Question

The values of $\alpha$, for which the system of equation
$x+y+z=1$
$x+2y+4z=\alpha$
$x+4y+10z=\alpha$ is consistent, are

• A. $1,-2$
• B. $-1,2$
• C. $1,2$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

$x+y=1-z$

$x+2y=1-z+y=\alpha -4z$

$⟹y=\alpha -1-3z$

$⟹x=1-z-(\alpha -1-3z)=2+2z-\alpha$

Apply above values on Third equation

$(2+2z-\alpha )+4(\alpha -1-3z)+10z=\alpha$

$-2-10z+3\alpha +10z=\alpha$

$⟹\alpha =1$

only $\alpha =1$ satisfies the consistency

So answer is Option D