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provide solution for rd sharma maths class 12 chapter Solution of simultaneous linear equation exercise 7.2 question 6

Answers (1)

Answer:

$x= \frac{k}{3},y= \frac{2k}{3},z= k$

Hint:

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

Given:

$x+y-z=0$

$x-2y+z=0$

$3x+6y-5z=0$

Explanation:

The given homogeneous system can be written in matrix form

i.e. $\begin{bmatrix} 1 & 1 & -1\\ 1& -2 & 1\\ 3& 6 & -5 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0\\0 \end{bmatrix}$

$A=\begin{bmatrix} 1 & 1 & -1\\ 1& -2 & 1\\ 3& 6 & -5 \end{bmatrix}$

let

$\left | A \right |=\begin{vmatrix} 1 & 1 & -1\\ 1& -2 & 1\\ 3& 6 & -5 \end{vmatrix}$

$= 1(10-6)-1(-5-3)-1(6+6)$

$= 4+8-12$

$= 0$

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

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

$x+y=z$

$x-2y=-z$

Let $z=k$ , then we have

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

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

Where,

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

So $A^{-1}$ exists

^Now,

Put in third equation, so these values satisfies it.

Hence ^, where

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