Question

Let $A=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$. Then the number of $3\times 3$ matrices $B$ with entries from the set $\{1,2,3,4,5\}$ and satisfying $AB=BA$ is_______.

Answer 1

Answer: 3125

Let matrix $B=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$
$\because AB=BA$
$\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$
$\left[\begin{array}{ccc}d& e& f\\ a& b& c\\ g& h& i\end{array}\right]=\left[\begin{array}{ccc}b& a& c\\ e& d& f\\ h& g& i\end{array}\right]$
$\Rightarrow d=b,e=a,f=c,g=h$
$\therefore$ Matrix $B=\left[\begin{array}{ccc}a& b& c\\ b& a& c\\ g& g& i\end{array}\right]$
No. of ways of selecting $a,b,c,g,i$
$=5\times 5\times 5\times 5\times 5$
$=5^5=3125$
$\therefore$ No. of Matrices $B=3125$