If   then A3 = A–1 What is adj. A ? from Mathematics Det

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 Multiple Choice QuestionsLong Answer Type

211.

Let A = open square brackets table row 4 cell space space space minus 6 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space minus 1 end cell cell space space 1 end cell row cell negative 4 end cell cell space space space 11 end cell cell negative 1 end cell end table close square brackets.
Show that A is inevitable. Find adj. A and A Also verify that AA–1 = A–1 A = I.

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212.

Verify AA–1 = A–1 A = I where
straight A space equals open square brackets table row 0 cell space space space 0 end cell cell space space minus 1 end cell row 3 cell space space space space 4 end cell cell space space space space space 5 end cell row cell negative 2 end cell cell negative 4 end cell cell space space minus 7 end cell end table close square brackets.

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 Multiple Choice QuestionsShort Answer Type

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213.

If straight A space equals space open square brackets table row 3 cell space space minus 3 end cell cell space space space 4 end cell row 2 cell space space minus 3 end cell cell space space 4 end cell row 0 cell space minus 1 end cell cell space space 1 end cell end table close square brackets.  then A3 = A–1 What is adj. A ?


straight A space equals space open square brackets table row 3 cell space space space minus 3 end cell cell space space 4 end cell row 2 cell space space space minus 3 end cell cell space space 4 end cell row 0 cell space space minus 1 end cell cell space space 1 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 3 cell space minus 3 end cell cell space space space 4 end cell row 2 cell space minus 3 end cell cell space space space 4 end cell row 0 cell space minus 1 end cell cell space space space 1 end cell end table close vertical bar space equals 3 open vertical bar table row cell negative 3 end cell cell space space space 4 end cell row cell negative 1 end cell cell space space space 1 end cell end table close vertical bar minus left parenthesis negative 3 right parenthesis space open vertical bar table row 2 cell space space 4 end cell row 0 cell space space 1 end cell end table close vertical bar plus 4 open vertical bar table row 2 cell space space space minus 3 end cell row 0 cell space space 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 equals 3 left parenthesis negative 3 plus 4 right parenthesis plus 3 left parenthesis 2 minus 0 right parenthesis plus 4 left parenthesis negative 2 plus 0 right parenthesis space equals space 3 plus 6 minus 8 space equals space 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 negative 3 end cell cell space space 4 end cell row cell negative 1 end cell cell space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space space 4 end cell row 0 cell space space space space 1 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space minus 3 end cell row 0 cell space space space minus 1 end cell end table close vertical bar
i.e.,   1, – 2, – 2 respectively
Co-factors of the elements of second row of | A | are
negative open vertical bar table row cell negative 3 end cell cell space space space space 4 end cell row cell negative 1 end cell cell space space space space 1 end cell end table close vertical bar comma space space open vertical bar table row 3 cell space space space space space space 4 end cell row 0 cell space space space space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 3 cell space space space minus 3 end cell row 0 cell space space space space minus 1 end cell end table close vertical bar
i.e.. – 1, 3, 3 respectively
Co-factors of the elements of third row of | A | are
open vertical bar table row cell negative 3 end cell cell space space 4 end cell row cell negative 3 end cell cell space space 4 end cell end table close vertical bar comma space space minus open vertical bar table row 3 cell space space space 4 end cell row 2 cell space space space 4 end cell end table close vertical bar comma space open vertical bar table row 3 cell space space minus 3 end cell row 2 cell space space space minus 3 end cell end table close vertical bar
i.e., 0, – 4, – 3 respectively

