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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 $straight A to the power of negative 1 end exponent.$

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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.

straight A space equals space open square brackets table row 3 cell space space space space space 1 end cell row cell negative 1 end cell cell space space space space space 2 end cell end table close square brackets therefore space space space space straight A squared space equals space space open square brackets table row 3 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 space space open square brackets table row 3 cell space space space 1 end cell row cell negative 1 end cell cell space space space 2 end cell end table close square brackets space equals space open square brackets table row cell 9 minus 1 end cell cell space space space space space 3 plus 2 end cell row cell negative 3 minus 2 end cell cell space space minus 1 plus 4 end cell end table close square brackets space equals space open square brackets table row 8 cell space space 5 end cell row cell negative 5 end cell cell space space 3 end cell end table close square brackets therefore space space space space straight A squared minus 5 straight A plus 7 straight I space equals open square brackets table row 8 cell space space space 5 end cell row cell negative 5 end cell cell space space 3 end cell end table close square brackets minus 5 open square brackets table row 3 cell space space space 1 end cell row cell negative 1 end cell cell space space space 2 end cell end table close square brackets plus 7 space open square brackets table row 1 cell space space space space 0 end cell row 0 cell space 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 equals open square brackets table row cell space space 8 end cell cell space space space 5 end cell row cell negative 5 end cell cell space space space space 3 end cell end table close square brackets plus space open square brackets table row cell negative 15 end cell cell space space minus 5 end cell row 5 cell space space minus 10 end cell end table close square brackets space plus space open square brackets table row 7 cell space space space 0 end cell row 0 cell space space space 7 end cell end table close square brackets

equals space open square brackets table row cell 8 minus 15 plus 7 end cell cell space space space space space space space 5 minus 5 plus 0 end cell row cell negative 5 plus 5 plus 0 end cell cell space space space space space space space 3 minus 10 plus 7 end cell end table close square brackets space equals space open square brackets table row 0 cell space space 0 end cell row 0 cell space space 0 end cell end table close square brackets space equals space straight O therefore space space space space straight A squared minus 5 straight A space plus space 7 straight I space equals space straight O rightwards double arrow space space space space space space 7 straight I space equals space minus straight A squared plus 5 straight A space space space space space space rightwards double arrow space space space space space 7 straight A to the power of negative 1 end exponent space equals space minus straight A plus 5 thin space straight I rightwards double arrow space space space space space space space 7 straight A to the power of negative 1 end exponent space equals space minus open square brackets table row cell space space space 3 end cell cell space space space 1 end cell row cell negative 1 end cell cell space space space 2 end cell end table close square brackets space space plus space 5 space open square brackets table row 1 cell space space space 0 end cell row 0 cell space space 1 end cell end table close square brackets space rightwards double arrow space space space space space space space 7 straight A to the power of negative 1 end exponent space equals space open square brackets table row cell negative 3 end cell cell space space space minus 1 end cell row 1 cell space space space minus 2 end cell end table close square brackets space plus space open square brackets table row 5 cell space space 0 end cell row 0 cell space space 5 end cell end table close square brackets rightwards double arrow space space space space space space space space 7 straight A to the power of negative 1 end exponent space equals space open square brackets table row cell negative 3 plus 5 end cell cell space space space space space space minus 1 plus 0 end cell row cell 1 plus 0 end cell cell space space space space minus 2 plus 5 end cell end table close square brackets space space space space space space space rightwards double arrow space space space 7 straight A to the power of negative 1 end exponent space equals space open square brackets table row 2 cell space space space minus 1 end cell row 1 cell space space space space space space space space 3 end cell end table close square brackets therefore space space space space straight A to the power of negative 1 end exponent space equals space 1 over 7 open square brackets table row 2 cell space space space space space minus 1 end cell row 1 cell space space space space space space space space 3 end cell end table close square brackets

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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