Solve each of the following system of homogeneous linear equations.

Solution:

Given: $3 x+y+z=0$

$x-4 y+3 z=0$

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

$D=\left|\begin{array}{rrr}3 & 1 & 1 \\ 1 & -4 & 3 \\ 2 & 5 & -2\end{array}\right|=0$

The system has infinitely many solutions. Putting $z=k$ in the first two equations, we get

$3 x+y=-k$

$x-4 y=-3 k$

Solving these equations by Cramer's rule, we get

$x=\frac{D_{1}}{D}=\frac{\left|\begin{array}{rr}-k & 1 \\ -3 k & -4\end{array}\right|}{\left|\begin{array}{rr}3 & 1 \\ 1 & -4\end{array}\right|}=-\frac{7 k}{13}$

$y=\frac{D_{2}}{D}=\frac{\left|\begin{array}{rr}3 & -k \\ 1 & -3 k\end{array}\right|}{\left|\begin{array}{rr}3 & 1 \\ 1 & -4\end{array}\right|}=\frac{8 k}{13}$

$z=k$

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

or $x=-7 k, y=8 k$ and $z=13 k$

Clearly, these values satisfy the third equation. Thus,

$x=-7 k$

$y=8 k$

$z=13 k$$\quad[k \in R]$