Question

The number of solution(s) of the system of equations $x+4y-z=0,3x-4y-z=0$ and $x-3y+z=0$ is

Answer 1

$x+4y-z=0$

$3x-4y-z=0$

$x-3y+z=0$

$\Delta =\left|\begin{array}{ccc}1& 4& -1\\ 3& -4& -1\\ 1& -3& 1\end{array}\right|$

$=1(-4-3)-4(3+1)-1(-9+4)$

$=-7-16+5$

$\Delta =-18$

$\Delta x=\left|\begin{array}{ccc}0& 4& -1\\ 0& -4& -1\\ 0& -3& 1\end{array}\right|=0$

$\Delta y=0,\Delta z=0$

Hence, system of equation has unique solution, i.e., ${}^{\prime }{1}^{\prime }$ solution.