Question

The system $\left[\begin{array}{ccc}1& -1& 2\\ 3& 5& -3\\ 2& 6& a\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}3\\ b\\ 2\end{array}\right]$ has no solution if

• A. $a\ne -5,b\ne 5$
• B. $a=-5,b\ne 5$
• C. $a=-5,b=5$
• D. $a\ne -5,b=5$

Answer 1

The correct option is B $a=-5,b\ne 5$

For the system of equations to have no solution,
$\Delta =0$
$\Rightarrow a=-5$
and at least one of ${\Delta }_1,{\Delta }_2,{\Delta }_3$ is non-zero.
${\Delta }_3=\left|\begin{array}{ccc}1& -1& 3\\ 3& 5& b\\ 2& 6& 2\end{array}\right|\ne 0$

$\Rightarrow b\ne 5$

Therefore, the system has no solution if $a=-5,b\ne 5$