#### Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 1 subquestion (i) maths textbook solution

$x=-1\: \: and\: \: y=4$

Given:

$\left[\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 5 \end{array}\right]$

Hint:

X=A-1B  is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A

Solution:

$\left[\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 5 \end{array}\right]$

$A X=B\\ |A|=\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right]=10-6=4 \neq 0\\ This\; has\; a\; unique\; sol\! ution\; given\; by\; X=A^{-1} B.\\ C_{i j}\; be\; the\; co-\! factor\; o\! f\; the\; elements\; a_{i j}\; in\; A=\left[a_{i j}\right]. \; Then,$

\begin{aligned} &C_{11}=(-1)^{1+1}(2)=2 , \quad C_{12}=(-1)^{1+2}(3)=-3 \\ &C_{21}=(-1)^{2+1}(2)=-2 , \quad C_{22}=(-1)^{2+2}(5)=5 \\ &\operatorname{adj} A=\left[\begin{array}{cc} 2 & -3 \\ -2 & 5 \end{array}\right]^{T} \\ &\quad=\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right] \end{aligned}

\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right] \\ X &=A^{-1} B \\ &=\frac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right]\left[\begin{array}{l} 3 \\ 5 \end{array}\right] \\ \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{4}\left[\begin{array}{c} 6-10 \\ -9+25 \end{array}\right]=\left[\begin{array}{c} -\frac{4}{4} \\ \frac{16}{4} \end{array}\right]} \\ &\therefore x=-1 \quad, \quad \mathrm{y}=4 \end{aligned}