Question

The system of linear equations $x+y+z=6,x+2y+3z=10$ and $x+2y+az=b$ has no solutions when

• A. $a=2,b\ne 3$
• B. $a=3,b\ne 10$
• C. $b=2,a=3$
• D. $b=3,a\ne 10$

Answer 1

The correct option is B $a=3,b\ne 10$

The given systems of linear equations has no solutions if $\Delta =0.$
$\text{Consider}\Delta =0$
$\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 2& a\end{array}\right|=0\Rightarrow a=3$

For $a=3,$ the last equation becomes
$x+2y+3z=b$
If $b=10,$ the second and the third equation coincide and hence, there will be infinite solutions.
So, $b\ne 10$