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

$x=7\; \; and\; \; y=-2$

Given:

$\left[\begin{array}{ll} 3 & 1 \\ 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 19 \\ 23 \end{array}\right]$

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:

$\left[\begin{array}{ll} 3 & 1 \\ 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 19 \\ 23 \end{array}\right]$

\begin{aligned} &A X=B \\ &|A|=\left[\begin{array}{cc} 3 & 1 \\ 3 & -1 \end{array}\right]=-3-3=-6 \neq 0 \end{aligned}

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

$\begin{array}{ll} C_{11}=(-1)^{1+1}(-1)=-1 & , \quad C_{12}=(-1)^{1+2}(3)=-3 \\ C_{21}=(-1)^{2+1}(1)=-1 & , \quad C_{22}=(-1)^{2+2}(3)=3 \end{array}$

\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{cc} -1 & -3 \\ -1 & 3 \end{array}\right]^{T} \\ &=\left[\begin{array}{cc} -1 & -1 \\ -3 & 3 \end{array}\right] \\ A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{-6}\left[\begin{array}{cc} -1 & -1 \\ -3 & 3 \end{array}\right] \end{aligned}

\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{-6}\left[\begin{array}{cc} -1 & -1 \\ -3 & 3 \end{array}\right]\left[\begin{array}{l} 19 \\ 23 \end{array}\right] \\ &=\frac{1}{-6}\left[\begin{array}{l} -19-23 \\ -57-69 \end{array}\right] \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{-6}\left[\begin{array}{l} -19-23 \\ -57-69 \end{array}\right]=\left[\begin{array}{c} \frac{-42}{-6} \\ \frac{12}{-6} \end{array}\right]} \\ &\therefore x=7 \quad, \quad \mathrm{y}=-2 \end{aligned}