If $A=\left[\begin{array}{lll}n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n\end{array}\right]$ and $B=\left[\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c^{1} & c_{2} & c_{3}\end{array}\right]$, then $A B$ is equal to
(a) $B$
(b) $n B$
(c) $B^{n}$
(d) $A+B$
(b) $n B$
Here,
$A=\left[\begin{array}{lll}n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n\end{array}\right]$ and $B=\left[\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{array}\right]$
$\therefore A B=\left[\begin{array}{ccc}n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n\end{array}\right]\left[\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{lll}n a_{1} & n a_{2} & n a_{3} \\ n b_{1} & n b_{2} & n b_{3} \\ n c_{1} & n c_{2} & n c_{3}\end{array}\right]$
$\Rightarrow A B=n\left[\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{array}\right]$
$\Rightarrow A B=n B$
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