Mathematics - Determinant Question with Solution | TestHub

Given equations, $x-y+2z=4$ $3x+y+4z=6$ $x+y+z=1$ Let $\Delta =\left|\begin{array}{ccc}1& -1& 2\\ 3& 1& 4\\ 1& 1& 1\end{array}\right|$ $=1(1-4)+1(3-4)+2(3-1)$ $=-3-1+4=0$ and ${\Delta }_1=\left|\begin{array}{ccc}4& -1& 2\\ 6& 1& 4\\ 1& 1& 1\end{array}\right|$ $=4(1-4)+1(6-4)+2(6-1)$ $=-12+2+10=0$ Now, $\Delta =0$ and ${\Delta }_1=0$ $\therefore$ These equations have infinitely many solutions.