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Let A = $open square brackets table row 4 cell space space space minus 6 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space minus 1 end cell cell space space 1 end cell row cell negative 4 end cell cell space space space 11 end cell cell negative 1 end cell end table close square brackets.$
Show that A is inevitable. Find adj. A and A Also verify that AA^–1 = A^–1 A = I.

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212.
Verify AA^–1 = A^–1 A = I where
$straight A space equals open square brackets table row 0 cell space space space 0 end cell cell space space minus 1 end cell row 3 cell space space space space 4 end cell cell space space space space space 5 end cell row cell negative 2 end cell cell negative 4 end cell cell space space minus 7 end cell end table close square brackets.$

Here $straight A space equals open square brackets table row 0 cell space space space space 0 end cell cell space space space minus 1 end cell row 3 cell space space space space space 4 end cell cell space space space space space space 5 end cell row cell negative 2 end cell cell space space minus 4 end cell cell space space minus 7 end cell end table close square brackets$

$therefore space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 0 cell space space 0 end cell cell space space minus 1 end cell row 3 cell space space 4 end cell cell space space space space space 5 end cell row cell negative 2 end cell cell space space minus 4 end cell cell space space minus 7 end cell end table close vertical bar space equals space minus open vertical bar table row 3 cell space space space space space 4 end cell row cell negative 2 end cell cell space space space minus 4 end cell end table close vertical bar space equals space minus left parenthesis negative 12 plus 8 right parenthesis space equals space 4 not equal to 0 therefore space space space straight A to the power of negative 1 end exponent space exists. space$
Co-factors of the elements of the first row of | A | are
$open vertical bar table row cell space space 4 end cell cell space space space space space 5 end cell row cell negative 4 end cell cell space space space minus 7 end cell end table close vertical bar comma space space space space minus open vertical bar table row 3 cell space space space space space space space 5 end cell row cell negative 2 end cell cell space space minus 7 end cell end table close vertical bar comma space space open vertical bar table row 3 cell space space space space space space 4 end cell row cell negative 2 end cell cell space space minus 4 end cell end table close vertical bar$
i.e., – 8, 11, –4 respectively
Co-factors of the elements of the second row of | A | are
$negative open vertical bar table row 0 cell space space minus 1 end cell row cell negative 4 end cell cell space space minus 7 end cell end table close vertical bar comma space open vertical bar table row 0 cell space space minus 1 end cell row cell negative 2 end cell cell space space minus 7 end cell end table close vertical bar comma space space minus open vertical bar table row cell space 0 end cell cell space space space space space space 0 end cell row cell negative 2 end cell cell space space minus 4 end cell end table close vertical bar$
i.e., 4, – 2, 0 respectively.
Co-factors of the elements of the third row of | A | are
$open vertical bar table row 0 cell space space minus 1 end cell row 4 cell space space space 5 end cell end table close vertical bar comma space space space minus open vertical bar table row 0 cell space space space minus 1 end cell row 3 cell space space space space space 5 end cell end table close vertical bar comma space space minus open vertical bar table row 0 cell space space space space 0 end cell row cell negative 2 end cell cell space space minus 4 end cell end table close vertical bar$
i.e., 4, – 3, 0 respectively.

