Mathematics - Matrices Question with Solution | TestHub

Let, the matrix $B=\left[\begin{array}{lll}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$

$\because AB=BA$

$\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]=\left[\begin{array}{lll}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\left[\begin{array}{lll}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{lll}d& e& f\\ a& b& c\\ g& h& i\end{array}\right]=\left[\begin{array}{lll}b& a& c\\ e& d& f\\ h& g& i\end{array}\right]$

On equating the matrices, we get

$\Rightarrow d=b, e=a, f=c, g=h$

$\therefore$ Matrix $B=\left[\begin{array}{lll}a& b& c\\ b& a& c\\ g& g& i\end{array}\right]$

No. of ways of selecting each of $a, b, c, g, i$ from the set $\left\{1, 2, 3, 4, 5\right\}$ are $5.$

Thus, the total number of ways are $=5\times 5\times 5\times 5\times 5$

$=5^5=3125$

$\therefore$ No. of Matrices $B=3125.$