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

$x=-15\; \; ,\; \; y=7$

Given:

$\left[\begin{array}{ll} 3 & 7 \\ 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 4 \\ -1 \end{array}\right]$

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:

$\left[\begin{array}{ll} 3 & 7 \\ 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 4 \\ -1 \end{array}\right]$

\begin{aligned} &A X=B \\ &A=\left[\begin{array}{cc} 3 & 7 \\ 1 & 2 \end{array}\right]\\ &|A|=\left[\begin{array}{cc} 3 & 7 \\ 1 & 2 \end{array}\right]=6-7=-1 \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}(2)=2 \quad, \quad C_{12}=(-1)^{1+2}(1)=-1 \\ &C_{21}=(-1)^{2+1}(7)=-7 \quad, \quad C_{22}=(-1)^{2+2}(3)=3 \\ &\operatorname{adj} A=\left[\begin{array}{cc} 2 & -1 \\ -7 & 3 \end{array}\right]^{T} \\ &\quad=\left[\begin{array}{cc} 2 & -7 \\ -1 & 3 \end{array}\right] \end{aligned}

\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{-1}\left[\begin{array}{cc} 2 & -7 \\ -1 & 3 \end{array}\right] \end{aligned}

\begin{aligned} X &=A^{-1} B \\ &=\left[\begin{array}{cc} -2 & 7 \\ 1 & -3 \end{array}\right]\left[\begin{array}{c} 4 \\ -1 \end{array}\right] \\ &=\left[\begin{array}{c} -8-7 \\ 4+3 \end{array}\right] \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} -15 \\ 7 \end{array}\right]} \\ &\therefore x=-15 \quad, \quad \mathrm{y}=7 \end{aligned}