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. $b=2a=3$
• D. $b=3a\ne 10$

Answer 1

Answer: (B)

For system of linear equations has no solution,
$\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 2& a\end{array}\right|=0$
$\Rightarrow 1(2a-6)-1(a-3)+1(2-2)=0$
$\Rightarrow 2a-6-a+3=0$
$\Rightarrow a-3=0$
$\Rightarrow a=3$
Now, options (a) and (d) satisfy $a=3$. If $b=10$, then equations (ii) and (iii) become identical and system will have infınite solutions. Hence, $a=3$, $b\ne 10$