Question

Let $A=\left[\begin{array}{cc}1& 2\\ 3& -5\end{array}\right]\mathrm{and}B=\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]$ and X be a matrix such that A = BX, then X is equal to

(a) $\frac{1}{2}\left[\begin{array}{cc}2& 4\\ 3& -5\end{array}\right]$

(b) $\frac{1}{2}\left[\begin{array}{cc}-2& 4\\ 3& 5\end{array}\right]$

(c) $\left[\begin{array}{cc}2& 4\\ 3& -5\end{array}\right]$

(d) none of these.

Answer 1

(a) $\frac{1}{2}\left[\begin{array}{cc}2& 4\\ 3& -5\end{array}\right]$

$$\begin{gathered} A=BX \\ \Rightarrow B^{-1}A=B^{-1}BX \\ \Rightarrow B^{-1}A=IX \\ \Rightarrow X=B^{-1}A...\left(1\right) \\ \mathrm{Now}, \\ B=\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right] \\ \mathrm{adj}B=\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right] \\ \left|B\right|=2 \\ \therefore B^{-1}=\frac{1}{\left|B\right|}\mathrm{adj}B=\frac{1}{2}\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right] \\ \mathrm{On}\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}{B}^{-1}\mathrm{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ X=\frac{1}{2}\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& -5\end{array}\right] \\ \Rightarrow X=\frac{1}{2}\left[\begin{array}{cc}2& 4\\ 3& -5\end{array}\right] \end{gathered}$$