Mathematics - Determinant Question with Solution | TestHub

Given,

System of questions

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

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

$3x+\alpha y+2z=\beta$

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

$\Rightarrow \alpha (6-\alpha )-2(4\alpha -3)+1(2{\alpha }^2-9)=0$

$\Rightarrow \left(\alpha -3\right)(\alpha +1)=0$

$\Rightarrow \alpha =-1,3$

$D_x=\left|\begin{array}{lll}2& 1& 1\\ 3& 1& 1\\ \alpha & 2& \beta \end{array}\right|=0\Rightarrow \beta =2$

$\Rightarrow 2\left(\beta -2\right)-1(3\beta -\alpha )+(6-\alpha )=0$

$\Rightarrow \beta =2$

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

$\Rightarrow \alpha (\beta -2)-1(2\alpha \beta -3)+1(4\alpha -3)=0$

$\Rightarrow \alpha (2-\beta )=0$

$\Rightarrow \alpha =0,\beta =2$

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

$\Rightarrow {\alpha }^2-\alpha \beta -3=0$

Now when $\beta =2,\alpha =-1$ the system of equations will have infinite solutions,

When $\alpha =3, \beta =2$ the system has infinite solution,

And when $\alpha =3, \beta \ne 2$ the system has no solution as $D_x \& D_y\ne 0$ for $\beta \ne 2$,

Hence, only option $\left(B\right)$ is not correct