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181.
If $straight A space equals open square brackets table row 2 cell space space space minus 3 end cell row 3 cell space space space space space space 4 end cell end table close square brackets comma space$ show that $straight A squared minus 6 straight A plus 17 space straight I space equals space straight O.$ Hence find $straight A to the power of negative 1 end exponent.$

straight A space equals space open square brackets table row 2 cell space space space minus 3 end cell row 3 cell space space space space space 4 end cell end table close square brackets therefore space space space straight A squared space equals space open square brackets table row 2 cell space space minus 3 end cell row 3 cell space space space space space 4 end cell end table close square brackets space open square brackets table row 2 cell space space minus 3 end cell row 3 cell space space space space space space 4 end cell end table close square brackets space equals space open square brackets table row cell 4 minus 9 end cell cell space space space minus 6 minus 12 end cell row cell 6 plus 12 end cell cell negative 9 plus 16 end cell end table close square brackets space space equals space open square brackets table row cell negative 5 end cell cell space space minus 18 end cell row 18 cell space space space space space space 7 end cell end table close square brackets therefore space space space straight A squared minus 6 straight A plus 17 space straight I space equals space open square brackets table row cell negative 5 end cell cell space space space minus 18 end cell row 18 cell space space space space space space space 7 end cell end table close square brackets space minus space 6 open square brackets table row 2 cell space space minus 3 end cell row 3 cell space space space space 4 end cell end table close square brackets space plus space 17 open square brackets table row 1 cell space space 0 end cell row 0 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 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

equals space open square brackets table row cell negative 5 end cell cell space space space space minus 18 end cell row 18 cell space space space space space space space 7 end cell end table close square brackets space plus space open square brackets table row cell negative 12 end cell cell space space space space space space 18 end cell row cell negative 18 end cell cell space space space minus 24 end cell end table close square brackets space plus space open square brackets table row 17 cell space space space 0 end cell row 0 cell space space space 17 end cell end table close square brackets equals space open square brackets table row cell negative 5 minus 12 plus 17 end cell cell space space space space space minus 18 plus 18 plus 0 end cell row cell 18 minus 18 plus 0 end cell cell space space space space space space space space 7 minus 24 plus 17 end cell end table close square brackets space equals space open square brackets table row 0 cell space space space 0 end cell row 0 cell space space space space 0 end cell end table close square brackets

therefore space space space straight A squared minus 6 straight A space plus space 17 space straight I space equals space straight O space space space space space space rightwards double arrow space space space space 17 space straight I space equals space minus straight A squared plus 6 straight A rightwards double arrow space space space space space 17 space straight I thin space straight A to the power of negative 1 end exponent space equals space minus straight A squared straight A to the power of negative 1 end exponent space plus space 6 straight A thin space straight A to the power of negative 1 end exponent space space space space space space rightwards double arrow space space space space space 17 space straight A to the power of negative 1 end exponent space equals space minus straight A space plus space 6 straight I rightwards double arrow space space space space space 17 space straight A to the power of negative 1 end exponent space equals space minus open square brackets table row 2 cell space space space minus 3 end cell row 3 cell space space space space space 4 end cell end table close square brackets space plus space open square brackets table row 1 cell space space space 0 end cell row 0 cell space space space 1 end cell end table close square brackets space space space rightwards double arrow space space space 17 space straight A to the power of negative 1 end exponent space equals space open square brackets table row cell negative 2 end cell cell space space space space space space space space 3 end cell row cell negative 3 end cell cell space space space minus 4 end cell end table close square brackets space plus space open square brackets table row 6 cell space space space 0 end cell row 0 cell space space space 6 end cell end table close square brackets rightwards double arrow space space space space space space 17 space straight A to the power of negative 1 end exponent space equals space open square brackets table row cell negative 2 plus 6 end cell cell space space space space space space 3 plus 0 end cell row cell negative 3 plus 0 end cell cell space space space minus 4 plus 6 end cell end table close square brackets space space space space space space rightwards double arrow space space space 17 space straight A to the power of negative 1 end exponent space equals space open square brackets table row 4 cell space space space space 3 end cell row cell negative 3 end cell cell space space space space 2 end cell end table close square brackets rightwards double arrow space space space space space space space space straight A to the power of negative 1 end exponent space equals space 1 over 17 open square brackets table row 4 cell space space space space 3 end cell row cell negative 3 end cell cell space space space space 2 end cell end table close square brackets

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182.

For the matrix $straight A space equals space open square brackets table row 2 cell space space space minus 1 end cell row 3 cell space space space space space space 2 end cell end table close square brackets comma$ show that A² – 4 A + 7 I = O. Hence obtain A^–1.

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Long Answer Type

183.

Show that the matrix $straight A space equals space open square brackets table row 2 cell space space space space space 3 end cell row 1 cell space space space space 2 end cell end table close square brackets$ satisfies the equation A² – 4A + I = O and hence find A^–1.

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184.

If $straight A space equals space open square brackets table row cell space space 3 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space space space space 2 end cell end table close square brackets comma$ show that A² – 5A + 7 I = O. Hence find A ^–1.

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185.

For the matrix $straight A space equals space open square brackets table row 3 cell space space space 2 end cell row 1 cell space space space 1 end cell end table close square brackets comma$ find the numbers a and b such that A² + aA + bI = O. Hence find A^–1.

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Short Answer Type

186. If A is square matrix such that A³ = I, prove that A is non-singular and find adj. A and prove that A^–1 = A².

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187. For the matrix

straight A space equals space open square brackets table row 3 cell space space space space space space 1 end cell row 7 cell space space space space space 5 end cell end table close square brackets comma

find x and y so that A² + xI = yA. Hence find A^–1 .

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Long Answer Type

188.

If $straight A space equals space open square brackets table row 1 cell space space space space space tanx end cell row cell negative tanx end cell cell space space space 1 end cell end table close square brackets comma space space space space then space straight A apostrophe straight A to the power of negative 1 end exponent space equals space open square brackets table row cell cos space 2 straight x end cell cell space space space space space minus sin space 2 straight x end cell row cell sin space 2 straight x end cell cell space space space space space space space space cos space 2 straight x end cell end table close square brackets$

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189.

Find the inverse of the matrix $straight A space equals space open square brackets table row straight a cell space space straight b end cell row straight c cell space space fraction numerator 1 plus bc over denominator straight a end fraction end cell end table close square brackets$ and show that a A ^-1 = (a² + b c + 1) I – a A.

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190.

If $straight A space equals space open square brackets table row 1 cell space space space space 3 end cell row 2 cell space space space space 7 end cell end table close square brackets space space and space straight B space equals space open square brackets table row 3 cell space space space space 4 end cell row 6 cell space space space space 2 end cell end table close square brackets comma$
verify (AB)^–1 = B ^–1 A^–1

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181. If show that Hence find

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