Let A be the matrix Find A –1 and verify that where I is 2 × 2 unit matrix. - Wired Faculty

Question

Let A be the matrix $open square brackets table row 3 cell space space 8 end cell row 2 cell space space 1 end cell end table close square brackets.$ Find A ^–1 and verify that $straight A to the power of negative 1 end exponent space equals space 1 over 13 straight A space minus space 4 over 13 straight I$
where I is 2 × 2 unit matrix.

Solution

Here space straight A space equals space open square brackets table row 3 cell space space space 8 end cell row 2 cell space space 1 end cell end table close square brackets therefore 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 8 end cell row 2 cell space space space space space 1 end cell end table close vertical bar space equals space 3 minus 16 space equals space minus 13 space not equal to space 0 space space space space space rightwards double arrow 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 1, – 2 respectively
Co-factors of the elements of second row of | A | are – 8, 3 respectively
$therefore space space space space adj. space straight A space equals space open square brackets table row 1 cell space space space minus 2 end cell row cell negative 8 end cell cell 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 1 cell space space space space space minus 8 end cell row cell negative 2 end cell cell space 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 negative 1 over 13 open square brackets table row cell space space space 1 end cell cell space space space space space minus 8 end cell row cell negative 2 end cell cell space space space space space space space space 3 end cell end table close square brackets Now space 1 over 13 straight A minus 4 over 13 straight I space equals space 1 over 13 open square brackets table row 3 cell space space space 8 end cell row 2 cell space space space 1 end cell end table close square brackets minus 4 over 13 open square brackets table row 1 cell space space 0 end cell row 0 cell space space 1 end cell end table close square brackets space equals space 1 over 13 open square brackets table row 3 cell space 8 end cell row 2 cell space 1 end cell end table close square brackets minus 1 over 13 open square brackets table row 4 cell space 0 end cell row 0 cell space 4 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 space space space equals 1 over 13 open square brackets table row cell 3 minus 4 end cell cell space space space space 8 minus 0 end cell row cell 2 minus 0 end cell cell space space space space 1 minus 4 end cell end table close square brackets space equals space 1 over 13 open square brackets table row cell negative 1 end cell cell space space space 8 end cell row 2 cell space space minus 3 end cell end table close square brackets equals negative 1 over 13 open square brackets table row 1 cell space space minus 8 end cell row cell negative 2 end cell cell space space space space space 3 end cell end table close square brackets therefore space space space space space space straight A to the power of negative 1 end exponent space equals space 1 over 13 straight A minus 4 over 13 straight I.$

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