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Maths-

General

Easy

Question

If A = $open square brackets table row 0 1 2 row 1 2 3 row 3 a 1 end table close square brackets$ and A^–1 = $open square brackets table row cell 1 divided by 2 end cell cell negative 1 divided by 2 end cell cell 1 divided by 2 end cell row cell negative 4 end cell 3 c row cell 5 divided by 2 end cell cell negative 3 divided by 2 end cell cell 1 divided by 2 end cell end table close square brackets$ then value of a and c is -

1,1
1, – 1
1,2
– 1,1

The correct answer is: 1, – 1

I = AA^–1 = $fraction numerator 1 over denominator 2 end fraction$ $open square brackets table row 0 1 2 row 1 2 3 row 3 a 1 end table close square brackets$ $open square brackets table row 1 cell negative 1 end cell 1 row cell negative 8 end cell 6 cell 2 c end cell row 5 cell negative 3 end cell 1 end table close square brackets$
= $open square brackets table row 1 0 cell c plus 1 end cell row 0 1 cell 2 left parenthesis c plus 1 right parenthesis end cell row cell 4 left parenthesis 1 minus a right parenthesis end cell cell 3 left parenthesis a minus 1 right parenthesis end cell cell 2 plus a c end cell end table close square brackets$
comparing the elements of AA^–1 with those of I,
c + 1 = 0 $rightwards double arrow$ c = –1
and a – 1 = 0 $rightwards double arrow$ a = 1

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