#### Provide solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (ii)

\begin{aligned} &x=3, y=2, z=-1 \end{aligned}

Given:

$A=\left[\begin{array}{ccc} 3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{array}\right]$

Hint:

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

Solution:

\begin{aligned} A &=\left[\begin{array}{ccc} 3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{array}\right] \\ |A| &=3(3-0)+4(2-5)+2(0-3) \\ &=9-12-6 \\ &=-9 \end{aligned}

$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$

\begin{aligned} &C_{11}=-1^{1+1}\left|\begin{array}{ll} 3 & 5 \\ 0 & 1 \end{array}\right|=3 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} -4 & 2 \\ 0 & 1 \end{array}\right|=4 \\ &C_{12}=-1^{1+2}\left|\begin{array}{ll} 2 & 5 \\ 1 & 1 \end{array}\right|=3 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right|=1 \end{aligned}

\begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{ll} 2 & 3 \\ 1 & 0 \end{array}\right|=-3 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{cc} 3 & -4 \\ 1 & 0 \end{array}\right|=-4 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} -4 & 2 \\ 3 & 5 \end{array}\right|=-26 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{ll} 3 & 2 \\ 2 & 5 \end{array}\right|=-11 \\ &C_{33}=-1^{3+3}\left|\begin{array}{cc} 3 & -4 \\ 2 & 3 \end{array}\right|=17 \end{aligned}

\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17 \end{array}\right] \\ A^{-1} &=\frac{1}{|A|} \operatorname{adjA} \\ &=\frac{1}{-9}\left[\begin{array}{ccc} 3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17 \end{array}\right] \end{aligned}

\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{-9}\left[\begin{array}{ccc} 3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17 \end{array}\right]\left[\begin{array}{c} -1 \\ 7 \\ 2 \end{array}\right] \\ &=\frac{1}{-9}\left[\begin{array}{c} -3+28-52 \\ -3+7-22 \\ 3-28+34 \end{array}\right]=\frac{1}{-9}\left[\begin{array}{c} -27 \\ -18 \\ 9 \end{array}\right] \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 3 \\ 2 \\ -1 \end{array}\right]} \\ &x=3, y=2, z=-1 \end{aligned}