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અહી $A=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] $ છે. તો શ્રેણિક $\mathrm{B}$ કે જેની કક્ષા $3 \times 3$ હોય અને તેના ઘટકો ગણ $\{1,2,3,4,,5\}$ માંથી હોય અને જે $A B=B A$ નું સમાધાન કરે તેવા શ્રેણીકની સંખ્યા મેળવો.
$3500$
$3125$
$4500$
$6000$
Solution
Let matrix $B=\left[\begin{array}{lll}a & b & c \\ d & e & f \\ g & n & i\end{array}\right]$
$\therefore A B=B A$
$\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right]=\left[\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right]\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\left[\begin{array}{lll}d & e & f \\ a & b & c \\ g & h & i\end{array}\right]=\left[\begin{array}{lll}b & a & c \\ e & d & f \\ h & g & i\end{array}\right]$
$\Rightarrow d=b, e=a, f=c, g=h$
$\therefore$ Matrix $B=\left[\begin{array}{lll}a & b & c \\ b & a & c \\ g & g & i\end{array}\right]$
No. of ways of selecting $a, b, c, g$,
$\mathrm{i}=5 \times 5 \times 5 \times 5 \times 5$
$=5^{5}=3125$
$\therefore$ No. of Matrices $B=3125$
