જો $A=\left[\begin{array}{rrr}0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0\end{array}\right]$,$B=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0\end{array}\right]$,અને $C=\left[\begin{array}{c}2 \\ -2 \\ 3\end{array}\right]$ હોય,તો $AC$,$BC$ અને $(A + B)C$ ની ગણતરી કરો. વળી,ચકાસો કે $(A+B)C = AC + BC$.

(N/A) સૌ પ્રથમ,આપણે $A+B$ ની ગણતરી કરીએ:
$A+B = \left[\begin{array}{rrr}0+0 & 6+1 & 7+1 \\ -6+1 & 0+0 & 8+2 \\ 7+1 & -8+2 & 0+0\end{array}\right] = \left[\begin{array}{rrr}0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0\end{array}\right]$.
હવે,$(A+B)C$ ની ગણતરી કરીએ:
$(A+B)C = \left[\begin{array}{rrr}0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0\end{array}\right] \left[\begin{array}{r}2 \\ -2 \\ 3\end{array}\right] = \left[\begin{array}{r}0(2) + 7(-2) + 8(3) \\ -5(2) + 0(-2) + 10(3) \\ 8(2) + (-6)(-2) + 0(3)\end{array}\right] = \left[\begin{array}{r}0 - 14 + 24 \\ -10 + 0 + 30 \\ 16 + 12 + 0\end{array}\right] = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$.
આગળ,$AC$ ની ગણતરી કરીએ:
$AC = \left[\begin{array}{rrr}0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0\end{array}\right] \left[\begin{array}{r}2 \\ -2 \\ 3\end{array}\right] = \left[\begin{array}{r}0(2) + 6(-2) + 7(3) \\ -6(2) + 0(-2) + 8(3) \\ 7(2) + (-8)(-2) + 0(3)\end{array}\right] = \left[\begin{array}{r}0 - 12 + 21 \\ -12 + 0 + 24 \\ 14 + 16 + 0\end{array}\right] = \left[\begin{array}{r}9 \\ 12 \\ 30\end{array}\right]$.
આગળ,$BC$ ની ગણતરી કરીએ:
$BC = \left[\begin{array}{rrr}0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0\end{array}\right] \left[\begin{array}{r}2 \\ -2 \\ 3\end{array}\right] = \left[\begin{array}{r}0(2) + 1(-2) + 1(3) \\ 1(2) + 0(-2) + 2(3) \\ 1(2) + 2(-2) + 0(3)\end{array}\right] = \left[\begin{array}{r}0 - 2 + 3 \\ 2 + 0 + 6 \\ 2 - 4 + 0\end{array}\right] = \left[\begin{array}{r}1 \\ 8 \\ -2\end{array}\right]$.
છેલ્લે,$AC + BC$ ની ગણતરી કરીએ:
$AC + BC = \left[\begin{array}{r}9 \\ 12 \\ 30\end{array}\right] + \left[\begin{array}{r}1 \\ 8 \\ -2\end{array}\right] = \left[\begin{array}{r}9+1 \\ 12+8 \\ 30-2\end{array}\right] = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$.
આમ,$(A+B)C = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$ અને $AC+BC = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$ હોવાથી,આપણે ચકાસ્યું છે કે $(A+B)C = AC + BC$.