Question

Solve the following non-homogeneous equations of three unknowns by using determinants:
$2x+2y+z=5$
$x-y+z=1$
$3x+y+2z=4$

Answer 1

Now,

$\Delta =\left|\begin{array}{ccc}2& 2& 1\\ 1& -1& 1\\ 3& 1& 2\end{array}\right|=2(-2-1)-2(2-3)+1(1+3)$$=-6+2+4=0$

${\Delta }_1=\left|\begin{array}{ccc}5& 2& 1\\ 1& -1& 1\\ 4& 1& 2\end{array}\right|=5(-2-1)-2(2-4)+1(1+4)$$=-15+4+5=-6$

${\Delta }_2=\left|\begin{array}{ccc}2& 5& 1\\ 1& 1& 1\\ 3& 4& 2\end{array}\right|=2(-2-4)-5(2-3)+1(4-3)$$=-4+5+1=2$

${\Delta }_3=\left|\begin{array}{ccc}2& 2& 5\\ 1& -1& 1\\ 3& 1& 4\end{array}\right|=2(-4-1)-2(4-3)+5(1+3)$$=-10-2+20=8$

Here $\Delta =0$ ${\Delta }_1=-6$

${\Delta }_2=2$ & ${\Delta }_3=8$

According to cramer's rule if $\Delta =0$ at least any one ${\Delta }_1,{\Delta }_2$ and ${\Delta }_3$ is non-zero then the system of eqn is inconsistent.

$\therefore$ Hence, no solution will exist.