#### Provide solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 1 subquestion (vi)

$x=\frac{9}{4}\; \; ,\; \; y=\frac{1}{4}$

Given:

$\left[\begin{array}{ll} 3 & 1 \\ 5 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 7 \\ 12 \end{array}\right]$

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:

$\left[\begin{array}{ll} 3 & 1 \\ 5 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 7 \\ 12 \end{array}\right]$

\begin{aligned} &A X=B \\ &|A|=\left[\begin{array}{cc} 3 & 1 \\ 5 & 3 \end{array}\right]=9-5=4 \neq 0 \end{aligned}

$This\; has\; a\; unique \; solution\; given\; by\; X=A^{-1} B.\\ C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$

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

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

\begin{aligned} X =A^{-1} B \\ =\frac{1}{4}\left[\begin{array}{cc} 3 & -1 \\ -5 & 3 \end{array}\right]\left[\begin{array}{c} 7 \\ 12 \end{array}\right] \\ =\frac{1}{4}\left[\begin{array}{c} 21-12 \\ -35+36 \end{array}\right] \\ \left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{9}{4} \\ \frac{1}{4} \end{array}\right] \\ \therefore x =\frac{9}{4} \quad, \quad \mathrm{y}=\frac{1}{4} \end{aligned}