Verify that (AB)^–1 = B^–1 A^–1 for the matrices A and B where $straight A space equals space open square brackets table row 3 cell space space 2 end cell row 7 cell space space 5 end cell end table close square brackets comma space space straight B space equals space open square brackets table row 4 cell space space 6 end cell row 3 cell space space 2 end cell end table close square brackets$

Solution

straight A space equals space open square brackets table row 3 cell space space space 2 end cell row 7 cell space space 5 end cell end table close square brackets comma space straight B space equals space open square brackets table row 4 cell space space space space 6 end cell row 3 cell space space space space 2 end cell end table close square brackets

open vertical bar straight A close vertical bar space equals open vertical bar table row 3 cell space space space 2 end cell row 7 cell space space space 5 end cell end table close vertical bar space equals space 15 minus 14 space equals space 1

Co-factors of the elements of first row of | A | are 5, – 7 respectively
Co-factors of the elements of second row of | A | are – 2, 3 respectively

therefore space space space space adj. space straight A space equals space open square brackets table row 5 cell space space minus 7 end cell row cell negative 2 end cell cell 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 5 cell space space space minus 2 end cell row cell negative 7 end cell cell space space space space space 3 end cell end table close square brackets 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 open square brackets table row 5 cell space space space minus 2 end cell row cell negative 7 end cell cell space space space space space space 3 end cell end table close square brackets open vertical bar straight B close vertical bar space equals space open vertical bar table row 4 cell space space space space 6 end cell row 3 cell space space space space 2 end cell end table close vertical bar space equals space 8 minus 18 space equals space minus 10

Co-factors of the elements of first row of | B | are 2, – 3 respectively
Co-factors of the elements of second row of | B | are – 6, 4 respectively
$therefore space space space space adj. space straight B space equals space open square brackets table row 2 cell space space space minus 3 end cell row cell negative 6 end cell cell space space space space space 4 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 2 cell space space space space minus 6 end cell row cell negative 3 end cell cell space space space space space space 4 end cell end table close square brackets therefore space space space space space straight B to the power of negative 1 end exponent space equals space fraction numerator adj. space straight B over denominator open vertical bar straight B close vertical bar end fraction space equals space minus 1 over 10 open square brackets table row cell space space 2 end cell cell space space space space minus 6 end cell row cell negative 3 end cell cell space space space space space 4 end cell end table close square brackets therefore space space space space straight B to the power of negative 1 end exponent straight A to the power of negative 1 end exponent space equals space minus 1 over 10 open square brackets table row 2 cell space space space minus 6 end cell row cell negative 3 end cell cell space space space space space 4 end cell end table close square brackets space open square brackets table row 5 cell space space space minus 2 end cell row cell negative 7 end cell cell space space space space space 3 end cell end table close square brackets space equals space minus 1 over 10 open square brackets table row cell 10 plus 42 end cell cell space minus 4 minus 18 end cell row cell negative 15 minus 28 end cell cell 6 plus 12 end cell end table close square brackets therefore space space space straight B to the power of negative 1 end exponent straight A to the power of negative 1 end exponent space equals space minus 1 over 10 open square brackets table row 52 cell space space space minus 22 end cell row cell negative 43 end cell cell space space space space 18 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... left parenthesis 1 right parenthesis space space space space space space space space space space AB space equals space open square brackets table row 3 cell space space space 2 end cell row 7 cell space space space 5 end cell end table close square brackets space open square brackets table row 4 cell space space 6 end cell row 3 cell space 2 end cell end table close square brackets space equals open square brackets table row cell 12 plus 6 end cell cell space space space 18 plus 4 end cell row cell 28 plus 15 end cell cell space space space 42 plus 10 end cell end table close square brackets space equals space open square brackets table row 18 cell space 22 end cell row 43 cell space 52 end cell end table close square brackets open vertical bar AB close vertical bar space equals space open vertical bar table row 18 cell space space space 22 end cell row 43 cell space space space 52 end cell end table close vertical bar space equals space 936 minus 946 space equals space minus 10$
Co-factors of the elements of first row of | AB | are 52, – 43 respectively.
Co-factors of the elements of second row of | AB | are – 22, 18 respectively.
$therefore space space adj. space left parenthesis AB right parenthesis space equals space open square brackets table row 52 cell space space space minus 43 end cell row cell negative 22 end cell cell space space space space space 18 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 52 cell space space minus 22 end cell row cell negative 43 end cell cell space space space 18 end cell end table close square brackets therefore space space space left parenthesis AB right parenthesis to the power of negative 1 end exponent space equals space fraction numerator adj. space left parenthesis AB right parenthesis over denominator open vertical bar AB close vertical bar end fraction space equals space minus 1 over 10 open square brackets table row 52 cell space space minus 22 end cell row cell negative 43 end cell cell space space space space 18 end cell end table close square brackets space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis$
From (1) and (2), (AB)^–1 B^–1 A^–1.

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