## Precalculus (6th Edition) Blitzer

$\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{21}} & {{a}_{22}} \\ \end{matrix} \right]$ Nothing happens to the element in the first matrix.
We show that $AI=A$. Consider the matrix: $\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{12}} & {{a}_{22}} \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ Let the provided matrix be denoted by, \begin{align} & A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{12}} & {{a}_{22}} \\ \end{matrix} \right] \\ & I=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \end{align} Now, compute the matrix as $B=A\cdot I$ \begin{align} & A\cdot I=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{21}} & {{a}_{22}} \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} {{a}_{11}}\times 1+{{a}_{12}}\times 0 & {{a}_{11}}\times 0+{{a}_{12}}\times 1 \\ {{a}_{21}}\times 1+{{a}_{22}}\times 0 & {{a}_{21}}\times 0+{{a}_{22}}\times 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{21}} & {{a}_{22}} \\ \end{matrix} \right] \\ & =\left[ A \right] \end{align} Thus, the element is the first matrix remains the same.