Question

Solve the equations
$2x+3y-3z=03x-3y+z=03x-2y-3z=0$
Find xyz?

Answer 1

We have
$2x+3y-3z=03x-3y+z=03x-2y-3z=0$
The given system of equations in the matrix form are written as below:
$\left[\begin{array}{ccc}2& 3& -3\\ 3& -3& 1\\ 3& -2& -3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
$AX=0...\left(1\right)$
where
$A=\left[\begin{array}{ccc}2& 3& -3\\ 3& -3& 1\\ 3& -2& -3\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $0=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
$\therefore |A|=2(9+2)-3(-9-3)-3(-6+9)=22+36-9=49\ne 0$
Hence, the equations have a unique trivial solution $x=0,y=0$ and $z=0$.