Mathematics - Determinant Question with Solution | TestHub

Given: 

System of linear equations,

$x+y+z=4\mu ,$

$x+2y+2\lambda z=10\mu ,$

$x+3y+4\lambda {}^{2}z={\mu }^2+15$.

$\Rightarrow \Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 2\lambda \\ 1& 3& 4{\lambda }^2\end{array}\right|$

$\Rightarrow \Delta =\left(8{\lambda }^2-6\lambda \right)-\left(4{\lambda }^2-2\lambda \right)+\left(3-2\right)$

$\Rightarrow \Delta =8{\lambda }^2-6\lambda -4{\lambda }^2+2\lambda +1$

$\Rightarrow \Delta =4{\lambda }^2-4\lambda +1$

$\Rightarrow \Delta =(2\lambda -1{)}^2$

For unique solution, $\Delta \ne 0$.

$\Rightarrow 2\lambda -1\ne 0$

$\Rightarrow \lambda \ne \frac{1}{2}$

Let, $\Delta =0, \lambda =\frac{1}{2}$

We know that, for infinite solutions, ${\Delta }_x={\Delta }_y={\Delta }_z=0$

$\Rightarrow {\Delta }_x=\left|\begin{array}{ccc}4\mu & 1& 1\\ 10\mu & 2& 1\\ {\mu }^2+15& 3& 1\end{array}\right|$

$\Rightarrow {\Delta }_x=4\mu \left(2-3\right)-\left(10\mu -{\mu }^2-15\right)+\left(30\mu -2{\mu }^2-30\right)$

$\Rightarrow {\Delta }_x=-4\mu -10\mu +{\mu }^2+15+30\mu -2{\mu }^2-30$

$\Rightarrow {\Delta }_x=-{\mu }^2+16\mu -15$

$\Rightarrow (\mu -15)(\mu -1)=0$

So, for infinite solution $\lambda =\frac{1}{2}, \mu =1$ or $15$.