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If $straight A space equals open square brackets table row 2 cell space space space space space space space 3 end cell row 1 cell space space minus 4 end cell end table close square brackets comma space straight B space equals space open square brackets table row cell space space space 1 end cell cell space space space space minus 2 end cell row cell negative 1 end cell cell space space space space space space 3 end cell end table close square brackets comma$
verify that (AB)^–1 = B^–1 A^–1

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192.

Let $straight A space equals space open square brackets table row 3 cell space space space 7 end cell row 2 cell space space 5 end cell end table close square brackets space and space straight B space equals space open square brackets table row 6 cell space space 8 end cell row 7 cell space space 9 end cell end table close square brackets comma$ verify that (AB)^–1 = B^–1 A^–1.

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193. 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

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194. Verify that (AB)^–1 = B^–1 A^–1 for the matrices A and B where

straight A space equals 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 6 cell space space space space 7 end cell row 8 cell space space space space 9 end cell end table close square brackets

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195. 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 2 cell space space space 1 end cell row 5 cell space space space 3 end cell end table close square brackets comma space space space straight B space equals space open square brackets table row 4 cell space space space 5 end cell row 3 cell space space 4 end cell end table close square brackets

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Short Answer Type

196.

Find (AB)^–1 if $straight A space equals space open square brackets table row 3 cell space 4 end cell row 1 cell space 1 end cell end table close square brackets comma space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 4 cell space space 3 end cell row 2 cell space space 1 end cell end table close square brackets$

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197.

Find (AB)^–1 if $straight A space equals space open square brackets table row 5 cell space space 0 end cell row 2 cell space space space 3 end cell end table close square brackets comma space space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 1 cell space space space 2 end cell row 1 cell space space 4 end cell end table close square brackets$

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Long Answer Type

198.
Find the inverse of the matrix
$straight A space equals space open square brackets table row cell negative 1 end cell cell space space 1 end cell cell space space 2 end cell row cell space space space 3 end cell cell negative 1 end cell cell space space 1 space end cell row cell space minus 1 end cell cell space space 3 end cell cell space 4 end cell end table close square brackets.$

straight A space equals space open square brackets table row cell negative 1 end cell cell space space 1 end cell cell space space space space 2 end cell row cell space space 3 end cell cell negative 1 end cell cell space space space 1 end cell row cell negative 1 end cell cell space space 3 end cell cell space space 4 end cell end table close square brackets

therefore space space open vertical bar straight A close vertical bar space equals space open vertical bar table row cell negative 1 end cell cell space space 1 end cell cell space space 2 end cell row 3 cell space minus 1 end cell cell space space 1 end cell row cell negative 1 end cell 3 cell space space 4 end cell end table close vertical bar space equals space left parenthesis negative 1 right parenthesis space open vertical bar table row cell negative 1 end cell cell space space 1 end cell row cell space 3 end cell cell space space 4 end cell end table close vertical bar minus 1 space open vertical bar table row 3 cell space space 1 end cell row cell negative 1 end cell cell space space 4 end cell end table close vertical bar space plus space 2 space open vertical bar table row 3 cell space minus 1 end cell row cell negative 1 end cell cell space space space space 3 end cell end table close vertical bar space space space space space space space space space space space equals space minus left parenthesis 4 minus 3 right parenthesis space minus space left parenthesis 12 plus 1 right parenthesis space plus space 2 thin space left parenthesis 9 minus 1 right parenthesis space equals space 7 minus 13 plus 16 space equals space 10 space not equal to space 0 therefore space space space space straight A to the power of negative 1 end exponent space exists. space space space space space space space space space space space straight A subscript 11 space equals space open vertical bar table row cell negative 1 end cell cell space space 1 end cell row cell space space 3 end cell cell space space 4 end cell end table close vertical bar space equals space minus 4 minus 3 space equals space minus 7 space space space space space space space space space space space straight A subscript 12 space equals space minus open vertical bar table row 3 cell space space 1 end cell row cell negative 1 end cell cell space space 4 end cell end table close vertical bar space equals negative left parenthesis 12 plus 1 right parenthesis space equals space minus 13 space space space space space space space space space space space straight A subscript 13 space equals space open vertical bar table row 3 cell space space minus 1 end cell row 1 cell space space space space 3 end cell end table close vertical bar space equals space 9 minus 1 space equals space 8 space space space space space space space space space space straight A subscript 21 space equals space minus open vertical bar table row 1 cell space space 2 end cell row 3 cell space space 4 end cell end table close vertical bar space equals space minus left parenthesis 4 minus 6 right parenthesis space equals space 2 space space space space space space space space space space straight A subscript 22 space equals space open vertical bar table row cell negative 1 end cell cell space space 2 end cell row cell negative 1 end cell cell space 4 end cell end table close vertical bar space equals space minus 4 plus 2 space equals space minus 2 space space space space space space space space space space straight A subscript 23 space equals space minus open vertical bar table row cell negative 1 end cell cell space 1 space space end cell row cell negative 1 end cell 3 end table close vertical bar space equals space minus left parenthesis negative 3 plus 1 right parenthesis space equals space 2 space space space space space space space space space space space straight A subscript 31 space equals space open vertical bar table row 1 cell space space 2 end cell row cell negative 1 end cell cell space space 1 end cell end table close vertical bar space equals space 1 plus 2 space equals space 3 space space space space space space space space space space space straight A subscript 32 space equals space minus open vertical bar table row cell negative 1 end cell cell space 2 end cell row 3 1 end table close vertical bar space equals space minus left parenthesis negative 1 minus 6 right parenthesis space equals space 7 space space space space space space space space space space space space straight A subscript 33 space equals space open vertical bar table row cell negative 1 end cell cell space space 1 end cell row 3 cell space space minus end cell end table close vertical bar space equals space 1 minus 3 space equals space minus 2

therefore space space adj. space straight A space equals space open square brackets table row cell negative 7 end cell cell negative 13 end cell cell space space 8 end cell row 2 cell negative 2 end cell cell space space 2 end cell row 3 cell space 7 end cell cell negative 2 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 7 end cell cell space space 2 end cell cell space space 3 end cell row cell negative 13 end cell cell space space minus 2 end cell cell space space 7 end cell row 8 2 cell negative 2 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 1 over 10 open square brackets table row cell negative 7 end cell cell space space space 2 end cell cell space space space 3 end cell row cell negative 13 end cell cell space minus 2 end cell cell space space 7 end cell row 8 2 cell space minus 2 end cell end table close square brackets

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199. 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.

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200.

If $straight A space equals space open square brackets table row 0 cell space space space 0 end cell cell space space 1 end cell row 0 cell space space 1 end cell cell space space 0 end cell row 1 cell space space 0 end cell cell space space 0 end cell end table close square brackets comma$ show that A^–1 = A.

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198. Find the inverse of the matrix