$A B=\left[\begin{array}{ll}{a} & {2 b} \\ {3 a} & {4 b}\end{array}\right]$
$B A=\left[\begin{array}{ll}{a} & {0} \\ {0} & {b}\end{array}\right]\left[\begin{array}{ll}{1} & {2} \\ {3} & {4}\end{array}\right]$$=\left[\begin{array}{ll}{a} & {2 a} \\ {3 b} & {4 b}\end{array}\right]$
Hence, $A B=B A$ only when $a=b$
$\therefore$ There can be infinitely many $B^{\prime} s$
for which $A B=B A$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$T_p=\left\{A=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{a}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1, \ldots ., \mathrm{p}-1\}\right\}$
$1.$ The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and $\operatorname{det}(\mathrm{A})$ divisible by $\mathrm{p}$ is
$(A)$ $(\mathrm{p}-1)^2$ $(B)$ $2(\mathrm{p}-1)$
$(C)$ $(\mathrm{p}-1)^2+1$ $(D)$ $2 \mathrm{p}-1$
$2.$ The number of $A$ in $T_p$ such that the trace of $A$ is not divisible by $p$ but det $(A)$ is divisible by $p$ is [Note: The trace of a matrix is the sum of its diagonal entries.]
$(A)$ $(\mathrm{p}-1)\left(\mathrm{p}^2-\mathrm{p}+1\right)$ $(B)$ $\mathrm{p}^3-(\mathrm{p}-1)^2$
$(C)$ $(\mathrm{p}-1)^2$ $(D)$ $(p-1)\left(p^2-2\right)$
$3.$ The number of $A$ in $T_p$ such that det $(A)$ is not divisible by $p$ is
$(A)$ $2 \mathrm{p}^2$ $(B)$ $p^3-5 p$ $(C)$ $p^3-3 p$ $(D)$ $p^3-p^2$
Give the answer question $1,2$ and $3.$
$f (\theta)=\left|\begin{array}{ccc}-\sin ^{2} \theta & -1-\sin ^{2} \theta & 1 \\ -\cos ^{2} \theta & -1-\cos ^{2} \theta & 1 \\ 12 & 10 & -2\end{array}\right|$ are $m$ and $M$ respectively, then the ordered pair $( m , M )$ is equal to