## Precalculus (6th Edition) Blitzer

The inverse of the matrix $A$ is, ${{A}^{-1}}=\left[ \begin{matrix} -\frac{3}{5} & \frac{1}{5} \\ 1 & 0 \\ \end{matrix} \right]$
At first we will find, \begin{align} & \det A=0\left( 3 \right)-1\left( 5 \right) \\ & =0-5 \\ & =-5 \end{align} The inverse of the matrix $A$ is calculated as below: \begin{align} & {{A}^{-1}}=\frac{1}{\det A}\left[ \begin{matrix} 3 & -1 \\ -5 & 0 \\ \end{matrix} \right] \\ & =\frac{1}{-5}\left[ \begin{matrix} 3 & -1 \\ -5 & 0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -\frac{3}{5} & \frac{1}{5} \\ 1 & 0 \\ \end{matrix} \right] \end{align} Now we will consider, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 0 & 1 \\ 5 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} -\frac{3}{5} & \frac{1}{5} \\ 1 & 0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0\left( -\frac{3}{5} \right)+1 & 0\left( \frac{1}{5} \right)+1\left( 0 \right) \\ 5\left( -\frac{3}{5} \right)+3\left( 1 \right) & 5\left( \frac{1}{5} \right)+3\left( 0 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \end{align} Next we will consider, \begin{align} & {{A}^{-1}}A=\left[ \begin{matrix} -\frac{3}{5} & \frac{1}{5} \\ 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 1 \\ 5 & 3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -\frac{3}{5}\left( 0 \right)+\frac{1}{5}\left( 5 \right) & -\frac{3}{5}\left( 1 \right)+\frac{1}{5}\left( 3 \right) \\ 1\left( 0 \right)+0\left( 5 \right) & 1\left( 1 \right)+0\left( 3 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \end{align} Clearly, $A{{A}^{-1}}={{A}^{-1}}A=I$ Hence, ${{A}^{-1}}=\left[ \begin{matrix} -\frac{3}{5} & \frac{1}{5} \\ 1 & 0 \\ \end{matrix} \right]$.