Question

If the system of linear equations
$x+ay+z=3$
$x+2y+2z=6$
$x+5y+3z=b$
has no solution, then :

• A. $a=-1,b\ne 91$
• B. $a=-1,b\ne 9$
• C. $a\ne -1,b=9$
• D. $a=1,b\ne 9$

Answer 1

Answer: (D)

$a=1,b\ne 9$

If the system of equations has no solution then

$\Delta =0$ and at least one of ${\Delta }_1,{\Delta }_2$ and ${\Delta }_3$ is not zero.

$\Delta =\left|\begin{array}{ccccc}1& & a& & 1\\ 1& & 2& & 2\\ 1& & 5& & 3\end{array}\right|=0$

$⟹-a-1=0⟹a=-1$

${\Delta }_2=\left|\begin{array}{ccccc}1& & 1& & 3\\ 1& & 2& & 6\\ 1& & 3& & b\end{array}\right|\ne 0$

$⟹b\ne 9$