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#### Need solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 5

$x=2 \quad , \quad y=-1 \quad , \quad z=4$

Given:

\begin{aligned} &A=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right] \quad, \quad B=\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \backslash \\ &x-y=3, \quad 2 x+3 y+4 z=17 \quad, \quad y+2 z=7 \end{aligned}

Hint:

For matrix multiply matrix A with matrix B, Then X=A-1B  is formula for which is used to solve this problem.

Solution:

\begin{aligned} A B &=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \\ &=\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \\ A B &=6\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=6 I_{3} \end{aligned}

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

\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{6}\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right]\left[\begin{array}{c} 3 \\ 17 \\ 7 \end{array}\right] \\ &=\frac{1}{6}\left[\begin{array}{c} 6+34-28 \\ -12+34-28 \\ 6-17+35 \end{array}\right] \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{6}\left[\begin{array}{c} 12 \\ -6 \\ 24 \end{array}\right]} \\ &x=2, y=-1, z=4 \end{aligned}