If $straight A space equals open square brackets table row 2 cell space space space space space space space 3 end cell row 5 cell space space space minus 2 end cell end table close square brackets$ , show that $straight A to the power of negative 1 end exponent space equals space 1 over 19 straight A$ .

Solution

straight A space equals space open square brackets table row 2 cell space space space space space space space 3 end cell row 5 cell space space space minus 2 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 2 cell space space space space space space 3 end cell row 5 cell space space space space minus 2 end cell end table close vertical bar space equals negative 4 minus 15 space equals space minus 19 space not equal to 0 space 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 – 2, – 5 respectively
Co-factors of the elements of second row of | A | are -3, 2 respectively

therefore space space space adj. space straight A space equals space open square brackets table row cell negative 2 end cell cell space space space space minus 5 end cell row cell negative 3 end cell 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 2 end cell cell space space space space space minus 3 end cell row cell negative 5 end cell cell 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 19 open square brackets table row cell negative 2 end cell cell space space space space space minus 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 1 over 19 open square brackets table row 2 cell space space space space space space space 3 end cell row 5 cell space space space minus 2 end cell end table close square brackets space equals space 1 over 19 straight A therefore space space space straight A to the power of negative 1 end exponent space equals space 1 over 19 straight A.

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