Q.$A=\begin{bmatrix}1&-1&3\\2&3&2\end{bmatrix}$, $B=\begin{bmatrix}1&0\\-2&3\\4&3\end{bmatrix}$ and $C=\begin{bmatrix}1&2\\-2&0\\4&-3\end{bmatrix}$.

Given $A=\begin{bmatrix}1&-1&3\\2&3&2\end{bmatrix}$, $B=\begin{bmatrix}1&0\\-2&3\\4&3\end{bmatrix}$, $C=\begin{bmatrix}1&2\\-2&0\\4&-3\end{bmatrix}$.

Left side — A(B+C):

$B+C=\begin{bmatrix}1+1&0+2\\-2-2&3+0\\4+4&3-3\end{bmatrix}=\begin{bmatrix}2&2\\-4&3\\8&0\end{bmatrix}$

$A(B+C)$: Row 1: $[1(2)+(-1)(-4)+3(8),\ 1(2)+(-1)(3)+3(0)]=[30,\ -1]$

Row 2: $[2(2)+3(-4)+2(8),\ 2(2)+3(3)+2(0)]=[8,\ 13]$

$A(B+C)=\begin{bmatrix}30&-1\\8&13\end{bmatrix}$

Right side — AB+AC (the correct left-distributive law):

$AB$: Row 1: $[1(1)+(-1)(-2)+3(4),\ 1(0)+(-1)(3)+3(3)]=[15,\ 6]$

Row 2: $[2(1)+3(-2)+2(4),\ 2(0)+3(3)+2(3)]=[4,\ 15]$

$AB=\begin{bmatrix}15&6\\4&15\end{bmatrix}$

$AC$: Row 1: $[1(1)+(-1)(-2)+3(4),\ 1(2)+(-1)(0)+3(-3)]=[15,\ -7]$

Row 2: $[2(1)+3(-2)+2(4),\ 2(2)+3(0)+2(-3)]=[4,\ -2]$

$AC=\begin{bmatrix}15&-7\\4&-2\end{bmatrix}$

$AB+AC=\begin{bmatrix}15+15&6-7\\4+4&15-2\end{bmatrix}=\begin{bmatrix}30&-1\\8&13\end{bmatrix}$

This matches A(B+C) exactly. …

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