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

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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

straight A space equals space open square brackets table row 1 cell space space 3 end cell row 2 cell space space 7 end cell end table close square brackets comma space space straight B space equals space open square brackets table row 3 cell space space space 4 end cell row 6 cell space space space 2 end cell end table close square brackets open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space 3 end cell row 2 cell space space 7 end cell end table close vertical bar space equals space 7 minus 6 space equals space 1

Co-factors of elements of first row of |A| are 7 and – 2 respectively
Co-factors of elements of second row of |A| are – 3 and 1 respectively
$therefore space space adj. space straight A space equals space open square brackets table row 7 cell space space space minus 2 end cell row cell negative 3 end cell cell space space space space space 1 end cell end table close square brackets to the power of apostrophe space space equals space open square brackets table row cell space 7 end cell cell space space space minus 3 end cell row cell negative 2 end cell cell space space space space space 1 end cell end table close square brackets space space space space space space straight A to the power of negative 1 end exponent space equals space fraction numerator adj space straight A over denominator open vertical bar straight A close vertical bar end fraction equals space open square brackets table row cell space space space space 7 end cell cell space space space space minus 3 end cell row cell negative 2 end cell cell space space space space space 1 end cell end table close square brackets space space space space space space space space open vertical bar straight B close vertical bar space equals space open vertical bar table row 3 cell space space space space 4 end cell row 6 cell space space space space 2 end cell end table close vertical bar space equals space 6 minus 24 space equals space minus 18$
Co-factors of elements of first row of | B | are 2 and – 6 respectively.
Co-factors of elements of second row of | B | are – 4 and 3 respectively.
$therefore space space space space space space adj. space straight B space equals space open square brackets table row cell space space space 2 end cell cell space space space space minus 6 end cell row cell negative 4 end cell cell space space space space space 3 end cell end table close square brackets to the power of apostrophe space space equals space open square brackets table row 2 cell space space space minus 4 end cell row cell negative 6 end cell cell space space space space space 3 end cell end table close square brackets space space space space space space space space space space space space space straight B to the power of negative 1 end exponent space equals space fraction numerator adj space straight B over denominator open vertical bar straight B close vertical bar end fraction space equals negative 1 over 18 open square brackets table row 2 cell space space space minus 4 end cell row cell negative 6 end cell cell space space space space space 3 end cell end table close square brackets therefore space space space thin space straight B to the power of negative 1 end exponent straight A to the power of negative 1 end exponent space equals space minus 1 over 18 open square brackets table row cell space space 2 end cell cell space space space space space minus 4 end cell row cell negative 6 end cell cell space space space space space space space 3 end cell end table close square brackets space space open square brackets table row 7 cell space space space space space minus 3 end cell row cell negative 2 end cell cell space 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 space space space space space equals negative 1 over 18 space open square brackets table row cell 14 plus 8 end cell cell space space space space minus 6 minus 4 end cell row cell negative 42 minus 6 end cell cell space space space space space space 18 plus 3 end cell end table close square brackets space equals space minus 1 over 18 space open square brackets table row 22 cell space space space minus 10 end cell row cell negative 48 end cell cell space space space space space 21 end cell end table close square brackets therefore space space space space space straight B to the power of negative 1 end exponent straight A to the power of negative 1 end exponent space space equals space minus 1 over 18 open square brackets table row 22 cell space space minus 10 end cell row cell negative 48 end cell cell space space space space space 21 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 space space space... left parenthesis 1 right parenthesis space space space space space space space space space space space AB equals space open square brackets table row 1 cell space space space 3 end cell row 2 cell space space space 7 end cell end table close square brackets space space open square brackets table row 3 cell space space space 4 end cell row 6 cell space space space 2 end cell end table close square brackets space equals space open square brackets table row cell 3 plus 18 end cell cell space space space space space space 4 plus 6 end cell row cell 6 plus 42 end cell cell space space space space space space 8 plus 14 end cell end table close square brackets space equals space open square brackets table row 21 cell space space space 10 end cell row 48 cell space space space 22 end cell end table close square brackets open vertical bar AB close vertical bar space equals space space open vertical bar table row 21 cell space space space space space space 10 end cell row 48 cell space space space space space 22 end cell end table close vertical bar space equals space 462 minus 480 space equals negative 18$

Co-factors of elements of first row of | AB | are 22 and – 48 respectively.
Co-factors of elements of second row of | AB | are – 10 and 21 respectively.
$therefore space space adj. space left parenthesis AB right parenthesis space equals space open square brackets table row 22 cell space space space minus 48 end cell row cell negative 10 end cell cell space space space space space 21 end cell end table close square brackets to the power of comma space equals space open square brackets table row 22 cell space space space minus 10 end cell row cell negative 48 end cell cell space space space space space space space 21 end cell end table close square brackets space space space space space space thin space left parenthesis AB right parenthesis to the power of negative 1 end exponent space equals space fraction numerator adj. space left parenthesis AB right parenthesis over denominator open vertical bar AB close vertical bar end fraction space equals space minus 1 over 18 open square brackets table row 22 cell space space space space minus 10 end cell row cell negative 48 end cell cell space space space space space space 21 end cell end table close square brackets space space space space space space space space space space space space... left parenthesis 2 right parenthesis$
From (1) and (2), we get, (AB) ^–1 = B ^–1 A^–1

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Long Answer Type

190. If verify (AB)–1 = B –1 A–1