Question

Solve the following homogeneous linear equations.
$x+2y-5z=0,3x+4y+6z=0,x+y+z=0$, find $xyz$.

Answer 1

Given,

$x+2y-5z=0,3x+4y+6z=0,x+y+z=0$

$\left[\begin{array}{c}x+2y-5z=0\\ 3x+4y+6z=0\\ x+y+z=0\end{array}\right]$

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

$\left[\begin{array}{c}3(-2y+5z)+4y+6z=0\\ -2y+5z+y+z=0\end{array}\right]$

substitute $3(-2y+5z)+4y+6z=0:y=\frac{21z}{2}$

$\left[\begin{array}{c}-2\cdot \frac{21z}{2}+5z+\frac{21z}{2}+z=0\end{array}\right]$

upon solving the above equation, we get,

$-2\frac{21z}{2}+5z+\frac{21z}{2}+z=0:z=0$

$y=\frac{21\cdot 0}{2}=0$

$x=-2\cdot 0+5\cdot 0=0$

$\therefore x=y=z=0$