$therefore space space space space space space adj. space straight A space equals space open square brackets table row cell negative 8 end cell cell space space 11 end cell cell space space minus 4 end cell row 4 cell space minus 2 end cell cell space space space space space 0 end cell row 4 cell negative 3 end cell cell space space space space space 0 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 8 end cell cell space space 4 end cell cell space space 4 end cell row 11 cell negative 2 end cell cell negative 3 end cell row cell negative 4 end cell cell space 0 end cell cell space space 0 end cell end table close square brackets Now comma 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 1 fourth open square brackets table row cell negative 8 end cell cell space space 4 end cell cell space space space space space 4 end cell row 11 cell space minus 2 end cell cell space space minus 3 end cell row cell negative 4 end cell cell space space 0 end cell cell space space space space 0 end cell end table close square brackets therefore space space space space space AA to the power of negative 1 end exponent space equals space 1 fourth open square brackets table row 0 cell space space 0 end cell cell space space minus 1 end cell row 3 cell space space 4 end cell cell space space space space space 5 end cell row cell negative 2 end cell cell negative 4 end cell cell space minus 7 end cell end table close square brackets space open square brackets table row cell negative 8 end cell cell space space space space 4 end cell cell space space space space space 4 end cell row 11 cell space minus 2 end cell cell space space minus 3 end cell row cell negative 4 end cell cell space space space 0 end cell cell space space space space space space 0 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 fourth open square brackets table row cell 0 plus 0 plus 4 end cell cell space space space space 0 plus 0 plus 0 end cell cell space space space space space space 0 plus 0 plus 0 end cell row cell negative 24 plus 44 minus 20 space end cell cell space 12 minus 8 plus 0 end cell cell space space space space space space 12 minus 12 plus 0 end cell row cell 16 minus 44 plus 28 end cell cell space minus 8 plus 8 plus 0 end cell cell space space space space minus 8 plus 12 plus 0 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 equals space 1 fourth open square brackets table row 4 cell space 0 end cell cell space space 0 end cell row 0 cell space space 4 end cell cell space 0 end cell row 0 cell space space 0 space end cell 4 end table close square brackets space equals space open square brackets table row 1 cell space space 0 end cell cell space space 0 end cell row 0 cell space 1 end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space 1 end cell end table close square brackets space equals space straight I straight A to the power of negative 1 end exponent straight A space equals space 1 fourth open square brackets table row cell negative 8 end cell cell space space space space 4 end cell cell space space space space 4 end cell row 11 cell space space minus 2 end cell cell space space minus 3 end cell row cell negative 4 end cell cell space space 0 end cell cell space space 0 end cell end table close square brackets space open square brackets table row 0 cell space space 0 end cell cell space space minus 1 end cell row 3 cell space space 4 end cell cell space space space 5 end cell row cell negative 2 end cell cell space space minus 4 end cell cell space space minus 7 end cell end table close square brackets space space space space space space space space space space space space equals space 1 fourth open square brackets table row cell 0 plus 12 minus 8 end cell cell space space space 0 plus 16 minus 16 end cell cell space space 8 plus 20 minus 28 end cell row cell 0 minus 6 plus 6 end cell cell 0 minus 8 plus 12 end cell cell negative 11 minus 10 plus 21 end cell row cell 0 plus 0 plus 0 end cell cell 0 plus 0 plus 0 end cell cell 4 plus 0 plus 0 end cell end table close square brackets space space space space space space space space space space space space space equals space 1 fourth open square brackets table row 4 cell space space space 0 end cell cell space space space 0 end cell row 0 cell space space 4 end cell cell space space 0 end cell row 0 cell space space 0 end cell cell space space 4 end cell end table close square brackets space equals space open square brackets table row 1 cell space space 0 end cell cell space space 0 end cell row 0 cell space space 1 end cell cell space space 0 end cell row 0 cell space space 0 end cell cell space space 1 end cell end table close square brackets space equals space straight I therefore space space straight A space straight A to the power of negative 1 end exponent space equals space straight A to the power of negative 1 end exponent straight A space space equals straight I$

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Short Answer Type

213.

If $straight A space equals space open square brackets table row 3 cell space space minus 3 end cell cell space space space 4 end cell row 2 cell space space minus 3 end cell cell space space 4 end cell row 0 cell space minus 1 end cell cell space space 1 end cell end table close square brackets.$ then A³ = A^–1What is adj. A ?

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Long Answer Type

214.

If $straight A space equals space open square brackets table row 1 cell space space minus 1 end cell cell space space 1 end cell row 2 cell space minus 1 end cell cell space space 0 end cell row 1 cell space space space space 0 end cell cell space space 0 end cell end table close square brackets comma$ find A^–1 and show that A^–1 = A².

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215.

If $straight A space equals space open square brackets table row 3 cell space space minus 3 end cell cell space space space 4 end cell row 2 cell space space minus 3 end cell cell space space space 4 end cell row 0 cell space space minus 1 end cell cell space space space 1 end cell end table close square brackets comma$ show that A⁴ = I. Hence find A^–1

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216.

If A = $open square brackets table row 2 cell space space 0 end cell cell space space minus 1 end cell row 5 cell space space 1 end cell cell space space space space 0 end cell row 0 cell space space 1 end cell cell space space space space 3 end cell end table close square brackets comma$ prove that A^–1 = A^–2 – 6 A + 111.

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217.

If $straight A space equals space open square brackets table row 1 cell space space 2 end cell cell space space 2 end cell row 2 cell space space 1 end cell cell space space 2 end cell row 2 cell space space 2 end cell cell space space 1 end cell end table close square brackets comma space$ Find A^–1 and hence prove that A² – 4A – 5 I = O.

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218.

For the matrix $straight A space equals space open square brackets table row 1 cell space space 1 end cell cell space space space space space 1 end cell row 1 cell space space space 2 end cell cell space minus 3 end cell row 2 cell negative 1 end cell cell space space space space 3 end cell end table close square brackets comma$ show that A³ + 6A² + 5A + 11 I = 0. Hence, find A^–1.

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219.

If $straight A space equals space open square brackets table row cell space space 2 end cell cell space space space minus 1 end cell cell space space space space space 1 end cell row cell negative 1 end cell cell space space space space space 2 end cell cell space minus 1 end cell row cell space space 1 end cell cell space space minus 1 end cell cell space space space space 2 end cell end table close square brackets comma$
verify that A¹ – 6A² + 9A – 4 I = O and hence find A^–1.

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220.

Compute (AB)^-1 where $straight A space equals space open square brackets table row 5 cell space space space 0 end cell cell space space 4 end cell row 2 cell space space 3 space end cell cell space 2 end cell row 1 cell space 2 end cell cell space 1 end cell end table close square brackets comma space space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 1 cell space space 2 end cell cell space space 3 end cell row 1 cell space space 4 end cell cell space space 3 end cell row 1 cell space space 3 end cell cell space space 4 end cell end table close square brackets$

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212. Verify AA–1 = A–1 A = I where

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