Q.If $A=\begin{bmatrix}1&2\\3&5\end{bmatrix}$, $B=\begin{bmatrix}0&4\\2&-1\end{bmatrix}$, show that $AB \neq BA$, but $|AB|=|A|.|B|$

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

Compute AB:

$(AB)_{11}=1(0)+2(2)=4$

$(AB)_{12}=1(4)+2(-1)=4-2=2$

$(AB)_{21}=3(0)+5(2)=10$

$(AB)_{22}=3(4)+5(-1)=12-5=7$

$AB=\begin{bmatrix}4&2\\10&7\end{bmatrix}$

Compute BA:

$(BA)_{11}=0(1)+4(3)=12$

$(BA)_{12}=0(2)+4(5)=20$

$(BA)_{21}=2(1)+(-1)(3)=2-3=-1$

$(BA)_{22}=2(2)+(-1)(5)=4-5=-1$

$BA=\begin{bmatrix}12&20\\-1&-1\end{bmatrix}$

Since $\begin{bmatrix}4&2\\10&7\end{bmatrix}\neq\begin{bmatrix}12&20\\-1&-1\end{bmatrix}$, $AB\neq BA$.

Now check the determinant relation:

$|A|=1(5)-2(3)=5-6=-1$

$|B|=0(-1)-4(2)=0-8=-8$ …

Compute AB and BA to show they differ, then separately compute |A|, |B|, …

• Assuming that because $AB\neq BA$, their determinants must also differ — determinant multiplicativity $|AB|=|A||B|$ (and by symmetry $|BA|=|B||A|$) holds regardless of whether AB=BA. …