Find the inverse of the matrix:
$open square brackets table row cell space space space 2 end cell cell space space space space minus 1 end cell cell space space space space space 1 end cell row cell negative 1 end cell cell space space space space space space space 2 end cell cell space space minus 1 end cell row cell space space 1 end cell cell space space space space minus 1 end cell cell space space space space space 2 end cell end table close square brackets$

Solution

straight A space equals space open square brackets table row cell space 2 end cell cell space space space minus 1 end cell cell space space space space space space 1 end cell row cell negative 1 end cell cell space space space space space space space 2 end cell cell space space minus 1 end cell row cell space 1 end cell cell space space minus 1 end cell cell space space space space space space 2 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 1 end cell cell space space space space space space 1 end cell row cell negative 1 end cell cell space space space space space 2 end cell cell space space minus 1 end cell row 1 cell space minus 1 end cell cell space space space space 2 end cell end table close vertical bar space equals space 2 open vertical bar table row 2 cell space space space minus 1 end cell row cell negative 1 end cell cell space space space space space 2 end cell end table close vertical bar minus left parenthesis negative 1 right parenthesis open vertical bar table row cell negative 1 end cell cell space space minus 1 end cell row cell space space 1 end cell cell space space space space 2 end cell end table close vertical bar plus 1 open vertical bar table row cell negative 1 end cell cell space space space space space 2 end cell row cell space 1 end cell cell space space minus 1 end cell end table close vertical bar space space space space space space space space space space space space space space space equals 2 left parenthesis 4 minus 1 right parenthesis space plus 1 left parenthesis negative 2 plus 1 right parenthesis plus 1 left parenthesis 1 minus 2 right parenthesis space equals space 6 minus 1 minus 1 space equals space 4 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 the first row of | A | are

open vertical bar table row 2 cell space space space minus 1 end cell row cell negative 1 end cell cell space space space space space 2 end cell end table close vertical bar comma space space minus open vertical bar table row cell negative 1 end cell cell space space space minus 1 end cell row 1 cell space space space space space space 2 end cell end table close vertical bar comma space space space space open vertical bar table row cell negative 1 end cell cell space space space space space 2 end cell row 1 cell space space minus 1 end cell end table close vertical bar

i.e. 3, 1, – 1 respectively
Co-factors of the elements of the second row of | A | are
$negative open vertical bar table row cell negative 1 end cell cell space space 1 end cell row cell negative 1 end cell cell space space 2 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space 1 end cell row 1 cell space space space 2 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space minus 1 end cell row 1 cell space space space minus 1 end cell end table close vertical bar$
i.e. I, 3, I respectively
Co-factors of the elements of the third row of | A | are
$open vertical bar table row cell negative 1 end cell cell space space space space space space 1 end cell row 2 cell space space minus 1 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space space space space space space 1 end cell row cell negative 1 end cell cell space space space minus 1 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space space minus 1 end cell row cell negative 1 end cell cell space space space space space space space 2 end cell end table close vertical bar$
i.e. 1, 1, 3 respectively
$therefore space space space adj. space straight A space equals space open square brackets table row 3 cell space space space space 1 end cell cell space space space minus 1 end cell row 1 cell space space space space 3 end cell cell space space space space space 1 end cell row cell negative 1 end cell cell space space space space 1 end cell cell space space space space space 3 end cell end table close square brackets to the power of apostrophe space equals space space open square brackets table row 3 cell space space 1 end cell cell space space minus 1 end cell row 1 cell space space 3 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space space 1 end cell cell space space space space 3 end cell end table close square brackets Now space straight A to the power of negative 1 end exponent space equals space fraction numerator adj. space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 fourth open square brackets table row cell space space 3 end cell cell space space 1 end cell cell space space minus 1 end cell row cell space space 1 end cell cell space space 3 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space space space 1 end cell cell space space space 3 end cell end table close square brackets.$

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