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Question
If \[A = \begin{bmatrix}n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n\end{bmatrix} \text {and B} = \begin{bmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c^1 & c_2 & c_3\end{bmatrix}\]then AB is equal to
Options
B
nB
`B^n`
A+B
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Solution
\[Here, \]
\[A = \begin{bmatrix}n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n\end{bmatrix} \text{and B} = \begin{bmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}\]
\[ \therefore AB = \begin{bmatrix}n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n\end{bmatrix}\begin{bmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}\]
\[ \Rightarrow AB = \begin{bmatrix}n a_1 & n a_2 & n a_3 \\ n b_1 & n b_2 & n b_3 \\ n c_1 & n c_2 & n c_3\end{bmatrix}\]
\[ \Rightarrow AB = n\begin{bmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}\]
\[ \Rightarrow AB = nB\]
