Mathematics - Determinant Question with Solution | TestHub

For infinite solution $∆={∆}_x={∆}_y={∆}_z=0$

$∆=\left|\begin{array}{lll}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& 1& \alpha \end{array}\right|=0$

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

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

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

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

$\Rightarrow \alpha =1,\alpha =-2,1$

If $\beta =1$, then all planes are overlapping

$\therefore$ option (d) is correct.

Put $\alpha =2, \beta =1$ the system equations as,

$2x+y+z=1$

$x+2y+z=1$

$x+y+2z=1$

Adding all three above equations we get,

$x+y+z=\frac{3}{4}$

$\therefore$ option (c) is correct.

Now, if $\alpha =-2, \beta =1$ then,

$∆=0, {∆}_x\ne 0$  (No solution)

$\therefore$ option (b) is correct.

Now, if $\alpha =2$

$∆\ne 0$ (unique solution exists)

$\therefore$ option (a) is incorrect