For what value of $a$, the pair of linear equations $a x + 3y = a - 3$ and $12x + a y = a$ has no solution?
Step-by-Step Solution
Key Concept: $\\dfrac{a}{12} = \\dfrac{3}{a} \<br>eq \\dfrac{a-3}{a} \Rightarrow a^2 = 36 \Rightarrow a = \pm 6$. For $a = -6$, $\\dfrac{-6}{12} = \\dfrac{3}{-6} = -\\dfrac{1}{2}$, but $\\dfrac{-6-3}{-6} = \\dfrac{9}{6} = \\dfrac{3}{2} <br>eq -1/2$. So $a = -6$.
$a^2 = 36 \Rightarrow a = \pm 6$. [1.0 Mark]
For $a = -6$: $a_1/a_2 = -1/2, b_1/b_2 = -1/2, c_1/c_2 = 3/2$. Thus $a_1/a_2 = b_1/b_2
eq c_1/c_2$. Value is $a = -6$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding $a = \pm 6$: 1.0 Mark
Verifying $a = -6$ for no solution condition: 1.0 Mark
Correct Answer: