Question

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

Answer 1

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

\[D = \begin{vmatrix}3 & 1 & 1 \\ 1 & - 4 & 3 \\ 2 & 5 & - 2\end{vmatrix} = 0\] 
The system has infinitely many solutions . Putting z = k in the first two equations, we get
\[3x + y = - k\] 
\[x - 4y = - 3k\] 
Solving these equations by Cramer's rule, we get
\[x = \frac{D_1}{D} = \frac{\begin{vmatrix}- k & 1 \\ - 3k & - 4\end{vmatrix}}{\begin{vmatrix}3 & 1 \\ 1 & - 4\end{vmatrix}} = - \frac{7k}{13}\] 
\[y = \frac{D_2}{D} = \frac{\begin{vmatrix}3 & - k \\ 1 & - 3k\end{vmatrix}}{\begin{vmatrix}3 & 1 \\ 1 & - 4\end{vmatrix}} = \frac{8k}{13} \] 
\[z = k\] 
\[ \Rightarrow x = - \frac{7k}{13}, y = \frac{8k}{13}\text{ and }z = k\] 
\[or x = - 7k, y = 8k\text{ and }z = 13k\] 
Clearly, these values satisfy the third equation . 
Thus, 
\[x = - 7k\] 
\[y = 8k    \left[ k \in R \right]\] 
\[z = 13k \]