Q.Verify that A(B+C)=AB+BC [printed verbatim; note: the correct left-distributive identity is $A(B+C)=AB+AC$ — see solution] in each of the following matrices $A=\begin{bmatrix}4&-2\\2&3\end{bmatrix}$, $B=\begin{bmatrix}-1&1\\3&-2\end{bmatrix}$ and $C=\begin{bmatrix}4&1\\2&-1\end{bmatrix}$

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

Left side — A(B+C):

$B+C=\begin{bmatrix}-1+4&1+1\\3+2&-2-1\end{bmatrix}=\begin{bmatrix}3&2\\5&-3\end{bmatrix}$

$A(B+C)=\begin{bmatrix}4&-2\\2&3\end{bmatrix}\begin{bmatrix}3&2\\5&-3\end{bmatrix}$

Row 1: $[4(3)+(-2)(5),\ 4(2)+(-2)(-3)]=[2,\ 14]$

Row 2: $[2(3)+3(5),\ 2(2)+3(-3)]=[21,\ -5]$

$A(B+C)=\begin{bmatrix}2&14\\21&-5\end{bmatrix}$

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

$AB=\begin{bmatrix}4(-1)+(-2)(3)&4(1)+(-2)(-2)\\2(-1)+3(3)&2(1)+3(-2)\end{bmatrix}=\begin{bmatrix}-10&8\\7&-4\end{bmatrix}$

$AC=\begin{bmatrix}4(4)+(-2)(2)&4(1)+(-2)(-1)\\2(4)+3(2)&2(1)+3(-1)\end{bmatrix}=\begin{bmatrix}12&6\\14&-1\end{bmatrix}$

$AB+AC=\begin{bmatrix}-10+12&8+6\\7+14&-4-1\end{bmatrix}=\begin{bmatrix}2&14\\21&-5\end{bmatrix}$

This matches A(B+C) exactly, confirming $A(B+C)=AB+AC$.

Checking the literally printed 'AB+BC': …

Compute B+C, then A(B+C); separately compute AB and AC and add; the printed 'AB+BC …

• Copying the book's printed 'AB+BC' literally without checking it against A(B+C) numerically — always verify with AB+AC for the left-distributive law.
• Sign errors when adding B and C entrywise before multiplying by A. …