If A=left[begin{array}{rrr}0 & 6 & 7 -6 & 0 | vedclass

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(N/A) First,we calculate $A+B$: $A+B = \left[\begin{array}{rrr}0+0 & 6+1 & 7+1 \ -6+1 & 0+0 & 8+2 \ 7+1 & -8+2 & 0+0\end{array}ight] = \left[\begi

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(N/A) First,we calculate $A+B$:
$A+B = \left[\begin{array}{rrr}0+0 & 6+1 & 7+1 \\ -6+1 & 0+0 & 8+2 \\ 7+1 & -8+2 & 0+0\end{array}\right] = \left[\begin{array}{rrr}0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0\end{array}\right]$.
Now,calculate $(A+B)C$:
$(A+B)C = \left[\begin{array}{rrr}0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0\end{array}\right] \left[\begin{array}{r}2 \\ -2 \\ 3\end{array}\right] = \left[\begin{array}{r}0(2) + 7(-2) + 8(3) \\ -5(2) + 0(-2) + 10(3) \\ 8(2) + (-6)(-2) + 0(3)\end{array}\right] = \left[\begin{array}{r}0 - 14 + 24 \\ -10 + 0 + 30 \\ 16 + 12 + 0\end{array}\right] = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$.
Next,calculate $AC$:
$AC = \left[\begin{array}{rrr}0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0\end{array}\right] \left[\begin{array}{r}2 \\ -2 \\ 3\end{array}\right] = \left[\begin{array}{r}0(2) + 6(-2) + 7(3) \\ -6(2) + 0(-2) + 8(3) \\ 7(2) + (-8)(-2) + 0(3)\end{array}\right] = \left[\begin{array}{r}0 - 12 + 21 \\ -12 + 0 + 24 \\ 14 + 16 + 0\end{array}\right] = \left[\begin{array}{r}9 \\ 12 \\ 30\end{array}\right]$.
Next,calculate $BC$:
$BC = \left[\begin{array}{rrr}0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0\end{array}\right] \left[\begin{array}{r}2 \\ -2 \\ 3\end{array}\right] = \left[\begin{array}{r}0(2) + 1(-2) + 1(3) \\ 1(2) + 0(-2) + 2(3) \\ 1(2) + 2(-2) + 0(3)\end{array}\right] = \left[\begin{array}{r}0 - 2 + 3 \\ 2 + 0 + 6 \\ 2 - 4 + 0\end{array}\right] = \left[\begin{array}{r}1 \\ 8 \\ -2\end{array}\right]$.
Finally,calculate $AC + BC$:
$AC + BC = \left[\begin{array}{r}9 \\ 12 \\ 30\end{array}\right] + \left[\begin{array}{r}1 \\ 8 \\ -2\end{array}\right] = \left[\begin{array}{r}9+1 \\ 12+8 \\ 30-2\end{array}\right] = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$.
Since $(A+B)C = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$ and $AC+BC = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$,we have verified that $(A+B)C = AC + BC$.

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