Showing That the Identity Matrix Acts as a 1

Given matrix A, show that $A I = I A = A .$

A = [\begin{matrix} 3 & 4 \\ −2 & 5 \end{matrix}]

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Use matrix multiplication to show that the product of $A$ and the identity is equal to the product of the identity and A.

$A I = [\begin{array}{r} 3 & 4 \\ −2 & 5 \end{array}] \begin{array}{r} \end{array} [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{r} 3 \cdot 1 + 4 \cdot 0 & 3 \cdot 0 + 4 \cdot 1 \\ −2 \cdot 1 + 5 \cdot 0 & −2 \cdot 0 + 5 \cdot 1 \end{array}] = [\begin{array}{r} 3 & 4 \\ −2 & 5 \end{array}]$
$I A = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] \begin{array}{r} \end{array} [\begin{array}{r} 3 & 4 \\ −2 & 5 \end{array}] = [\begin{array}{r} 1 \cdot 3 + 0 \cdot (−2) & 1 \cdot 4 + 0 \cdot 5 \\ 0 \cdot 3 + 1 \cdot (−2) & 0 \cdot 4 + 1 \cdot 5 \end{array}] = [\begin{array}{r} 3 & 4 \\ −2 & 5 \end{array}]$

How did it go?