Q.Show that AB=BA where, $A=\begin{bmatrix}-2&3&-1\\-1&2&-1\\-6&9&-4\end{bmatrix}$, $B=\begin{bmatrix}1&3&-1\\2&2&-1\\3&0&-1\end{bmatrix}$

Given $A=\begin{bmatrix}-2&3&-1\\-1&2&-1\\-6&9&-4\end{bmatrix}$, $B=\begin{bmatrix}1&3&-1\\2&2&-1\\3&0&-1\end{bmatrix}$.

Compute AB:

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

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

Row 3: $[-6(1)+9(2)+(-4)(3),\ -6(3)+9(2)+(-4)(0),\ -6(-1)+9(-1)+(-4)(-1)] = [0,\ 0,\ 1]$

$AB=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I$

Compute BA:

Row 1: $[1(-2)+3(-1)+(-1)(-6),\ 1(3)+3(2)+(-1)(9),\ 1(-1)+3(-1)+(-1)(-4)] = [1,\ 0,\ 0]$ …

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

• Stopping after computing only AB and assuming BA will automatically match without actually checking.
• Sign errors while multiplying the many negative entries in these matrices. …