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Cosine Rule

If A=n000n000...

Question

If $A = [\begin{matrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{matrix}] and B = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c^{1} & c_{2} & c_{3} \end{matrix}]$ , then AB is equal to
(a) B
(b) nB
(c) Bⁿ
(d) A + B

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Solution

(b) nB

$Here, A = [\begin{matrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{matrix}] and B = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}] ∴ A B = [\begin{matrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{matrix}] [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}] \Rightarrow A B = [\begin{matrix} n a_{1} & n a_{2} & n a_{3} \\ n b_{1} & n b_{2} & n b_{3} \\ n c_{1} & n c_{2} & n c_{3} \end{matrix}] \Rightarrow A B = n [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}] \Rightarrow A B = n B$

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