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Need solution for rd sharma maths class 12 chapter adjoint and inverse of a matrix exercise multiple choice question question 23

Answers (1)

option (a) \frac{1}{2}\begin{bmatrix} 2 &4 \\ 3&-5 \end{bmatrix}

Given: A=\begin{bmatrix} 1 &2 \\ 3&-5 \end{bmatrix} \& A=\begin{bmatrix} 1 &0 \\ 0&2 \end{bmatrix}  and

X be a matrix such that A = BX

Hint: put A & B in equation and square to find matrix X.

Solution:  – A= BX

\begin{aligned} &\Rightarrow \mathrm{B}^{-1} \mathrm{~A}=\mathrm{X} \\ &\mathrm{B}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right] \\ &|B|=2 \\ &B^{-1}=\frac{\mathrm{adj} \mathrm{B}}{|\mathrm{B}|} \\ &=\frac{1}{2}\left[\begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right] \\ &\text { Now, } \mathrm{B}^{-1} \mathrm{~A}=\mathrm{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]=\mathrm{X} \\ &\frac{1}{2}\left[\begin{array}{cc} 2 & 4 \\ 3 & -5 \end{array}\right]=\mathrm{X} \end{aligned}

Hence, option (a) is correct

 

 

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