If the system of linear equations $(a - 1) x + 3 y = 2, 6 x + (1 - 2 b) y = 6$ has an infinite number of solutions, then a and b respectively are

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$3, - 4$

$- 4, - 3$

$4, - 3$

$3, 4$

answer is A.

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Detailed Solution

Given equation are $(a - 1) x + 3 y = 2, 6 x + (1 - 2 b) y = 6$

Here

$a_{1} = a - 1, b_{1} = 3, c_{1} = - 2,$

$a_{2} = 6, b_{2} = (1 - 2 b), c_{2} = - 6$

Given equations has infinite number of solutions

Then $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$\frac{a - 1}{6} = \frac{3}{1 - 2 b} = \frac{- 2}{- 6}$

$\frac{a - 1}{6} = \frac{3}{1 - 2 b} = \frac{1}{3}$

$\frac{a - 1}{6} = \frac{1}{3}; \frac{3}{1 - 2 b} = \frac{1}{3}$

$a - 1 = 2$

$a = 3$

$1 - 2 b = 9$

$2 b = - 8$

$b = - 4$

$∴ a = 3; b = - 4$

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