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Standard X

Mathematics

Finite and Infinite Sets

If the system...

Question

If the system of equation, $a^{2} x - a y = 1 - a$ & $b x + (3 - 2 b) y = 3 + a$ possesses a unique solution $x = 1$ , $y = 1$ then:

a = 1

b = 1

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a = - 1

b = 1

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a = 0

b = 0

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None of the above

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Solution

The correct option is B $a = - 1$ , $b = 1$
$a^{2} x - a y = 1 - a$
$b x + (3 - 2 b) y = 3 + a$
put $x = 1, y = 1$
$a^{2} - a = 1 - a$
$a^{2} = 1$
$a = \pm 1$
$b + 3 - 2 b = + a$
$- b = a$
$b = - a$
So, $b = \mp 1$
$(- 1, 1)$ or $(1, - 1)$

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If the system of equation, a2x−ay=1−a & bx+(3−2b)y=3+a possesses a unique solution x=1, y=1 then:

The correct option is B a=−1, b=1a2x−ay=1−abx+(3−2b)y=3+aput x=1,y=1a2−a=1−aa2=1a=±1b+3−2b=+a−b=ab=−aSo, b=∓1(−1,1) or (1,−1)

If the system of equation, $a^{2} x - a y = 1 - a$ & $b x + (3 - 2 b) y = 3 + a$ possesses a unique solution $x = 1$ , $y = 1$ then:

The correct option is B $a = - 1$ , $b = 1$
$a^{2} x - a y = 1 - a$
$b x + (3 - 2 b) y = 3 + a$
put $x = 1, y = 1$
$a^{2} - a = 1 - a$
$a^{2} = 1$
$a = \pm 1$
$b + 3 - 2 b = + a$
$- b = a$
$b = - a$
So, $b = \mp 1$
$(- 1, 1)$ or $(1, - 1)$