If  then compute the inverse of A and verify that A–1 A =

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 Multiple Choice QuestionsLong Answer Type

201.

Find straight A to the power of negative 1 end exponent if straight A space equals space open square brackets table row 0 cell space space space space 1 end cell cell space space space space 1 end cell row 1 cell space space space space 0 end cell cell space space space 1 end cell row 1 cell space space space space 1 end cell cell space space space 0 end cell end table close square brackets Also show that straight A to the power of negative 1 end exponent space space equals fraction numerator straight A squared minus 3 straight I over denominator 2 end fraction.

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202.

If straight A space equals space open square brackets table row 1 cell space space 3 end cell cell space space space 3 end cell row 1 cell space space 4 end cell cell space space 3 end cell row 1 cell space 3 end cell cell space space space 4 end cell end table close square brackets comma then verify that A adj A = | A | I. Also find A .

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 Multiple Choice QuestionsShort Answer Type

203.

Find the inverse of the matrix open square brackets table row 1 cell space space 2 end cell cell space space 3 end cell row 0 cell space space 2 end cell cell space space 4 end cell row 0 cell space space 0 end cell cell space space 5 end cell end table close square brackets.

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 Multiple Choice QuestionsLong Answer Type

204.

Find the inverse of the matrix open square brackets table row 1 cell space space space space 0 end cell cell space space space space 0 end cell row 3 cell space space space 3 end cell cell space space space space space 0 end cell row 5 cell space space space space 2 end cell cell space space minus 1 end cell end table close square brackets

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205.

Find the inverse of the matrix:
open square brackets table row 2 cell space space space space space 1 end cell cell space space space space space 3 end cell row 4 cell space minus 1 end cell cell space space space space space 0 end cell row cell negative 7 end cell cell space space space space space 2 end cell cell space space space space 1 end cell end table close square brackets
 

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206.

Find the inverse of the matrix:
open square brackets table row cell space space space 2 end cell cell space 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 space space 2 end cell cell space space minus 1 end cell row cell space space 1 end cell cell space space space space minus 1 end cell cell space space space space space 2 end cell end table close square brackets

 

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 Multiple Choice QuestionsShort Answer Type

207.

Find the inverse of the matrix:
open square brackets table row 2 cell space space minus 3 end cell cell space space space space 3 end cell row 3 cell space space space space space 2 end cell cell space space space space space 3 end cell row 3 cell space space minus 2 end cell cell space space space space 2 end cell end table close square brackets

 

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 Multiple Choice QuestionsLong Answer Type

208.

Find the inverse of the matrix:
open square brackets table row 1 cell space space 2 end cell cell space space 2 end cell row 2 cell space 1 end cell cell space 2 end cell row 2 cell space 2 end cell cell space 1 end cell end table close square brackets


 

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209.

Find the inverse of the matrix:
open square brackets table row 1 cell space 0 end cell cell space 0 end cell row 0 cell space space cos space straight alpha end cell cell space space sin space straight alpha end cell row 0 cell space sin space straight alpha end cell cell space space minus cos space straight alpha end cell end table close square brackets



 

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210.

If straight A space equals space open square brackets table row 1 cell space 2 end cell cell space space 5 end cell row 2 cell space 3 end cell cell space 1 end cell row cell negative 1 end cell cell space 1 end cell cell space 1 end cell end table close square brackets comma then compute the inverse of A and verify that A–1 A = I = A A–1


