Question

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

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

Answer 1

The correct option is B

$a=3b\ne 10$

$\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 2& a\end{array}\right|=0\Rightarrow a=3$

$2^{nd}$ and $3^{rd}$ equations are same for a = 3. So there is no solution if constant is different i.e b $\ne$ 10