Mathematics - Determinant Question with Solution | TestHub

Given system of linear equations:

$x+y+z=6$

$\alpha x+\beta y+7z=3$

$x+2y+3z=14$

On taking determinant, we get

$\left|A\right|=\left|\begin{array}{ccc}1& 1& 1\\ \alpha & \beta & 7\\ 1& 2& 3\end{array}\right|$

$=(3\beta -14)-(3\alpha -7)+(2\alpha -\beta )$

$=2\beta -\alpha -7$

Here, if $\alpha =\beta =7$ then $\left|A\right|=0$ and the system has no solution. So, option A is correct.

If $\alpha =\beta , \alpha \ne 7$ then the system has a unique solution. So, option B is correct.

If $(\alpha , \beta )\ne (7, 7)$ then $\left|A\right|\ne 0$. So, system has not infinitely many solution. Therefore, option $D$ is false.