#### explain solution rd sharma class 12 chapter Solution of simultaneous linear equation exercise 7.2 question 4 maths

$x= 2k,y= 4k,z= k$

Hint:

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

Given:

$x+y-6z=0$

$x-y+2z=0$

$-3x+y+2z=0$

Explanation:

The given homogeneous system can be written in matrix form

i.e.        $\begin{bmatrix} 1 & 1 & -6\\ 1& -1 & 2\\ -3& 1 & 2 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0\\0 \end{bmatrix}$

$A=\begin{bmatrix} 1 & 1 & -6\\ 1& -1 & 2\\ -3& 1 & 2 \end{bmatrix}$

let

$\left | A \right |=\begin{vmatrix} 1 & 1 & -6\\ 1& -1 & 2\\ -3& 1 & 2 \end{vmatrix}$

$= 1(-2-2)-1(2+6)-6(1-3)$

$= -4-8+12$

$= 0$

So, the given system of equation has a non-trivial solution.

To find these solutions, we write the first two equations as

$x+y=6z$

$x-y=-2z$

Let $z=k$, then we have

$\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} 6k\\ -2k\end{bmatrix}$

$A=\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix},X=\begin{bmatrix} x\\ y\end{bmatrix},B=\begin{bmatrix} 6k\\ -2k\end{bmatrix}$

Where,

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

So $A^{-1}$ exists

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

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

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

Now,   $X=A^{-1}B$

$\begin{bmatrix} x\\ y\end{bmatrix}=\frac{1}{-2}\begin{bmatrix} -1 & -1\\ -1& 1\end{bmatrix}\begin{bmatrix} 6k\\ -2k\end{bmatrix}$

$\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} 2k\\ 4k\end{bmatrix}$

$x=2k, y=4k$

Put in third equation, so these values $x=2k, y=4k,z=k$ satisfies it.

Hence ,$x=2k, y=4k,z=k$, where $k\in R$