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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?

Let $straight A space equals open square brackets table row 4 cell space minus 3 end cell cell space space space space space minus 1 end cell row 2 cell space space space space space straight a end cell cell space space space space space space space space 6 end cell row 3 cell space space minus 5 end cell cell space space space minus 4 end cell end table close square brackets$
$therefore space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 4 cell space minus 3 end cell cell 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 6 end cell row 3 cell space space minus 5 end cell cell space space space minus 4 end cell end table close vertical bar space space space space space space space space space space space space equals space 4 open vertical bar table row cell space space space straight a end cell cell space space space space space space space 6 end cell row cell negative 5 end cell cell space space space minus 4 end cell end table close vertical bar minus left parenthesis negative 3 right parenthesis space open vertical bar table row 2 cell space space space space space space 6 end cell row 3 cell space space minus 4 end cell end table close vertical bar plus left parenthesis negative 1 right parenthesis thin space open vertical bar table row 2 cell space space space space space space space space straight a end cell row 3 cell space space space space minus 5 end cell end table close vertical bar space space space space space space space space space space space space equals 4 left parenthesis negative 4 straight a plus 30 right parenthesis plus 3 left parenthesis negative 8 minus 18 right parenthesis minus left parenthesis negative 10 minus 3 straight a right parenthesis space space space space space space space space space space space space equals negative 16 straight a plus 120 minus 78 plus 10 plus 3 straight a equals negative 13 straight a plus 52 For space straight A space be space to space singular comma space we space have space space space open vertical bar straight A close vertical bar space equals space 0 space space space space space space space space space rightwards double arrow space space space space space space minus 13 straight a space plus space 52 space equals space 0 space space space space space rightwards double arrow space space space 13 space straight a space equals space 52 space space space space rightwards double arrow space space space straight a space equals space 4$

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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.

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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 $straight A squared minus 6 straight A space plus space straight I space equals space straight O comma space hence space find space straight A to the power of negative 1 end exponent.$

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177. For what value of ‘a’ is the matrixsingular?

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