## Precalculus (6th Edition) Blitzer

Assuming the matrix, $A=\left[ \begin{matrix} 2 & 1 \\ 3 & 1 \\ \end{matrix} \right],B=\left[ \begin{matrix} 4 & 7 \\ 1 & 2 \\ \end{matrix} \right]$ Now, we will simplify the expression ${{\left[ A+B \right]}^{-1}}={{A}^{-1}}+{{B}^{-1}}$ as below, $A=\left[ \begin{matrix} 2 & 1 \\ 3 & 1 \\ \end{matrix} \right],B=\left[ \begin{matrix} 4 & 7 \\ 1 & 2 \\ \end{matrix} \right]$ For the expression, \begin{align} & \left[ A+B \right]=\left[ \begin{matrix} 2 & 1 \\ 3 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} 4 & 7 \\ 1 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 2+4 & 1+7 \\ 3+1 & 1+2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 6 & 8 \\ 4 & 3 \\ \end{matrix} \right] \end{align} Now, The inverse of matrix $\left[ A+B \right]$ is equal to: ${{\left[ A+B \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$ Now, compare the matrix to the original matrix. So, \begin{align} & a=6 \\ & b=8 \\ & c=4 \\ & d=3 \end{align} Now, the inverse is: ${{\left[ A+B \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$ Substitute the values to get, \begin{align} & {{\left[ A+B \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right] \\ & =\frac{1}{-6}\left[ \begin{matrix} 3 & -8 \\ -4 & 6 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -\frac{1}{2} & \frac{4}{3} \\ \frac{2}{3} & -1 \\ \end{matrix} \right] \end{align} Therefore, the inverse of the matrix $\left[ A+B \right]$ is $\left[ \begin{matrix} -\frac{1}{2} & \frac{4}{3} \\ \frac{2}{3} & -1 \\ \end{matrix} \right]$; invertible matrix. For the inverse $A$ and inverse $B$, consider the matrix, $A=\left[ \begin{matrix} 2 & 1 \\ 3 & 1 \\ \end{matrix} \right]$ $B=\left[ \begin{matrix} 4 & 7 \\ 1 & 2 \\ \end{matrix} \right]$ The inverse of the matrix is: ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$ Compare the value, to get, \begin{align} & {{A}^{-1}}={{\left[ \begin{matrix} 2 & 1 \\ 3 & 1 \\ \end{matrix} \right]}^{-1}} \\ & =\frac{1}{-1}\left[ \begin{matrix} 1 & -1 \\ -3 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & 1 \\ 3 & -2 \\ \end{matrix} \right] \end{align} \begin{align} & {{B}^{-1}}=\frac{1}{1}\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right] \end{align} Therefore the expression, \begin{align} & {{A}^{-1}}+{{B}^{-1}}=\left[ \begin{matrix} -1 & 1 \\ 3 & -2 \\ \end{matrix} \right]+\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & -6 \\ 2 & 2 \\ \end{matrix} \right] \end{align} The inverse of ${{A}^{-1}}+{{B}^{-1}}$ is $\left[ \begin{matrix} 1 & -6 \\ 2 & 2 \\ \end{matrix} \right]$ Hence, ${{\left[ A+B \right]}^{-1}}\ne {{A}^{-1}}+{{B}^{-1}}$ and $A+B$ is invertible. Hence, the statement is false.