Answer
$\left[ \begin{matrix}
{{a}_{11}} & {{a}_{12}} \\
{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right]$
Nothing happens to the element in the first matrix.
Work Step by Step
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.
