 0 \frac{- 5}{2} & 1 & 0 5 & - 2 & 2 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ A \Rightarrow ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 1 & 1 - 15 & 6 & - 5 5 & - 2 & 2 \end{matrix} ⎤ ⎥ ⎦ A$
Hence, $⎡ ⎢ ⎣ \begin{matrix} 3 & - 1 & 1 - 15 & 6 & - 5 5 & - 2 & 2 \end{matrix} ⎤ ⎥ ⎦$ is the inverse of given matrix A.

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