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Maths-

General

Easy

Question

Let A $open square brackets table row 1 2 row 3 4 end table close square brackets$ and B = $open square brackets table row a b row c d end table close square brackets$ are two matrices such that AB = BA and c ≠ 0, then value of $fraction numerator a minus d over denominator 3 b minus c end fraction$ is

0
2
–2
– 1

The correct answer is: – 1

AB = $open square brackets table row 1 2 row 3 4 end table close square brackets$ $open square brackets table row a b row c d end table close square brackets$ = $open square brackets table row cell a plus 2 c end cell cell b plus 2 d end cell row cell 3 a plus 4 c end cell cell 2 c plus 4 d end cell end table close square brackets$
BA = $open square brackets table attributes columnalign left left columnspacing 1em end attributes row cell a b end cell row cell c d end cell end table close square brackets$ $open square brackets table row 1 2 row 3 4 end table close square brackets$ = $open square brackets table row cell a plus 3 b end cell cell 2 a plus 4 b end cell row cell c plus 3 d end cell cell 2 c plus 4 d end cell end table close square brackets$
if AB = BA, then a + 2c = a + 3b
$rightwards double arrow$ 2c = 3b $rightwards double arrow$ b ≠ 0
b + 2d = 2a + 4b
$rightwards double arrow$ 2a – 2d = – 3b
$fraction numerator a minus d over denominator 3 b minus c end fraction equals fraction numerator negative fraction numerator 3 over denominator 2 end fraction b over denominator 3 b minus fraction numerator 3 over denominator 2 end fraction b end fraction$ = – 1

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