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221.
Compute (AB)¹ where:
$straight A space equals space open square brackets table row 1 cell space space space space 1 end cell cell space space space 2 end cell row 0 cell space space space 2 end cell cell space minus 3 end cell row 3 cell space minus 2 end cell cell space space space space 4 end cell end table close square brackets space space space space space space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 1 cell space space space space space 2 end cell cell space space space space space 0 end cell row 0 cell space space space 3 end cell cell space space minus 1 end cell row 1 cell space space 0 end cell cell space space space space 2 end cell end table close square brackets$

Here space straight A space equals space open square brackets table row 1 cell space space space space space 1 end cell cell space space space space space 2 end cell row 0 cell space space space space space space 2 end cell cell space space minus 3 end cell row 3 cell space space minus 2 end cell cell space space space space 4 end cell end table close square brackets comma 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 2 end cell cell space space 0 end cell row 0 cell space space space 3 end cell cell negative 1 end cell row 1 cell space space space 0 end cell cell space space 2 end cell end table close square brackets

therefore space space space space open vertical bar straight A close vertical bar space equals open vertical bar table row 1 cell space space space space 1 end cell cell space space space space 2 end cell row 0 cell space space space space 2 end cell cell space space minus 3 end cell row 3 cell space minus 2 end cell cell space space space space space 4 end cell end table close vertical bar space equals space 1 open vertical bar table row 2 cell space space minus 3 end cell row cell negative 2 end cell cell space space space space space 4 end cell end table close vertical bar minus 1 open vertical bar table row 0 cell space space minus 3 end cell row 3 cell space space space space space 4 end cell end table close vertical bar plus 2 open vertical bar table row 0 cell space space space space space space 2 end cell row 3 cell space space minus 2 end cell end table close vertical bar space space space space space space equals 1 left parenthesis 8 minus 6 right parenthesis minus 1 left parenthesis 0 plus 9 right parenthesis plus 2 left parenthesis 0 minus 6 right parenthesis space equals space 2 minus 9 minus 12 space equals space minus 19 therefore space space space space space straight A to the power of negative 1 end exponent space exists.

Co-factors of the elements of first row of | A | are

open vertical bar table row 2 cell space space space minus 3 end cell row cell negative 2 end cell cell space space space space space 4 end cell end table close vertical bar comma space space minus open vertical bar table row 0 cell space space minus 3 end cell row 3 cell space space space space 4 end cell end table close vertical bar comma space space open vertical bar table row 0 cell space space space space space 2 end cell row 3 cell space space minus 2 end cell end table close vertical bar

i.e., 2, – 9, – 6 respectively

Co-factors of the elements of first row of | A | are

negative open vertical bar table row 1 cell space space space 2 end cell row cell negative 2 end cell cell space space space 4 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space space space 2 end cell row 3 cell space space space space space 4 end cell end table close vertical bar comma space minus space open vertical bar table row 1 cell space space space space space space space 1 end cell row 3 cell space space minus 2 end cell end table close vertical bar

i.e.. — 8, – 2, 5 respectively
Co-factors of the elements of third row of | A | are
$open vertical bar table row 1 cell space space space space space 2 end cell row 2 cell space minus 3 end cell end table close vertical bar minus space open vertical bar table row 1 cell space space space space space space 2 end cell row 0 cell space space minus 3 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space 1 end cell row 0 cell space space space 2 end cell end table close vertical bar$
i.e., -7, 3, 2 respectively.
$therefore space space adj. space straight A space equals space open square brackets table row 2 cell space space minus 9 end cell cell space space minus 6 end cell row cell negative 8 end cell cell space space minus 2 end cell cell space space space space space 5 end cell row cell negative 7 end cell cell space space space space 3 end cell cell space space space space 2 end cell end table close square brackets to the power of apostrophe space equals open square brackets table row cell space space 2 end cell cell space space space minus 8 end cell cell space minus 7 end cell row cell negative 9 end cell cell space space minus 2 end cell cell space space space space 3 end cell row cell negative 6 end cell cell space space space space space 5 end cell cell space space space 2 end cell end table close square brackets$

$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 space equals space minus 1 over 19 open square brackets table row 2 cell space space space minus 8 end cell cell space space space minus 7 end cell row cell negative 9 end cell cell space space space minus 2 end cell cell space space space space space 3 space end cell row cell negative 6 end cell cell space space space space space 5 end cell cell space space space 2 end cell end table close square brackets space equals 1 over 19 open square brackets table row cell negative 2 end cell cell space space 8 end cell cell space space space space space space 7 end cell row 9 cell space space 2 end cell cell space minus 3 end cell row 6 cell space minus 5 end cell cell space space minus 2 end cell end table close square brackets space space space space space space space space space space left parenthesis AB right parenthesis to the power of negative 1 end exponent space equals space straight B to the power of minus to the power of 1 straight A to the power of negative 1 end exponent space equals space 1 over 19 open square brackets table row 1 cell space space 2 end cell cell space space space 0 end cell row 0 cell space space 3 end cell cell 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 space open square brackets table row cell negative 2 end cell cell space space 8 end cell cell space space space 7 end cell row 9 cell space space space 2 end cell cell space space minus 3 end cell row 6 cell space minus 5 end cell cell space space minus 2 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 equals space 1 over 19 open square brackets table row cell negative 2 plus 18 plus 0 end cell cell space space space 8 plus 4 plus 0 end cell cell space space space 7 minus 6 plus 0 end cell row cell 0 plus 27 minus 6 end cell cell 0 plus 6 plus 5 end cell cell 0 minus 9 plus 2 end cell row cell negative 2 plus 0 plus 12 end cell cell 8 plus 0 minus 10 end cell cell 7 plus 0 minus 4 end cell end table close square brackets space equals 1 over 19 open square brackets table row 16 cell space space 12 end cell cell space space 1 end cell row 21 cell space 11 end cell cell negative 7 end cell row 10 cell space minus 2 end cell cell space space space 3 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$

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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

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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

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