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

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

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

Row 1: $[2(2)+4(3)+3(-1),\ 2(-2)+4(3)+3(1)] = [13,\ 11]$

Row 2: $[-1(2)+3(3)+2(-1),\ -1(-2)+3(3)+2(1)] = [5,\ 13]$

$AB=\begin{bmatrix}13&11\\5&13\end{bmatrix}$

Step 2 — compute (AB)C ($2\times2$ times $2\times2$):

Row 1: $[13(3)+11(1),\ 13(1)+11(3)] = [50,\ 46]$

Row 2: $[5(3)+13(1),\ 5(1)+13(3)] = [28,\ 44]$

$(AB)C=\begin{bmatrix}50&46\\28&44\end{bmatrix}$

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

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

Row 2: $[3(3)+3(1),\ 3(1)+3(3)] = [12,\ 12]$

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

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

• Multiplying the rectangular matrices in the wrong order — with non-square matrices, order is strictly fixed by dimension compatibility. …