Determinants Matrices Question 103
Question: Let $ A=[ \begin{aligned} & \begin{matrix} 5 & 6 & 1 \\ \end{matrix} \\ & \begin{matrix} 2 & -1 & 5 \\ \end{matrix} \\ \end{aligned} ] $ . Let there exist a matrix B such that $ AB= \begin{bmatrix} 35 & 49 \\ 29 & 13 \\ \end{bmatrix} $ . What is B equal to-
Options:
A) $ [ \begin{aligned} & \begin{matrix} 5 & 1 & 4 \\ \end{matrix} \\ & \begin{matrix} 2 & 6 & 3 \\ \end{matrix} \\ \end{aligned} ] $
B) $ [ \begin{aligned} & \begin{matrix} 2 & 6 & 3 \\ \end{matrix} \\ & \begin{matrix} 5 & 1 & 4 \\ \end{matrix} \\ \end{aligned} ] $
C) $ [ \begin{aligned} & \begin{matrix} 5 & 2 \\ \end{matrix} \\ & \begin{matrix} 1 & 6 \\ \end{matrix} \\ & \begin{matrix} 4 & 3 \\ \end{matrix} \\ \end{aligned} ] $
D) $ [ \begin{aligned} & \begin{matrix} 2 & 5 \\ \end{matrix} \\ & \begin{matrix} 6 & 1 \\ \end{matrix} \\ & \begin{matrix} 3 & 4 \\ \end{matrix} \\ \end{aligned} ] $
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Answer:
Correct Answer: C
Solution:
- [c] $ A= \begin{bmatrix} 5 & 6 & 1 \\ 2 & -1 & 5 \\ \end{bmatrix} $ and let $ B= \begin{bmatrix} 5 & 2 \\ 1 & 6 \\ 4 & 3 \\ \end{bmatrix} $
$ \therefore AB= \begin{bmatrix} 5 & 6 & 1 \\ 2 & -1 & 5 \\ \end{bmatrix} \begin{bmatrix} 5 & 2 \\ 1 & 6 \\ 4 & 3 \\ \end{bmatrix} $
$ = \begin{bmatrix} 25+6+4 & 10+36+3 \\ 10-1+20 & 4-6+15 \\ \end{bmatrix} $
$ = \begin{bmatrix} 35 & 49 \\ 29 & 13 \\ \end{bmatrix} $
BETA
