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Determinants

Long Answer Type

191.

If $straight A space equals open square brackets table row 2 cell space space space space space space space 3 end cell row 1 cell space space minus 4 end cell end table close square brackets comma space straight B space equals space open square brackets table row cell space space space 1 end cell cell space space space space minus 2 end cell row cell negative 1 end cell cell space space space space space space 3 end cell end table close square brackets comma$
verify that (AB)^–1 = B^–1 A^–1

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192.

Let $straight A space equals space open square brackets table row 3 cell space space space 7 end cell row 2 cell space space 5 end cell end table close square brackets space and space straight B space equals space open square brackets table row 6 cell space space 8 end cell row 7 cell space space 9 end cell end table close square brackets comma$ verify that (AB)^–1 = B^–1 A^–1.

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193. Verify that (AB)^–1 = B^–1 A^–1 for the matrices A and B where

straight A space equals space open square brackets table row 3 cell space space 2 end cell row 7 cell space space 5 end cell end table close square brackets comma space space straight B space equals space open square brackets table row 4 cell space space 6 end cell row 3 cell space space 2 end cell end table close square brackets

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194. Verify that (AB)^–1 = B^–1 A^–1 for the matrices A and B where $straight A space equals open square brackets table row 3 cell space space 2 end cell row 7 cell space space 5 end cell end table close square brackets comma space space straight B space equals space open square brackets table row 6 cell space space space space 7 end cell row 8 cell space space space space 9 end cell end table close square brackets$ .

straight A space equals space open square brackets table row 3 cell space space 2 end cell row 7 cell space space 5 end cell end table close square brackets space space straight B space equals space open square brackets table row 6 cell space space 7 end cell row 8 cell space space 9 end cell end table close square brackets

open vertical bar straight A close vertical bar space equals space open vertical bar table row 3 cell space space space 2 end cell row 7 cell space space space space 5 space end cell end table close vertical bar space equals space 15 minus 14 space equals space 1

Co-factors of the elements of first row | A | are 5, – 7 respectively.
Co-factors of the elements of second row of | A | are – 2, 3 respectively.

therefore space space adj. space straight A space equals space open square brackets table row 5 cell space space minus 7 end cell row cell negative 2 end cell cell space space space 3 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 5 cell space space space minus 2 end cell row cell negative 7 end cell cell space space space space 3 end cell end table close square brackets 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 fraction numerator adj. space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space open square brackets table row cell space space space 5 end cell cell space space space space minus 2 end cell row cell negative 7 end cell cell space space space space space space 3 end cell end table close square brackets space space space space space space space space space space space space space open vertical bar straight B close vertical bar space equals space open vertical bar table row 6 cell space space space 7 end cell row 8 cell space space 9 end cell end table close vertical bar space equals space 54 minus 56 space equals space minus 2

