Consider the system of linear equations \(x+y+z=5\), \(x+2 y+\lambda^2 z=9, x+3 y+\lambda z=\mu\), where \(\lambda, \mu …

\(∆=\left|\begin{matrix}1 & 1 & 1 \\ 1 & 2 & {\lambda }^{2} \\ 1 & 3 & \lambda \end{matrix}\right|=1\left(3-2\right)-{\lambda }^{2}\left(3-1\right)+\lambda \left(2-1\right)\\ =-\left(2{\lambda }^{2}-\lambda -1\right)\\ =-\left(\lambda -1\right)\left(2\lambda +1\right)\)

For unique solution \(\lambda \neq 1\) and \(\lambda \neq-\frac{1}{2}\)

If \(\lambda = 1\)

\({\Delta }_{x}=\left|\begin{matrix}5 & 1 & 1 \\ 9 & 2 & 1 \\ \mu & 3 & 1\end{matrix}\right|=-\mu +13\)

\({\Delta }_{y}=\left|\begin{matrix}1 & 5 & 1 \\ 1 & 9 & 1 \\ 1 & \mu & 1\end{matrix}\right|=0\)

\({\Delta }_{z}=\left|\begin{matrix}1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & \mu \end{matrix}\right|=\mu -13\)

Infinite solution \(\lambda =1&\mu =13\)
For no solution\(\lambda =1&\mu \neq 13\)
when \(\lambda =-\frac{1}{2}\) and \(\mu \neq 13\) gives no solution.