Find the sum of $open square brackets table row 2 cell space space space space minus 3 end cell row 5 cell space space space minus 7 end cell end table close square brackets$ and its multiplication inverse.

Solution

Let $straight A space equals space open square brackets table row 2 cell space space space space minus 3 end cell row 5 cell space space space minus 7 end cell end table close square brackets$
$therefore space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space minus 3 end cell row 5 cell space space space space minus 7 end cell end table close vertical bar space equals space minus 14 plus 15 space equals space 1 space not equal to 0 therefore space space space space straight A to the power of negative 1 end exponent space exists. space$

Co-factors of the elements of first row of | A | are – 7, – 5 respectively.
Co-factors of the elements of second row of | A | are 3, 2 respectively.
$space therefore space space space space adj. space straight A space equals space open square brackets table row cell negative 7 end cell cell space space space minus 5 end cell row 3 cell space space space space space space 2 end cell end table close square brackets space equals space open square brackets table row cell negative 7 end cell cell space space space space space 3 end cell row cell negative 5 end cell cell space space space space space 2 end cell end table close square brackets space space space space space space space 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 1 over 1 open square brackets table row cell negative 7 end cell cell space space space space 3 end cell row cell negative 5 end cell cell space space space space 2 end cell end table close square brackets space equals space open square brackets table row cell negative 7 end cell cell space space space space space 3 end cell row cell negative 5 end cell cell space space space space space 2 end cell end table close square brackets space therefore space space space space straight A plus straight A to the power of negative 1 end exponent space equals space open square brackets table row 2 cell space space space space space space minus 3 end cell row 5 cell space space space space space space space minus 7 end cell end table close square brackets plus space open square brackets table row cell negative 7 end cell cell space space space space 3 end cell row cell negative 5 end cell cell space space space 2 end cell end table close square brackets space equals space open square brackets table row cell 2 minus 7 end cell cell space space minus 3 plus 3 end cell row cell 5 minus 5 end cell cell space space minus 7 plus 2 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 equals space open square brackets table row cell negative 5 end cell cell space space space space space space space space 0 end cell row 0 cell space space space space minus 5 end cell end table close square brackets$

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Determinants

Find the sum of $open square brackets table row 2 cell space space space space minus 3 end cell row 5 cell space space space minus 7 end cell end table close square brackets$ and its multiplication inverse.

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