If the system $(\lambda-1)x+(\lambda-4)y+\lambda z=5,\ \lambda x+(\lambda-1)y+(\lambda-4)z=7,\ (\lambda+1)x+(\lambda+2)y-(\lambda+2)z=9$ has infinitely many solutions, then $\lambda^{2}+\lambda$ is equal to:
Step-by-Step Solution
Key Concept: Set $D=0$ to get candidate values of $\lambda$, then check $D_{x}=0$ (or any other $D_{i}=0$) to pick the value(s) giving infinitely many solutions.
Compute $D=\begin{vmatrix}\lambda-1&\lambda-4&\lambda\\\lambda&\lambda-1&\lambda-4\\\lambda+1&\lambda+2&-(\lambda+2)\end{vmatrix}$. Expansion (after row/column reductions) gives $(\lambda-3)(2\lambda+1)=0$, so $\lambda\in\{3,-\tfrac{1}{2}\}.$
Now $D_{x}=\begin{vmatrix}5&\lambda-4&\lambda\\7&\lambda-1&\lambda-4\\9&\lambda+2&-(\lambda+2)\end{vmatrix}=0$ yields $(3-\lambda)(23-2\lambda)=0$, so $\lambda\in\{3,\,\tfrac{23}{2}\}.$
Common value: $\lambda=3.$ Hence $\lambda^{2}+\lambda=9+3=12.$
Correct Answer: 4