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Question
Let \[A = \begin{bmatrix}1 & 2 \\ 3 & - 5\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}\] and X be a matrix such that A = BX, then X is equal to _____________ .
Options
\[\frac{1}{2}\begin{bmatrix}2 & 4 \\ 3 & - 5\end{bmatrix}\]
\[\frac{1}{2}\begin{bmatrix}- 2 & 4 \\ 3 & 5\end{bmatrix}\]
\[\begin{bmatrix}2 & 4 \\ 3 & - 5\end{bmatrix}\]
none of these
Solution
\[\frac{1}{2}\begin{bmatrix}2 & 4 \\ 3 & - 5\end{bmatrix}\]
\[A = BX\]
\[ \Rightarrow B^{- 1} A = B^{- 1} BX\]
\[ \Rightarrow B^{- 1} A = IX\]
\[ \Rightarrow X = B^{- 1} A . . . \left( 1 \right)\]
Now,
\[B = \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}\]
\[adjB = \begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}\]
\[\left| B \right| = 2\]
\[ \therefore B^{- 1} = \frac{1}{\left| B \right|}adjB = \frac{1}{2}\begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}\]
On putting the value of B-1 in eq. (1), we get
\[X = \frac{1}{2}\begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & 2 \\ 3 & - 5\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{2}\begin{bmatrix}2 & 4 \\ 3 & - 5\end{bmatrix}\]
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