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

\begin{aligned} &x=4, y=-3, z=1 \end{aligned}

Given:

\begin{aligned} &A=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right] \quad, \quad B=\left[\begin{array}{ccc} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 1 & 5 \end{array}\right] \\ &x-2 y=10 \\ &2 x+y+3 y=8 \\ &-2 y+z=7 \end{aligned}

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 B &=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right]\left[\begin{array}{ccc} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 1 & 5 \end{array}\right] \\ &=\left[\begin{array}{ccc} 7+4+0 & 2-2+0 & -6+6+0 \\ 14-2-12 & 4+1+6 & -12-3+15 \\ 0+4-4 & 0-1+2 & 0+6+5 \end{array}\right] \\ &=\left[\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}\right] \\ A B &=11\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}

\begin{aligned} &A B=11 I_{3} \\ &{\left[\frac{1}{11}\right] A B=I_{3}} \\ &{\left[\frac{1}{11} B\right] A=I_{3}} \\ &A^{-1}=\frac{1}{11} B \\ &\quad=\frac{1}{11}\left[\begin{array}{ccc} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 1 & 5 \end{array}\right] \end{aligned}

\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{11}\left[\begin{array}{ccc} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 1 & 5 \end{array}\right]\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right] \\ &=\frac{1}{11}\left[\begin{array}{c} 70+16-42 \\ -20+8-21 \\ -40+16+35 \end{array}\right]=\frac{1}{11}\left[\begin{array}{c} 44 \\ -33 \\ 11 \end{array}\right] \end{aligned}

\begin{aligned} &x=4, y=-3, z=1 \end{aligned}