#### Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 13 maths textbook solution

Given:

As per information the following equation,

$x+y+z=12,2 x+3 y+z=33, x+z-2 y=0$

Hint:

X=A-1B is used to solve this problem.

And 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:

$x+y+z=12 \\2 x+3 y+z=33 \\ x+z-2 y=0$

The three equations can be represented in the form of a matrix as

\begin{aligned} &\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 12 \\ 33 \\ 0 \end{array}\right]\\ &A X=B\\ &|A|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1 \end{array}\right|=3 \neq 0, A \text { is nonsingular its inverse exists } \end{aligned}

\begin{aligned} \operatorname{adjA} &=\left[\begin{array}{ccc} 9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{array}\right], A^{-1}=\frac{1}{|A|} \operatorname{adjA} \\ &=\frac{1}{3}\left[\begin{array}{ccc} 9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{array}\right] \end{aligned}

\begin{aligned} &X=A^{-1} B=\frac{1}{3}\left[\begin{array}{ccc} 9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{array}\right]\left[\begin{array}{c} 12 \\ 33 \\ 0 \end{array}\right]=\left[\begin{array}{l} 3 \\ 4 \\ 5 \end{array}\right] \\ &x=3, y=4, z=5 \end{aligned}