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#### need solution for rd sharma maths class 12 chapter Solution of simultaneous linear equation exercise 7.2 question 3

$x= \frac{-9k}{13},y= \frac{-k}{13},z= k$

Hint:

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

Given:

$3x-y+2z=0$

$4x+3y+3z=0$

$5x+7y+4z=0$

Explanation:

The given homogeneous system can be written in matrix form

i.e.        $\begin{bmatrix} 3 & -1 & 2\\ 4& 3 & 3\\ 5& 7 & 4 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0\\0 \end{bmatrix}$

$A=\begin{bmatrix} 3 & -1 & 2\\ 4& 3 & 3\\ 5& 7 & 4 \end{bmatrix}$

let

$\left | A \right |=\begin{vmatrix} 3 & -1 & 2\\ 4& 3 & 3\\ 5& 7 & 4 \end{vmatrix}$

$= 3(12-21)+1(16-15)+2(28-15)$

$= -27+1+26$

$= 0$

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

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

$3x-y=-2z$

$4x+3y=-3z$

Let $z=k$, then we have

$\begin{bmatrix} 3 & -1\\ 4 & 3\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} -2k\\ -3k\end{bmatrix}$

$A=\begin{bmatrix} 3 & -1\\ 4 & 3\end{bmatrix},X=\begin{bmatrix} x\\ y\end{bmatrix},B=\begin{bmatrix} -2k\\ -3k\end{bmatrix}$

Where,

$\left | A \right |=\begin{vmatrix} 3 & -1\\ 4 & 3\end{vmatrix}=13\neq 0$

So $A^{-1}$ exists

$adjA=\begin{bmatrix} 3 & 1\\ -4 & 3\end{bmatrix}$

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

$A^{-1}=\frac{1}{13}\begin{bmatrix} 3 & 1\\ -4& 3\end{bmatrix}$

Now,   $X=A^{-1}B$

$\begin{bmatrix} x\\ y\end{bmatrix}=\frac{1}{13}\begin{bmatrix} 3 & 1\\ -4& 3\end{bmatrix}\begin{bmatrix} -2k\\ -3k\end{bmatrix}$

$=\frac{1}{13}\begin{bmatrix} -9k\\ -k\end{bmatrix}$

$x=\frac{-9k}{13}, y=\frac{-k}{13}$

Put in third equation, it satisfies it too.

Hence $x=\frac{-9k}{13}, y=\frac{-k}{13},z=k$, where $k\in R$