Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

$x+y+z=\alpha$
$\alpha x+2\alpha y+3z=-1$
$x+3\alpha y+5z=4$
Has inconsistent solution
$D=\left|\begin{array}{ccc}1& 1& 1\\ \alpha & 2\alpha & 3\\ 1& 3\alpha & 5\end{array}\right|=0$
$\Rightarrow (\alpha -1{)}^2=0$
$\alpha =1$
For $\alpha =1$
$D_1=\left|\begin{array}{ccc}1& 1& 1\\ -1& 2& 3\\ 4& 3& 5\end{array}\right|$
$=(10-9)-(-5-12)+(-3-8)$
$=1+17-11\ne 0$
For $\alpha =1$ the system of equation has Inconsistent solution