Co-factors of the elements of first row of | B | are 9, – 8 respectively
Co-factors of the-elements of second row of | B | are – 7, 6 respectively
$therefore space space space adj. space straight B space equals space open square brackets table row 9 cell space space space space minus 8 end cell row cell negative 7 end cell cell space space space space space space 6 end cell end table close square brackets to the power of apostrophe space space equals space open square brackets table row 9 cell space space space minus 7 end cell row cell negative 8 end cell cell space space space space space 6 end cell end table close square brackets therefore space space space space space space straight B to the power of negative 1 end exponent space equals space fraction numerator adj. space straight B over denominator open vertical bar straight B close vertical bar end fraction space equals space minus 1 half open square brackets table row 9 cell space space space space minus 7 end cell row cell negative 8 end cell cell space space space space space space space 6 end cell end table close square brackets therefore space space space space space straight B to the power of negative 1 end exponent straight A to the power of negative 1 end exponent space equals negative 1 half open square brackets table row 9 cell space space space minus 7 end cell row cell negative 8 end cell cell space space space space space space 6 end cell end table close square brackets space space space open square brackets table row 5 cell space space space minus 2 end cell row cell negative 7 end cell cell space space space 3 end cell end table close square brackets space equals space minus 1 half open square brackets table row cell 45 plus 49 end cell cell space space space space minus 18 minus 21 end cell row cell negative 40 minus 42 end cell cell space space space space space space space 16 plus 18 end cell end table close square brackets space space space space space space space space straight B to the power of negative 1 end exponent straight A to the power of negative 1 end exponent space equals space minus 1 half open square brackets table row 94 cell space space space space minus 39 end cell row cell negative 82 end cell cell space space space space space 34 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... left parenthesis 1 right parenthesis space space space space AB space equals space open square brackets table row 3 cell space space space 2 end cell row 7 cell space space space 5 end cell end table close square brackets space open square brackets table row 6 cell space space 7 end cell row 8 cell space space 9 end cell end table close square brackets space equals open square brackets table row cell 18 plus 16 end cell cell space space space 21 plus 18 end cell row cell 42 plus 40 end cell cell space space 49 plus 45 end cell end table close square brackets space equals space open square brackets table row 34 cell space space 39 space end cell row 82 cell space 94 end cell end table close square brackets space space space open vertical bar AB close vertical bar space equals space open vertical bar table row 34 cell space space 39 end cell row 82 cell space space 94 end cell end table close vertical bar space equals space 3196 minus 3198 space equals space minus 2$

Co-factors of the elements of first row of | AB | are 94, – 82 respectively.
Co-factors of the elements of second row of | AB | are – 39, 34 respectively.
$therefore space space adj. space left parenthesis AB right parenthesis space equals space open square brackets table row cell space space 94 end cell cell space space space minus 82 end cell row cell negative 39 end cell cell space space space space space space space 34 end cell end table close square brackets to the power of apostrophe space equals open square brackets table row cell space 94 end cell cell space space minus 39 end cell row cell negative 82 end cell cell space space space space 34 end cell end table close square brackets therefore space space left parenthesis AB right parenthesis to the power of negative 1 end exponent space equals space fraction numerator adj. space left parenthesis AB right parenthesis over denominator open vertical bar AB close vertical bar end fraction equals space minus 1 half open square brackets table row cell space space 94 end cell cell space space space minus 39 end cell row cell negative 82 end cell cell space space space space space 34 end cell end table close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis space space space space space From space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space get space left parenthesis AB right parenthesis to the power of negative 1 end exponent space equals space straight B to the power of negative 1 end exponent straight A to the power of negative 1 end exponent.$

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195. Verify that (AB)^–1 = B^–1 A^–1 for the matrices A and B where

straight A space equals space open square brackets table row 2 cell space space space 1 end cell row 5 cell space space space 3 end cell end table close square brackets comma space space space straight B space equals space open square brackets table row 4 cell space space space 5 end cell row 3 cell space space 4 end cell end table close square brackets

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Short Answer Type

196.

Find (AB)^–1 if $straight A space equals space open square brackets table row 3 cell space 4 end cell row 1 cell space 1 end cell end table close square brackets comma space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 4 cell space space 3 end cell row 2 cell space space 1 end cell end table close square brackets$

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197.

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

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Long Answer Type

198.

Find the inverse of the matrix
$straight A space equals space open square brackets table row cell negative 1 end cell cell space space 1 end cell cell space space 2 end cell row cell space space space 3 end cell cell negative 1 end cell cell space space 1 space end cell row cell space minus 1 end cell cell space space 3 end cell cell space 4 end cell end table close square brackets.$

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199. Find the inverse of the matrix:

open square brackets table row 1 cell space space space minus 1 end cell cell space space space space space space 2 end cell row 0 cell space space space space space space 2 end cell cell space space space minus 3 end cell row 3 cell space space space minus 2 end cell cell space space space space space space 4 end cell end table close square brackets.

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200.

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

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Long Answer Type

194. Verify that (AB)–1 = B–1 A–1 for the matrices A and B where .

Short Answer Type

Long Answer Type