Question

The number of values of $\alpha$ for which the system of equations:
$x+y+z=\alpha $
$\alpha x+2 \alpha y+3 z=-1$
$x+3 \alpha y+5 z=4$
is inconsistent, is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

$x+y+z=\alpha $
$\alpha x+2 \alpha y+3 z=-1 $
$x+3 \alpha y+5 z=4$
Has inconsistent solution
$D=\begin{vmatrix}1 & 1 & 1 \\ \alpha & 2 \alpha & 3 \\ 1 & 3 \alpha & 5\end{vmatrix}=0$
$\Rightarrow(\alpha-1)^{2}=0$
$\alpha=1$
For $\alpha=1$
$D_{1}=\begin{vmatrix}1 & 1 & 1 \\ -1 & 2 & 3 \\ 4 & 3 & 5\end{vmatrix}$
$=(10-9)-(-5-12)+(-3-8)$
$=1+17-11 \neq 0$
For $\alpha=1$ the system of equation has Inconsistent solution