therefore space adj. space straight A space equals space open square brackets table row cell space space space 1 end cell cell space space space minus 2 end cell cell space space space minus 2 end cell row cell negative 1 end cell cell space space space space space space space 3 end cell cell space space space space space space 3 end cell row cell space space 0 end cell cell space space space minus 4 end cell cell space space minus 3 end cell end table close square brackets space equals space open square brackets table row 1 cell space space minus 1 end cell cell space space space space space 0 end cell row cell negative 2 end cell cell space space space space space 3 end cell cell space space minus 4 end cell row cell negative 2 end cell cell space space 3 end cell cell negative 3 end cell end table close square brackets
space space space space space space space 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 1 cell space space space minus 1 end cell cell space space space space space 0 end cell row cell negative 2 end cell cell space space space space 3 end cell cell space space minus 4 end cell row cell negative 2 end cell cell space space space space 3 end cell cell space space minus 3 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 space space space space space... left parenthesis 1 right parenthesis
Now space space straight A squared space equals space straight A. space straight A space equals space open square brackets table row 3 cell space space minus 3 end cell cell space space 4 end cell row 2 cell negative 3 end cell cell space space 4 end cell row 0 cell negative 1 end cell cell space 1 end cell end table close square brackets space open square brackets table row 3 cell negative 3 end cell cell space space 4 end cell row 2 cell negative 3 end cell cell space space 4 end cell row 0 cell negative 1 end cell cell space space 1 end cell end table close square brackets
space space space space space space space space space space space space space space equals space open square brackets table row cell 9 minus 6 plus 0 end cell cell space space space minus 9 plus 9 minus 4 end cell cell space space space 12 minus 12 plus 4 end cell row cell 6 minus 6 plus 0 end cell cell space minus 6 plus 9 minus 4 end cell cell space 8 minus 12 plus 4 end cell row cell 0 minus 2 plus 0 end cell cell space space space space 0 plus 3 minus 1 end cell cell space 0 minus 4 plus 1 end cell end table close square brackets space equals space open square brackets table row 3 cell space space minus 4 end cell cell space space space space 4 end cell row 0 cell space space space space space 1 end cell cell space space space space 0 end cell row cell negative 2 end cell cell space space 2 end cell cell space space minus 3 end cell end table close square brackets
straight A cubed space equals space straight A squared space straight A space equals space open square brackets table row 3 cell space space minus 4 end cell cell space space space space space 4 end cell row 0 cell space space space space space 1 end cell cell space space space space space 0 space end cell row cell negative 2 end cell cell space space space space 2 end cell cell space space minus 3 end cell end table close square brackets space open square brackets table row 3 cell space minus 3 end cell cell space space 4 end cell row 2 cell space minus 3 end cell cell space space 4 end cell row 0 cell space minus 1 end cell cell space space 1 end cell end table close square brackets
equals space open square brackets table row cell 9 minus 8 plus 0 end cell cell space minus 9 plus 12 minus 4 end cell cell space space 12 minus 16 plus 4 end cell row cell 0 minus 2 plus 0 end cell cell 0 plus 3 plus 0 end cell cell 0 minus 4 plus 0 end cell row cell negative 6 plus 4 plus 0 end cell cell 6 minus 6 plus 3 end cell cell negative 8 plus 8 minus 3 end cell end table close square brackets space equals open square brackets table row 1 cell space minus 1 end cell cell space space space 0 end cell row cell negative 2 end cell cell space space space 3 end cell cell negative 4 end cell row cell negative 2 end cell cell space space 3 end cell cell negative 3 end cell end table close square brackets

From (1) and (2),
A3 = A–1


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 Multiple Choice QuestionsLong Answer Type

214.

If straight A space equals space open square brackets table row 1 cell space space minus 1 end cell cell space space 1 end cell row 2 cell space minus 1 end cell cell space space 0 end cell row 1 cell space space space space 0 end cell cell space space 0 end cell end table close square brackets comma find A–1 and show that A–1 = A2.

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215.

If straight A space equals space open square brackets table row 3 cell space space minus 3 end cell cell space space space 4 end cell row 2 cell space space minus 3 end cell cell space space space 4 end cell row 0 cell space space minus 1 end cell cell space space space 1 end cell end table close square brackets comma  show that A4 = I. Hence find A–1

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216.

If A = open square brackets table row 2 cell space space 0 end cell cell space space minus 1 end cell row 5 cell space space 1 end cell cell space space space space 0 end cell row 0 cell space space 1 end cell cell space space space space 3 end cell end table close square brackets comma  prove that A–1 = A–2 – 6 A + 111.

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217.

If straight A space equals space open square brackets table row 1 cell space space 2 end cell cell space space 2 end cell row 2 cell space space 1 end cell cell space space 2 end cell row 2 cell space space 2 end cell cell space space 1 end cell end table close square brackets comma space Find A–1 and hence prove that A2 – 4A – 5 I = O.

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218.

For the matrix straight A space equals space open square brackets table row 1 cell space space 1 end cell cell space space space space space 1 end cell row 1 cell space space space 2 end cell cell space minus 3 end cell row 2 cell negative 1 end cell cell space space space space 3 end cell end table close square brackets comma show that A3 + 6A2 + 5A + 11 I = 0. Hence, find A–1.

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219.

If straight A space equals space open square brackets table row cell space space 2 end cell cell 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 2 end cell cell space minus 1 end cell row cell space space 1 end cell cell space space minus 1 end cell cell space space space space 2 end cell end table close square brackets comma
verify that A1 – 6A2 + 9A – 4 I = O and hence find A–1.

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220.

Compute (AB)-1  where straight A space equals space open square brackets table row 5 cell space space space 0 end cell cell space space 4 end cell row 2 cell space space 3 space end cell cell space 2 end cell row 1 cell space 2 end cell cell space 1 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 2 end cell cell space space 3 end cell row 1 cell space space 4 end cell cell space space 3 end cell row 1 cell space space 3 end cell cell space space 4 end cell end table close square brackets

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