If A = $open square brackets table row 2 cell space space space 5 end cell row 1 cell space space space 6 end cell end table close square brackets comma$ find $straight A to the power of negative 1 end exponent$ and verify that $straight A to the power of negative 1 end exponent space equals space minus 1 over 7 straight A plus 8 over 7 straight I$

Solution

Here $straight A space equals space open square brackets table row 2 cell space space space 5 end cell row 1 cell space space space 6 end cell end table close square brackets$
$therefore space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space space space 5 end cell row 1 cell space space space space space space 6 end cell end table close vertical bar space equals space 12 minus 5 space equals space 7 space not equal to space 0 space space space space space space rightwards double arrow space space space space straight A to the power of negative 1 end exponent space exists$
Co-factors of the elements of first row of | A | are 6, – 1 respectively
Co-factors of the elements of second row of | A | are – 5, 2 respectively
$therefore space space space adj. space straight A space equals space open square brackets table row 6 cell space space space space space minus 1 end cell row cell negative 5 end cell cell space space space space space space space space 2 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 6 cell space space space space space space minus 5 end cell row cell negative 1 end cell cell space space space space space space space space space 2 end cell end table close square brackets therefore space space space space straight A to the power of negative 1 end exponent space equals space fraction numerator adj. space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 over 7 open square brackets table row 6 cell space space space space minus 5 end cell row cell negative 1 end cell cell space space space space space space 2 end cell end table close square brackets Now comma space minus 1 over 7 straight A plus 8 over 7 straight I space equals space minus 1 over 7 open square brackets table row 2 cell space space space 5 end cell row 1 cell space space space space 6 end cell end table close square brackets plus space 8 over 7 open square brackets table row 1 cell space space space 0 end cell row 0 cell space space space 1 end cell end table close square brackets space equals space 1 over 7 open square brackets table row cell negative 2 end cell cell space space minus 5 end cell row cell negative 1 end cell cell negative 6 end cell end table close square brackets plus 1 over 7 open square brackets table row 8 cell space 0 end cell row 0 cell space space space 8 space end cell end table close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 over 7 open square brackets table row cell negative 2 plus 8 end cell cell space space space space space space space space space minus 5 plus 0 end cell row cell negative 1 plus 0 end cell cell space space space space space space space space minus 6 plus 8 end cell end table close square brackets space equals space 1 over 7 open square brackets table row cell space space space 6 end cell cell space space space space minus 5 end cell row cell negative 1 end cell cell space space space space space space 2 end cell end table close square brackets therefore space space space straight A to the power of negative 1 end exponent space equals space minus 1 over 7 straight A plus 8 over 7 straight I.$

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