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Maths-

General

Easy

Question

The value of x for which the matrix A = $open square brackets table row cell 2 divided by x end cell cell negative 1 end cell 2 row 1 x cell 2 x to the power of 2 end exponent end cell row 1 cell 1 divided by x end cell 2 end table close square brackets$ is singular is -

± 1
± 2
± 3
None of these

The correct answer is: ± 1

We have
|A| = $open parentheses fraction numerator 2 over denominator x end fraction close parentheses$ $open vertical bar table row x cell 2 x to the power of 2 end exponent end cell row cell 1 divided by x end cell 2 end table close vertical bar$ + $open vertical bar table row 1 cell 2 x to the power of 2 end exponent end cell row 1 2 end table close vertical bar$ + 2 $open vertical bar table row 1 x row 1 cell 1 divided by x end cell end table close vertical bar$
= $fraction numerator 2 over denominator x end fraction$ (0) + 2 –2x² + 2 $open parentheses fraction numerator 1 over denominator x end fraction minus x close parentheses$
= $fraction numerator 2 x left parenthesis 1 minus x to the power of 2 end exponent right parenthesis plus 2 left parenthesis 1 minus x to the power of 2 end exponent right parenthesis over denominator x end fraction$ = $fraction numerator 2 left parenthesis x plus 1 right parenthesis to the power of 2 end exponent left parenthesis 1 minus x right parenthesis over denominator x end fraction$
Now, |A| = 0 $rightwards double arrow$ x = ± 1

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