#### provide solution for rd sharma maths class 12 chapter Solution of simultaneous linear equation exercise 7.2 question 2

$x=\frac{-5k}{11}, y=\frac{12k}{11}, z=k$

Hint:

If $\left | A \right |= 0$, then the system of equation has non-trivial solution.

Given:

$2x-y+2z=0$

$5x+3y-z=0$

$x+5y-5z=0$

Explanation:

The given homogeneous system can be written in matrix form

i.e.      $\begin{bmatrix} 2& -1& 2\\ 5 & 3& -1\\ 1& 5& -5 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$

Let     $A=\begin{bmatrix} 2& -1& 2\\ 5 & 3& -1\\ 1& 5& -5 \end{bmatrix},X=\begin{bmatrix} x\\ y\\ z\end{bmatrix},B=\begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$

i.e.     $AX=B$

$\left | A \right |=\begin{vmatrix} 2& -1& 2\\ 5 & 3& -1\\ 1& 5& -5 \end{vmatrix}$

$=2(-15+5)+1(-25+1)+2(25-3)$

$=-20-24+ 44$

$=0$

So, the given system has non-trivial solution.

To find the solutions we write

$2x-y=-2z$

$5x+3y=z$

let          $z=k$

$\begin{bmatrix} 2 &-1 \\ 5 & 3 \end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} -2k\\ k\end{bmatrix}$

$\left | A \right |=\begin{vmatrix} 2 &-1 \\ 5 & 3 \end{vmatrix}=11\neq 0$

So A-1 exists

$adjA=\begin{bmatrix} 3 & 1\\ -5 & 2\end{bmatrix}$

$A^{-1}=\frac{1}{\left | A \right |}adjA$

$A^{-1}=\frac{1}{11}\begin{bmatrix} 3 & 1\\ -5 &2 \end{bmatrix}$

Now,           $X=A^{-1}B$

$\begin{bmatrix} x\\ y\end{bmatrix}=\frac{1}{11}\begin{bmatrix} 3 & 1\\ -5 &2 \end{bmatrix}\begin{bmatrix} -2k\\ k\end{bmatrix}$

$=\frac{1}{11}\begin{bmatrix} -5k\\ 12k\end{bmatrix}$

$x=\frac{-5k}{11},y=\frac{12k}{11}$

x,z and y satisfy the third equation

$\frac{-5k}{11}+5\left ( \frac{12k}{11} \right )-5z=0$

$5z=\frac{55k}{11}$

$z=k$

Hence $x=\frac{-5k}{11}, y=\frac{12k}{11}, z=k$, where $k\in R$