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Compute (AB)¹ where:
$straight A space equals space open square brackets table row 1 cell space space space space 1 end cell cell space space space 2 end cell row 0 cell space space space 2 end cell cell space minus 3 end cell row 3 cell space minus 2 end cell cell space space space space 4 end cell end table close square brackets space space space space space space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 1 cell space space space space space 2 end cell cell space space space space space 0 end cell row 0 cell space space space 3 end cell cell space space minus 1 end cell row 1 cell space space 0 end cell cell space space space space 2 end cell end table close square brackets$

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222.

If $straight A space equals space open square brackets table row 2 cell space space space 2 end cell cell space space space 1 end cell row cell negative 2 end cell cell space space space 1 end cell cell space space 2 end cell row 1 cell space minus 2 end cell cell space space 2 end cell end table close square brackets space space space and space straight B space equals space open square brackets table row 1 cell space space 3 end cell cell space space 2 end cell row 1 cell space space space 1 end cell cell space 1 end cell row 2 cell negative 3 end cell cell space 1 end cell end table close square brackets comma$ verify that (AB)^–1 = B"^–1 A^–1.

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223.

If $straight A to the power of negative 1 end exponent space equals space open square brackets table row 3 cell space space minus 1 end cell cell space space space space space 1 end cell row cell negative 15 end cell cell space space space space space 6 end cell cell space minus 5 end cell row 5 cell space minus 2 end cell cell space space space space 2 end cell end table close square brackets space space space and space straight B space equals space open square brackets table row cell space space space 1 end cell cell space space space space 2 end cell cell space space space minus 2 end cell row cell negative 1 end cell cell space space space space space 3 end cell cell space space space space space 0 end cell row cell space space 0 end cell cell space space minus 2 end cell cell space space space space space 1 end cell end table close square brackets comma$ find (AB)^–1.

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Multiple Choice Questions

224. Let A be a non-singular square matrix of order 3 × 3. Then | adj A| j is equal to

| A |
| A |²
| A |³
| A |³

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225.
If A is an invertible matrix of order 2, then det (A^–1) is equal to

det (A)

$fraction numerator 1 over denominator det space left parenthesis straight A right parenthesis end fraction$

1

1

fraction numerator 1 over denominator det space left parenthesis straight A right parenthesis end fraction

We know that
$AA to the power of negative 1 end exponent space equals space straight I$
$therefore space space space space space open vertical bar AA to the power of negative 1 end exponent close vertical bar space equals space space open vertical bar straight I close vertical bar space space space space space space space space space space space space space space rightwards double arrow space space space space space left parenthesis straight A right parenthesis space open vertical bar straight A to the power of negative 1 end exponent close vertical bar space equals space straight I therefore space space space space space open vertical bar straight A to the power of negative 1 end exponent close vertical bar space equals space fraction numerator 1 over denominator open vertical bar straight A close vertical bar end fraction therefore space space space space left parenthesis straight B right parenthesis space is space correct space answer.$

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226. If x, y, z are non-zero real numbers, then the inverse of matrix

straight A space equals open square brackets table row straight x cell space space space space 0 end cell cell space space space 0 end cell row 0 cell space space space straight y end cell cell space space space 0 end cell row 0 cell space space space 0 end cell cell space space space straight z end cell end table close square brackets space is

$open square brackets table row cell straight x to the power of negative 1 end exponent end cell cell space space 0 end cell cell space space 0 end cell row 0 cell space space straight y to the power of negative 1 space end exponent end cell cell space space 0 end cell row 0 0 cell space straight z to the power of negative 1 end exponent end cell end table close square brackets$
$xyz open square brackets table row cell straight x to the power of negative 1 end exponent end cell cell space space 0 end cell cell space space space 0 end cell row 0 cell space space straight y to the power of negative 1 end exponent space end cell cell space space 0 end cell row 0 0 cell space straight z to the power of negative 1 end exponent end cell end table close square brackets$
$1 over xyz open square brackets table row straight x cell space space 0 end cell cell space space 0 end cell row 0 cell space straight y end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space straight z end cell end table close square brackets$
$1 over xyz open square brackets table row straight x cell space space 0 end cell cell space space 0 end cell row 0 cell space straight y end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space straight z end cell end table close square brackets$