Question

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

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

Answer 1

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

For given system of linear equations which has no solution,
Case 1:
$△=0$
$\left|\begin{array}{ccc}1& a& 1\\ 1& 2& 2\\ 1& 5& 3\end{array}\right|=0$
$\Rightarrow 1(6-10)-a(3-2)+1(5-2)=0$
$\Rightarrow a=-1$

Case 2:
${△}_1\ne 0$
$\left|\begin{array}{ccc}1& -1& 3\\ 1& 2& 6\\ 1& 5& b\end{array}\right|\ne 0$
$\Rightarrow 1(2b-30)+1(b-6)+3(5-2)\ne 0$
$\Rightarrow b\ne 9$
Hence, $a=-1,b\ne 9$