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If $straight A space equals open square brackets table row 2 cell space space space space space space space 3 end cell row 5 cell space space space minus 2 end cell end table close square brackets$ , show that $straight A to the power of negative 1 end exponent space equals space 1 over 19 straight A$ .

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172.

Let A be the matrix $open square brackets table row 3 cell space space 8 end cell row 2 cell space space 1 end cell end table close square brackets.$ Find A ^–1 and verify that $straight A to the power of negative 1 end exponent space equals space 1 over 13 straight A space minus space 4 over 13 straight I$
where I is 2 × 2 unit matrix.

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173.

If A = $open square brackets table row 2 cell space space space 5 end cell row 1 cell space space space 6 end cell end table close square brackets comma$ find $straight A to the power of negative 1 end exponent$ and verify that $straight A to the power of negative 1 end exponent space equals space minus 1 over 7 straight A plus 8 over 7 straight I$

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174.

Given $straight A space equals space open square brackets table row 2 cell space space space space space minus 3 end cell row cell negative 4 end cell cell space space space space space space space 7 end cell end table close square brackets comma$ compute A^–1 and show that 2A^–1 = 9 I – A.

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175.

Find the inverse of $straight A space equals space open square brackets table row 3 cell space space space space space space space 5 end cell row 7 cell space space minus 11 end cell end table close square brackets$ and verify that AA^–1 = I.

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176.

For what values of x, is the following matrix singular?
$open square brackets table row cell 3 minus 2 straight x end cell cell space space space space straight x plus 1 end cell row 2 cell space space space space 4 end cell end table close square brackets$

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177. For what value of ‘a’ is the matrix

open square brackets table row 4 cell space minus 3 end cell cell space space space space space minus 1 end cell row 2 cell space space space space space straight a end cell cell space space space space space space space space 6 end cell row 3 cell space space minus 5 end cell cell space space space minus 4 end cell end table close square brackets

singular?

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Long Answer Type

178.

If $straight A space equals space open square brackets table row 3 cell space space space space space 1 end cell row 7 cell space space space space space 5 end cell end table close square brackets comma$ find x and y find that A² + xI = y A. Hence find A^–1

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179.
If $straight A space equals space open square brackets table row 6 cell space space space space 5 end cell row 7 cell space space space space 6 end cell end table close square brackets comma$ show that A² – 12A + I = O. Hence find A ^–1.

straight A space equals open square brackets table row 6 cell space space space 5 end cell row 7 cell space space space 6 end cell end table close square brackets therefore space space space space straight A squared space equals space open square brackets table row 6 cell space space space 5 end cell row 7 cell space space space 6 end cell end table close square brackets space open square brackets table row 6 cell space space space 5 end cell row 7 cell space space space 6 end cell end table close square brackets space equals space open square brackets table row cell 36 plus 35 end cell cell space space space space 30 plus 30 end cell row cell 42 plus 42 end cell cell space space space space 35 plus 36 end cell end table close square brackets space equals space open square brackets table row 71 cell space space space 60 end cell row 84 cell space space 71 end cell end table close square brackets therefore space space space space space straight A squared minus 12 straight A space plus space straight I space equals space open square brackets table row 71 cell space space space 60 end cell row 84 cell space space space 71 end cell end table close square brackets space minus 12 space open square brackets table row 6 cell space space space 5 end cell row 7 cell space space space 6 end cell end table close square brackets plus 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 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 71 cell space space 60 end cell row 84 cell space space 71 end cell end table close square brackets space plus space open square brackets table row cell negative 72 end cell cell space space space minus 60 end cell row cell negative 84 end cell cell space space minus 72 end cell end table close square brackets plus 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 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 71 minus 72 plus 1 end cell cell space space space 60 minus 60 plus 0 end cell row cell 84 minus 84 plus 0 end cell cell space space space 71 minus 72 plus 1 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 therefore space space space straight A squared minus 12 space straight A space plus space straight I space equals space straight O Post space multiplying space both space sides space by space straight A to the power of negative 1 end exponent comma space we space get space space space space space space space space space space space space space space space space space straight A squared straight A to the power of negative 1 end exponent space minus 12 AA to the power of negative 1 end exponent plus space IA to the power of negative 1 end exponent space equals space straight O or space space space space space space space space straight A minus 12 straight I plus straight A to the power of negative 1 end exponent space equals space straight O therefore space space space space space space space space space space space space 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 minus straight A plus 12 space straight I space equals space minus open square brackets table row 6 cell space space 5 end cell row 7 cell space space 6 end cell end table close square brackets space plus 12 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 space space space space space space space

equals space open square brackets table row cell negative 6 end cell cell space space space space space minus 5 end cell row cell negative 7 end cell cell space space space minus 6 end cell end table close square brackets space plus space open square brackets table row 12 cell space space space 0 end cell row 0 cell space space space 12 end cell end table close square brackets equals space open square brackets table row cell negative 6 plus 12 end cell cell space space space space minus 5 plus 0 end cell row cell negative 7 plus 0 end cell cell space minus 6 plus 12 end cell end table close square brackets space equals space open square brackets table row 6 cell space space space minus 5 end cell row cell negative 7 end cell cell space space space space space 6 end cell end table close square brackets

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Short Answer Type

180.

If $straight A space equals space open square brackets table row 2 cell space space space 7 end cell row 1 cell space space space 4 end cell end table close square brackets comma space space$ show that $straight A squared minus 6 straight A space plus space straight I space equals space straight O comma space hence space find space straight A to the power of negative 1 end exponent.$