#### need solution for rd sharma maths class 12 chapter determinants exercise 5.5 question 3

Answer: $x = -7k, y = 8k, z = 13k$, where $k \in R$

Hint: : If $\left | A \right |= 0$, then the system has an infinite number of solutions.

Given:

$3x+y+z=0$

$x - 4y + 3z = 0$

$2 x + 5y - 2z = 0$

Solution:

The above linear equations can be represented in matrix form i.e.

$A = \begin{bmatrix} 3 & 1 & 1\\ 1& -4 & 3\\ 2& 5& -2 \end{bmatrix}, X = \begin{bmatrix} x\\ y\\ z\end{bmatrix}, B = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$

Now,

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

$= 3(8 -15) -1(-2 -6) + 1(5+8)$

$= 3(-7) + 8 +15$

$= - 21 + 21$

$= 0$

So, the system of equations has non – trivial solutions.

Consider, first two equation and put z = k

$3 x + y = -k$

$x - 4y = -3k$

Now, By Crammer’s rule:

$x = \frac{D_{1}}{D} = \frac{\begin{vmatrix} -k & 1\\ -3k & -4 \end{vmatrix}}{\begin{vmatrix} 3&1 \\ 1& -4 \end{vmatrix}} = \frac{7k}{-13} = \frac{-7k}{13}$

$y = \frac{D_{2}}{D} = \frac{\begin{vmatrix} 3 & -k\\ 1 & -3k \end{vmatrix}}{\begin{vmatrix} 3&1 \\ 1& -4 \end{vmatrix}} = \frac{-8k}{-13} = \frac{8k}{13}$

$x=\frac{-7k}{13}, y = \frac{8k}{13}$  and $z=k$

or $x = -7k$, $y = 8k$ and $z = 13k$, where $k \in R$

It will satisfy the third equation.

Hence, $x = -7k$, $y = 8k$ and $z = 13k$, where $k \in R$