## Precalculus (6th Edition) Blitzer

The inverse of the matrix $A$ is, ${{A}^{-1}}=\left[ \begin{matrix} 3 & 1 \\ 2 & 1 \\ \end{matrix} \right]$ and $A{{A}^{-1}}={{A}^{-1}}A=I$
At first we will find, \begin{align} & \det A=3\left( 1 \right)-1\left( 2 \right) \\ & =3-2 \\ & =1 \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 \\ 2 & 1 \\ \end{matrix} \right] \\ & =\frac{1}{1}\left[ \begin{matrix} 3 & 1 \\ 2 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3 & 1 \\ 2 & 1 \\ \end{matrix} \right] \end{align} Now we will consider, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 1 & -1 \\ -2 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 1 \\ 2 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\left( 3 \right)-1\left( 2 \right) & 1\left( 1 \right)-1\left( 1 \right) \\ -2\left( 3 \right)+3\left( 2 \right) & -2\left( 1 \right)+3\left( 1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3-2 & 1-1 \\ -6+6 & -2+3 \\ \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} 3 & 1 \\ 2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -1 \\ -2 & 3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3\left( 1 \right)+1\left( -2 \right) & 3\left( -1 \right)+1\left( 3 \right) \\ 2\left( 1 \right)+1\left( -2 \right) & 2\left( -1 \right)+1\left( 3 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3-2 & -3+3 \\ 2-2 & -2+3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \end{align} Clearly, $A{{A}^{-1}}={{A}^{-1}}A=I$ Hence, ${{A}^{-1}}=\left[ \begin{matrix} 3 & 1 \\ 2 & 1 \\ \end{matrix} \right]$.