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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.

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

Here, $straight A space equals space open square brackets table row 1 cell space space space space tanx end cell row cell negative tanx end cell cell space space space 1 end cell end table close square brackets$
$therefore space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space space tanx end cell row cell negative tanx end cell cell space space space 1 end cell end table close vertical bar space equals space 1 plus tan squared straight x space equals space sec squared straight x therefore space space space straight A to the power of negative 1 end exponent space exists. space$

Co-factors of the elements of first row are 1, tan x respectively.
Co-factors of the elements of second row are – tan x, 1 respectively.
$therefore space space space adj. space straight A space equals space open square brackets table row 1 cell 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 to the power of apostrophe space equals space open square brackets table row 1 cell negative tanx end cell row tanx cell space space space 1 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 fraction numerator adj. space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space fraction numerator 1 over denominator sec squared straight x end fraction open square brackets table row 1 cell space space space space space minus tanx end cell row tanx cell space space space space space 1 end cell end table close square brackets Also comma space space space straight A apostrophe space equals space open square brackets table row 1 cell space space space space minus tanx end cell row tanx cell space space space space space space 1 end cell end table close square brackets therefore space space space straight A apostrophe straight A to the power of negative 1 end exponent space equals space open square brackets table row 1 cell space space space space minus tanx end cell row tanx cell space space space space 1 end cell end table close square brackets space fraction numerator 1 over denominator sec squared straight x end fraction open square brackets table row 1 cell space space space minus tanx end cell row tanx cell space 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 equals space fraction numerator 1 over denominator sec squared straight x end fraction open square brackets table row 1 cell space space space space minus tanx end cell row tanx cell space space space space space space space 1 end cell end table close square brackets space space open square brackets table row 1 cell space space space minus tanx end cell row tanx cell space 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 equals fraction numerator 1 over denominator sec squared straight x end fraction open square brackets table row cell 1 minus tanx end cell cell space space space space space 1 minus 2 space tanx end cell row cell 2 space tanx end cell cell space space 1 minus tan squared straight x end cell end table close square brackets space space space space space space space space space space space space space space space space space equals space cos squared straight x open square brackets table row cell fraction numerator cos squared straight x minus sin squared straight x over denominator cos space straight x end fraction end cell cell negative fraction numerator 2 space sinx space over denominator cos space straight x end fraction end cell row cell fraction numerator 2 space sinx space over denominator cos space straight x end fraction end cell cell fraction numerator cos squared straight x minus sin squared straight x over denominator cosx end fraction end cell end table close square brackets 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 cos squared straight x minus sin squared straight x end cell cell space space space space space minus 2 space sin space straight x space cos space straight x end cell row cell 2 space sinx space cosx end cell cell space space space space space space cos squared straight x minus sin squared straight x end cell end table close square brackets therefore space space space 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 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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188. If