Q.If $A=\begin{bmatrix}-1&1&1\\2&3&0\\1&-3&1\end{bmatrix}$, $B=\begin{bmatrix}2&1&4\\3&0&2\\1&2&1\end{bmatrix}$. State whether AB=BA? Justify your answer.

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

Compute AB (row of A times column of B):

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

Row 2: $[2(2)+3(3)+0(1),\ 2(1)+3(0)+0(2),\ 2(4)+3(2)+0(1)] = [13,\ 2,\ 14]$

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

$AB=\begin{bmatrix}2&1&-1\\13&2&14\\-6&3&-1\end{bmatrix}$

Compute BA (row of B times column of A):

Row 1: $[2(-1)+1(2)+4(1),\ 2(1)+1(3)+4(-3),\ 2(1)+1(0)+4(1)] = [4,\ -7,\ 6]$

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

Compute AB and BA (both 3×3) using row-by-column multiplication …

• Assuming AB=BA is guaranteed for two square matrices of the same order — it is not, in general.
• A single arithmetic slip that goes unnoticed because both matrices are 3×3 and 'look' similar. …