Question

Sol. [Key Concept: Infinitely many solutions require $D = D_{1} = D_{2} = D_{3} = 0$; solving gives $\lambda = 6,\ \mu = 16$, so $\lambda + \mu = 22$.]

$$\ D = \left| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda \end{matrix} \right| = \lambda - 6$$

$$D_{1} = \left| \begin{matrix} 6 & 1 & 1 \\ 10 & 2 & 5 \\ \mu & 3 & \lambda \end{matrix} \right| = 2\lambda + 3\mu - 60$$

$$D_{2} = \left| \begin{matrix} 1 & 6 & 1 \\ 1 & 10 & 5 \\ 2 & \mu & \lambda \end{matrix} \right| = 4(\lambda - \mu + 10)$$

$$D_{3} = \left| \begin{matrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 2 & 3 & \mu \end{matrix} \right| = \mu - 16$$

For infinite many solution

$${D = D_{1} = D_{2} = D_{3} = 0 }{\Rightarrow \lambda = 6\&\mu = 16 }{\Rightarrow \lambda + \mu = 22 }$$