Question

If the system of linear equations $x+ay+z=2,x+2y+2z=3,x+5y+3z=4$ has no solution, then

• A. $a=1,b\ne 9$
• B. $a\ne -1,b=9$
• C. $a=-1,b=9$
• D. $a=-1,b\ne 9$

Answer 1

The correct option is A $a=1,b\ne 9$

We have,

$x+ay+z=2.......\left(1\right)$

$x+2y+2z=3.......\left(2\right)$

$x+5y+3z=4......\left(3\right)$

has no solution.

Then,

We know that

$\Rightarrow 1(6-10)-a(3-2)+1(5-2)=0$

$\Rightarrow -4-a+3=0$

$\Rightarrow -a-1=0$

$\Rightarrow a=-1$

Then, ${\Delta }_1,{\Delta }_2,{\Delta }_3$ is non zero.