# Solve this

Question:

If $A=\left[\begin{array}{ccc}0 & 2 & 0 \\ 0 & 0 & 3 \\ -2 & 2 & 0\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & -4 & 0\end{array}\right]$, then $(A B)_{33}=$______

Solution:

The given matrices are $A=\left[\begin{array}{ccc}0 & 2 & 0 \\ 0 & 0 & 3 \\ -2 & 2 & 0\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & -4 & 0\end{array}\right]$.

$\therefore A B=\left[\begin{array}{ccc}0 & 2 & 0 \\ 0 & 0 & 3 \\ -2 & 2 & 0\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & -4 & 0\end{array}\right]$

$=\left[\begin{array}{ccc}0 \times 1+2 \times 3+0 \times 5 & 0 \times 2+2 \times 4+0 \times(-4) & 0 \times 3+2 \times 5+0 \times 0 \\ 0 \times 1+0 \times 3+3 \times 5 & 0 \times 2+0 \times 4+3 \times(-4) & 0 \times 3+0 \times 5+3 \times 0 \\ (-2) \times 1+2 \times 3+0 \times 5 & (-2) \times 2+2 \times 4+0 \times(-4) & (-2) \times 3+2 \times 5+0 \times 0\end{array}\right]$

$=\left[\begin{array}{ccc}6 & 8 & 10 \\ 15 & -12 & 0 \\ 4 & 4 & 4\end{array}\right]$

So, $(A B)_{3 \times 3}=\left[\begin{array}{ccc}6 & 8 & 10 \\ 15 & -12 & 0 \\ 4 & 4 & 4\end{array}\right]$

If $A=\left[\begin{array}{ccc}0 & 2 & 0 \\ 0 & 0 & 3 \\ -2 & 2 & 0\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & -4 & 0\end{array}\right]$, then $(A B)_{3 \times 3}=\left[\begin{array}{ccc}6 & 8 & 10 \\ 15 & -12 & 0 \\ 4 & 4 & 4\end{array}\right]$