Solve each of the following system of homogeneous linear equations.

Solution:

Given: $x+y-2 z=0$

$2 x+y-3 z=0$

$5 x+4 y-9 z=0$

$D=\left|\begin{array}{rrr}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right|$

$=1(-9+12)-1(-18+15)-2(8-5)$

$=0$

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

$x+y=2 k$

$2 x+y=3 k$

Using Cramer's rule, we get

$x=\frac{D_{1}}{D}=\frac{\left|\begin{array}{ll}2 k & 1 \\ 3 k & 1\end{array}\right|}{\left|\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right|}=\frac{-k}{-1}=k$

$y=\frac{D_{2}}{D}=\frac{\left|\begin{array}{cc}1 & 2 k \\ 2 & 3 k\end{array}\right|}{\left|\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right|}=\frac{-k}{-1}=k$

$z=k$

Clearly, these values satisfy the third equation. Thus,

$x=y=z=k \quad[k \in R]$