Q.Verify $A(BC)=(AB)C$ in each of the following cases. $A=\begin{bmatrix}1&0&1\\2&3&0\\0&4&5\end{bmatrix}$, $B=\begin{bmatrix}2&-2\\-1&1\\0&3\end{bmatrix}$ and $C=\begin{bmatrix}3&2&-1\\2&0&-2\end{bmatrix}$

Given $A=\begin{bmatrix}1&0&1\\2&3&0\\0&4&5\end{bmatrix}$, $B=\begin{bmatrix}2&-2\\-1&1\\0&3\end{bmatrix}$, $C=\begin{bmatrix}3&2&-1\\2&0&-2\end{bmatrix}$.

Step 1 — compute BC ($3\times2$ times $2\times3$ = $3\times3$):

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

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

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

$BC=\begin{bmatrix}2&4&2\\-1&-2&-1\\6&0&-6\end{bmatrix}$

Step 2 — compute A(BC) ($3\times3$ times $3\times3$):

Row 1: $[1(2)+0(-1)+1(6),\ 1(4)+0(-2)+1(0),\ 1(2)+0(-1)+1(-6)] = [8,\ 4,\ -4]$

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

Row 3: $[0(2)+4(-1)+5(6),\ 0(4)+4(-2)+5(0),\ 0(2)+4(-1)+5(-6)] = [26,\ -8,\ -34]$

$A(BC)=\begin{bmatrix}8&4&-4\\1&2&1\\26&-8&-34\end{bmatrix}$

Step 3 — compute AB ($3\times3$ times $3\times2$):

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

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

Row 3: $[0(2)+4(-1)+5(0),\ 0(-2)+4(1)+5(3)] = [-4,\ 19]$ …

Compute BC then A(BC); separately compute AB then (AB)C; show both ro …

• Mixing up multiplication order (computing CB instead of BC, or BA instead of AB).
• Losing track of intermediate entries in a long chain of products — write out each intermediate matrix fully before proceeding to the next multiplication. …