Let A and B be two square matrices of same order. If A B is a null...

We have, $\mathrm{AB}=0$ Let A=\left[$\begin{array}{ll}a & b \\ c & d\end{array}$\right] and B=\left[$\begin{array}{ll}p & q \\ r & s\end{array}$\right] So, \mathrm{AB}=\left[$\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}$\right]\left[$\begin{array}{ll}\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{s}\end{array}$\right]=\left[$\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}$\right]\Rightarrow\left[$$\begin{array}{ll} \mathrm{ap}+\mathrm{br} & \mathrm{aq}+\mathrm{bs} \\ \mathrm{cp}+\mathrm{dr} & \mathrm{cq}+\mathrm{ds} \end{array}$$\right]=\left[$\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}$\right] Comparing both sides, we have $$\begin{aligned} & \mathrm{ap}+\mathrm{br}=0 ....(i)\\ & \mathrm{aq}+\mathrm{bs}=0 ....(ii)\\ & \mathrm{cp}+\mathrm{dr}=0 ....(iii)\\ & \mathrm{cq}+\mathrm{ds}=0 ....(iv) \end{aligned}$$ Solving equation (i) and (iii), we get $(\mathrm{ad}-\mathrm{bc}) \mathrm{p}=0$ and $(\mathrm{ad}-\mathrm{bc}) \mathrm{r}=0$ Solving equation (ii) amd (iv), we get $(\mathrm{ad}-\mathrm{bc}) \mathrm{q}=0 \text{ and }(\mathrm{ad}-\mathrm{bc}) \mathrm{s}=0$ Now, If A is non-singular $\Rightarrow \mathrm{ad}-\mathrm{bc} \neq 0$. So, we get, $p=q=r=s=0$ So, B is a null matrix.