Question ID: #252
If the system of equations
$(\lambda-1)x+(\lambda-4)y+\lambda z=5$
$\lambda x+(\lambda-1)y+(\lambda-4)z=7$
$(\lambda+1)x+(\lambda+2)y-(\lambda+2)z=9$
has infinitely many solutions, then $\lambda^{2}+\lambda$ is equal to
- (1) 10
- (2) 12
- (3) 6
- (4) 20
Solution:
For infinitely many solutions, $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$.
$$\Delta = \begin{vmatrix}\lambda-1&\lambda-4&\lambda\\ \lambda&\lambda-1&\lambda-4\\ \lambda+1&\lambda+2&-(\lambda+2)\end{vmatrix} = 0$$
Solving $\Delta = 0 \implies (\lambda-3)(2\lambda+1)=0 \implies \lambda = 3, -\frac{1}{2}$.
Now check $\Delta_x = 0$:
$$\Delta_x = \begin{vmatrix}5&\lambda-4&\lambda\\ 7&\lambda-1&\lambda-4\\ 9&\lambda+2&-(\lambda+2)\end{vmatrix} = 0$$
Solving $\Delta_x = 0 \implies 2(3-\lambda)(23-2\lambda)=0 \implies \lambda = 3, \frac{23}{2}$.
The common value is $\lambda = 3$.
Required value:
$$\lambda^2 + \lambda = (3)^2 + 3 = 9 + 3 = 12$$
Ans. (2)
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