Question

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

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

Answer 1

Answer: (B)

For system of linear equations has no solution,
$\begin{vmatrix}1 & 1 & 1 \\1 & 2 & 3 \\1 & 2 & a\end{vmatrix}=0$
$\Rightarrow 1(2 a-6)-1(a-3)+1(2-2)=0$
$\Rightarrow 2 a-6-a+3=0$
$\Rightarrow a-3=0$
$\Rightarrow a=3$
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, $a=3$, $b \neq 10$