Question

What is the value of a for which the system of linear equations ax + 3y = a - 3 ; 12x + ay = a has no solution

• A. $a=6$
• B. $a=-3$
• C. $a=3$
• D. $a=-6$

Answer 1

Answer: (D)

$a=-6$

$ax+3y=a-3$
$\to ax+3y-a+3=0$
$a_1=a,b_1=3,c_1=-a+3$
$12x+ay=a$
$\to 12x+ay-a=0$
$a_2=12,b_2=a,c_2=-a$
Condition for no solution is $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_1}$
$\frac{a}{12}=\frac{3}{a}\ne \frac{a-3}{a}$
$\to a^2-36$
$\to a=\pm 6$
$a=+6$ is not possibel because it gives
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\therefore a=-6$ is the answer