Find the inverse of the matrix:
$open square brackets table row 1 cell space space space minus 1 end cell cell space space space space space space 2 end cell row 0 cell space space space space space space 2 end cell cell space space space minus 3 end cell row 3 cell space space space minus 2 end cell cell space space space space space space 4 end cell end table close square brackets.$

Solution

Let $straight A space equals space open square brackets table row 1 cell space space minus 1 end cell cell space space space space space space 2 end cell row 0 cell space space space space space 2 end cell cell space space minus 3 end cell row 3 cell space space minus 2 end cell cell space space space space 4 end cell end table close square brackets$
$therefore space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space minus 1 end cell cell space space space space space space 2 end cell row 0 cell space space space space space 2 end cell cell space space minus 3 end cell row 3 cell space space space minus 2 end cell cell space space space space space 4 end cell end table close vertical bar space equals space 1 space open vertical bar table row cell space space 2 end cell cell space space minus 3 end cell row cell negative 2 end cell cell space space space space space 4 end cell end table close vertical bar minus left parenthesis negative 1 right parenthesis space open vertical bar table row 0 cell space space minus 3 end cell row 3 cell space space space space space 4 end cell end table close vertical bar space plus 2 space open vertical bar table row 0 cell space space space space space space 2 end cell row 3 cell space space minus 2 end cell end table close vertical bar space space space space space space space space space space space space space space equals space 1 left parenthesis 8 minus 6 right parenthesis plus 1 left parenthesis 0 plus 9 right parenthesis plus 2 left parenthesis 0 minus 6 right parenthesis space equals space 2 plus 9 plus 12 space equals space minus 1 space not equal to space 0. therefore space space space straight A to the power of negative 1 end exponent space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row cell space space 2 end cell cell space space space minus 3 end cell row cell negative 2 end cell cell space space space space space space 4 end cell end table close vertical bar comma space space space space space minus open vertical bar table row 0 cell space space space minus 3 end cell row 3 cell space space space space space space 4 end cell end table close vertical bar comma space space open vertical bar table row 0 cell space space space space space space space 2 end cell row 3 cell space space space minus 2 end cell end table close vertical bar$
i.e. 2, –9, — 6 respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row cell negative 1 end cell cell space space space 2 end cell row cell negative 2 end cell cell space space 4 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space space space space 2 end cell row 3 cell space space space space space space space 4 end cell end table close vertical bar comma space space space space minus open vertical bar table row 1 cell space space space space minus 1 end cell row 3 cell space space space space space minus space 2 end cell end table close vertical bar$
i.e. 0, – 2, – 1 respectively.
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell negative 1 end cell cell space space space space space space space space 2 end cell row cell space 2 end cell cell space space space minus 3 end cell end table close vertical bar comma space space space space space minus open vertical bar table row 1 cell space space space space space space space 2 end cell row 0 cell space space space minus 3 end cell end table close vertical bar comma space space space space space open vertical bar table row 1 cell space space space minus 1 end cell row 0 cell space space space space space 2 end cell end table close vertical bar$
i.e. – 1, 3, 2 respectively.
$therefore space space space adj. space straight A space equals space open square brackets table row 2 cell space space space minus 9 end cell cell space space space space minus 6 end cell row 0 cell space space minus 2 end cell cell space space space minus 1 end cell row cell negative 1 end cell cell space space space 3 end cell cell space space space space space space 2 end cell end table close square brackets space equals space open square brackets table row cell space space 2 end cell cell space space space space space space 0 end cell cell space space space minus 1 end cell row cell negative 9 end cell cell space space space minus 2 end cell cell space space space space 3 end cell row cell negative 6 end cell cell space space space minus 1 end cell cell space space space space 2 end cell end table close square brackets therefore space space straight A to the power of negative 1 end exponent space equals space fraction numerator adj. space straight A over denominator open vertical bar straight A close vertical bar end fraction equals space fraction numerator 1 over denominator negative 1 end fraction space open square brackets table row cell space space space 2 end cell cell space space space space space space 0 end cell cell space space minus 1 end cell row cell negative 9 end cell cell space space space minus 2 end cell cell space space space space 3 end cell row cell negative 6 end cell cell space space space minus 1 end cell cell space space space space space 2 end cell end table close square brackets space equals space open square brackets table row cell negative 2 end cell cell space space space 0 end cell cell space space space space space 1 end cell row 9 cell space space 2 end cell cell space space minus 3 end cell row 6 cell space space space 1 end cell cell space space space minus 2 end cell end table close square brackets$

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