Mathematics - Determinant Question with Solution | TestHub

Given: 

$x+y+z=5$

$x+2y+{\lambda }^{2}z=9$

$x+3y+\lambda z=\mu$

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

$\Rightarrow △=2\lambda -3{\lambda }^2-\lambda +{\lambda }^2+1$

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

We know that for infinite solutions, $△=0 \& {△}_1={△}_2={△}_3=0$

$\Rightarrow -2{\lambda }^2+\lambda +1=0$

$\Rightarrow 2{\lambda }^2-\lambda -1=0$

$\Rightarrow \left(\lambda -1\right)\left(2\lambda +1\right)=0$

$\Rightarrow \lambda =1, \frac{-1}{2}$

Now, finding ${△}_1$ for $\lambda =1$ we get,

${\mathrm{\Delta }}_1=\left|\begin{array}{lll}5& 1& 1\\ 9& 2& 1\\ \mu & 3& 1\end{array}\right|$

$\Rightarrow {\mathrm{\Delta }}_1=\mu -13$

And for $\lambda =\frac{-1}{2}$ we get,

${\mathrm{\Delta }}_1=\left|\begin{array}{lll}5& 1& 1\\ 9& 2& \frac{1}{4}\\ \mu & 3& \frac{-1}{2}\end{array}\right|=\frac{1}{4}\left|\begin{array}{lll}5& 1& 4\\ 9& 2& 1\\ \mu & 3& -2\end{array}\right|=\frac{7}{4}\left(13-\mu \right)$

So, for $\lambda \ne 1 \& \mu \ne 13$ system has no solution,

Hence, option $\left(C\right)$ is the correct answer.