A) \[\left[ \begin{matrix} 2 & 4 \\ 3 & -5 \\ \end{matrix} \right]\]
B) \[\frac{1}{2}\left[ \begin{matrix} 2 & 4 \\ 3 & -5 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} -2 & 4 \\ 3 & 5 \\ \end{matrix} \right]\]
D) \[\frac{1}{2}\left[ \begin{matrix} -2 & 4 \\ 3 & 5 \\ \end{matrix} \right]\]
Correct Answer: B
Solution :
Given\[A=BX\]. Let\[X=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{y}_{1}} & {{y}_{2}} \\ \end{matrix} \right],\] \[\left[ \begin{matrix} 1 & 2 \\ 3 & -5 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{y}_{1}} & {{y}_{2}} \\ \end{matrix} \right]\] \[\Rightarrow \] \[\left[ \begin{matrix} 1 & 2 \\ 3 & -5 \\ \end{matrix} \right]=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ 2{{y}_{1}} & 2{{y}_{2}} \\ \end{matrix} \right]\] \[\Rightarrow \] \[{{x}_{1}}=1,{{x}_{2}}=2\] \[{{y}_{1}}=3/2,{{y}_{2}}=-5/2\] \[\therefore \] \[X=\left[ \begin{matrix} 1 & 2 \\ 3/2 & -5/2 \\ \end{matrix} \right]=\frac{1}{2}\left[ \begin{matrix} 2 & 4 \\ 3 & -5 \\ \end{matrix} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
