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Determinants

Question
CBSEENMA12034569
Wired Faculty App

Find the inverse of  straight A space equals space open square brackets table row 3 cell space space space space space space space 5 end cell row 7 cell space space minus 11 end cell end table close square brackets and verify that AA–1 = I.

Solution
straight A space equals space open square brackets table row 3 cell space space space space space space space 5 end cell row 7 cell space space space minus 11 end cell end table close square brackets therefore space space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 3 cell space space space space space space space space 5 end cell row 7 cell space space space minus 11 end cell end table close vertical bar space equals space minus 33 minus 35 space equals space minus 68 space not equal to 0 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 | A | are – 11, – 7 respectively.
Co-factors of the elements of second row of | A | are – 5, 3 respectively.
therefore space space space space space adj. space straight A space equals space open square brackets table row cell negative 11 end cell cell space space space space minus 7 end cell row cell negative 5 end cell cell space space space space space space 3 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 11 end cell cell space space space space minus 5 end cell row cell negative 7 end cell cell space space space space space space 3 end cell end table close square brackets therefore space 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 minus 1 over 68 open square brackets table row cell negative 11 end cell cell space space space space minus 5 end cell row cell negative 7 end cell cell space space space space space space space 3 end cell end table close square brackets Now comma space space space space AA to the power of negative 1 end exponent space equals space minus 1 over 68 open square brackets table row 3 cell space space space space space space space space 5 end cell row 7 cell space space space minus 11 end cell end table close square brackets space open square brackets table row cell negative 11 end cell cell space space space minus 5 end cell row cell negative 7 end cell cell space space space space space 3 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 equals space minus 1 over 68 open square brackets table row cell negative 33 minus 35 end cell cell space space space space space minus 15 plus 15 end cell row cell negative 77 plus 77 end cell cell space space space space space space space minus 35 minus 33 end cell end table close square brackets space equals space minus 1 over 68 open square brackets table row cell negative 68 end cell cell space space space space 0 end cell row 0 cell space space minus 68 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 equals 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 space equals space straight I therefore space space space space space AA to the power of negative 1 end exponent space space equals space space straight I space is space verified.
 

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