Mathematics - Determinant Question with Solution | TestHub

Given linear equations are 

$x+y+z=\alpha$

$\alpha x+2\alpha y+3z=-1$

$x+3\alpha y+5z=4$

For this system of linear equations to have inconsistent solution

$D=\left|\begin{array}{ccc}1& 1& 1\\ \alpha & 2\alpha & 3\\ 1& 3\alpha & 5\end{array}\right|=0$

$\Rightarrow {\left(\alpha -1\right)}^2=0$

$\Rightarrow \alpha =1$

Now for $\alpha =1$

$D_1=\left|\begin{array}{ccc}1& 1& 1\\ -1& 2& 3\\ 4& 3& 5\end{array}\right|$

$=\left(10-9\right)-\left(-5-12\right)+\left(-3-8\right)$

$=1+17-11\ne 0$

Hence, for $\alpha =1$ the system of equation has inconsistent solution.