Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 30

$X= \begin{bmatrix} -3 &-14 \\ 4& 17 \end{bmatrix}$

Hint:

Here, we use basic concept of determinant and inverse of matrix

$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$

Given:

$A=\left[\begin{array}{ll} 5 & 4 \\ 1 & 1 \end{array}\right], B=\left[\begin{array}{cc} 1 & -2 \\ 1 & 3 \end{array}\right]$

Solution:

\begin{aligned} &A X=B \\ &X=A^{-1} B \\ &|A|=1 \end{aligned}

Cofactor of A

\begin{aligned} &C_{11}=1, C_{21}=-4 \\ &C_{12}=-1, C_{22}=5 \\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} 1 & -1 \\ -4 & 5 \end{array}\right]^{T}=\left[\begin{array}{cc} 1 & -4 \\ -1 & 5 \end{array}\right] \end{aligned}

\begin{aligned} A^{-1} &=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ A^{-1} &=\frac{1}{1}\left[\begin{array}{cc} 1 & -4 \\ -1 & 5 \end{array}\right] \\ X &=\left[\begin{array}{cc} 1 & -4 \\ -1 & 5 \end{array}\right] \times\left[\begin{array}{cc} 1 & -2 \\ 1 & 3 \end{array}\right]=\left[\begin{array}{cc} -3 & -14 \\ 4 & 17 \end{array}\right] \end{aligned}