Question

Solve each of the following system of homogeneous linear equations.
x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0

Answer 1

Given: x + y − 2z = 0
            2x + y − 3z = 0              
            5x + 4y − 9z = 0

\[D = \begin{vmatrix}1 & 1 & - 2 \\ 2 & 1 & - 3 \\ 5 & 4 & - 9\end{vmatrix}\] 
\[ = 1( - 9 + 12) - 1( - 18 + 15) - 2(8 - 5)\] 
\[ = 0\] 
So, the system has infinitely many solutions . Putting z = k in the first two equations, we get
\[x + y = 2k\] 
\[2x + y = 3k\] 
Using Cramer's rule, we get
\[x = \frac{D_1}{D} = \frac{\begin{vmatrix}2k & 1 \\ 3k & 1\end{vmatrix}}{\begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix}} = \frac{- k}{- 1} = k\] 
\[y = \frac{D_2}{D} = \frac{\begin{vmatrix}1 & 2k \\ 2 & 3k\end{vmatrix}}{\begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix}} = \frac{- k}{- 1} = k \] 
\[z = k\] 
Clearly, these values satisfy the third equation . 
Thus, 
\[x = y = z = k \left[ k \in R \right]\]