Question

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

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

Answer 1

Answer: (B)

Given: $x+ay+z=3,x+2y+2z=6,x+5y+3z=b$
$\Delta =\left|\begin{array}{ccc}1& a& 1\\ 1& 2& 2\\ 1& 5& 3\end{array}\right|=1(6-10)-a(3-2)+1(5-2)$
$=-4-a+z=-(a+1)$
${\Delta }_1=\left|\begin{array}{ccc}3& a& 1\\ 6& 2& 2\\ b& 5& 3\end{array}\right|=3(6-10)-a(18-2b)+(30-2b)$
$=-12-18a+2ab+30-2b$
$=18-2b-18a+2ab$
$=18(1-a)-2b(1-a)$
$=2(g-b)(1-a)$
${\Delta }_2=\left|\begin{array}{ccc}1& 3& 1\\ 1& 6& 2\\ 1& b& 3\end{array}\right|=1(18-2b)-3(3-2)+1(b-6)$
$=18-2b-3+b-6$
$=(g-b)$
${\Delta }_3=\left|\begin{array}{ccc}1& a& 3\\ 1& 2& b\\ 1& 5& b\end{array}\right|=1(2b-30)-a(b-6)+3(5-2)$
$=2b-30-ab+6a+9$
$=b(2-a)-3(7-2a)$
For no solution, $\Delta =0$ and at least one of ${\Delta }_1,{\Delta }_2$ and ${\Delta }_3$ is non zero.
$\Delta =0\Rightarrow a=-1$
${\Delta }_1=0\Rightarrow a=1,b=9$
${\Delta }_2=0\Rightarrow b=9$
Therefore, for no solution, $a=-1$ and $b\ne 9$.