straight A space equals space open square brackets table row 1 cell space space space 2 end cell cell space space 5 end cell row 2 cell space space space 3 space end cell cell space 1 end cell row cell negative 1 end cell cell space 1 end cell cell space 1 end cell end table close square brackets
therefore space space space space open vertical bar straight A close vertical bar equals space open vertical bar table row 1 cell space space 2 end cell cell space space 5 space end cell row 2 cell space 3 end cell 1 row cell negative 1 end cell 1 1 end table close vertical bar space equals space 1 space open vertical bar table row 3 cell space space space space 1 end cell row 1 cell space space space space space 1 end cell end table close vertical bar minus 2 open vertical bar table row cell space space space 2 end cell cell space space space 1 end cell row cell negative 1 end cell cell space space space 1 end cell end table close vertical bar plus 5 space open vertical bar table row 2 cell space space space 3 end cell row cell negative 1 end cell cell space space space 1 end cell end table close vertical bar
space space space space space space space space space space space space space space equals space 1 left parenthesis 3 minus 1 right parenthesis space minus 2 left parenthesis 2 plus 1 right parenthesis plus 5 left parenthesis 2 plus 3 right parenthesis
space space space space space space space space space space space space space space space equals 2 minus 6 plus 25 space equals space 21 space not equal to space 0
space space space 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 3 cell space space space space 1 end cell row 1 cell space space space space space 1 end cell end table close vertical bar comma space minus open vertical bar table row 2 cell space space space 1 end cell row cell negative 1 end cell cell space space space 1 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space 3 end cell row cell negative 1 end cell cell space space space 1 end cell end table close vertical bar
i.e., 2, – 3,  5 respectively.
Co-factors of the elements of the second row of | A | are
negative open vertical bar table row 2 cell space space 5 end cell row 1 cell space space 1 end cell end table close vertical bar comma space space open vertical bar table row cell space space 1 end cell cell space space 5 end cell row cell negative 1 end cell cell space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row cell space space 1 end cell cell space space space 2 end cell row cell negative 1 end cell cell space space 1 end cell end table close vertical bar
i.e.,  3,   6, – 3 respectively.
Co-factors of the elements of the third row of | A | are
open vertical bar table row 2 cell space space space 5 end cell row 3 cell space space space 1 end cell end table close vertical bar comma space minus open vertical bar table row 1 cell space space space space 5 end cell row 2 cell space space space 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space 2 end cell row 2 cell space space space space 3 end cell end table close vertical bar
i.e., – 13, 9, – 1 respectively.
therefore space space space adj. space straight A space equals space open square brackets table row 2 cell space space space minus 3 end cell cell space space space space space space 5 end cell row 3 cell space space space space 6 end cell cell space space minus 3 end cell row cell negative 13 end cell cell space space space 9 end cell cell space space minus 1 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 2 cell space space 3 end cell cell space space minus 13 end cell row cell negative 3 end cell cell space space space 6 end cell cell space space space space space 9 end cell row 5 cell negative 3 end cell cell space space space minus 1 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 over 21 open square brackets table row 2 cell space space space space 3 end cell cell space space minus 13 end cell row cell negative 3 end cell cell space space space space 6 end cell cell space space space space space space space 9 end cell row 5 cell space minus 3 end cell cell space space space minus 1 end cell end table close square brackets
Also comma space space space straight A to the power of negative 1 end exponent straight A space equals space 1 over 21 open square brackets table row 2 cell space space space space 3 end cell cell space space minus 13 end cell row cell negative 3 end cell cell space space space space space space 6 end cell cell space space space space space space 9 end cell row 5 cell space minus 3 end cell cell space space minus 1 end cell end table close square brackets space open square brackets table row 1 cell space space 2 end cell cell space space space 5 end cell row 2 cell space space space 3 end cell cell space space space 1 end cell row cell negative 1 end cell cell space 1 end cell cell space space space 1 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 over 21 open square brackets table row cell 2 plus 6 plus 13 end cell cell space space space space space space 4 plus 9 minus 13 end cell cell space space space space space 10 plus 3 minus 13 end cell row cell negative 3 plus 12 minus 9 end cell cell space space space minus 6 plus 18 plus 9 end cell cell space space space minus 15 plus 6 plus 9 end cell row cell 5 minus 6 plus 1 end cell cell space space 10 minus 9 minus 1 end cell cell space space space space space space 25 minus 3 minus 1 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 over 21 open square brackets table row 21 cell space space 0 end cell cell space space 0 end cell row 0 cell space space 21 end cell cell space space 0 end cell row 0 0 cell space space 21 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 space 0 end cell row 0 cell space space 1 end cell cell space 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 space straight A to the power of negative 1 end exponent straight A space space equals space straight I.
space space space space space space space AA to the power of negative 1 end exponent space equals space 1 over 21 open square brackets table row 1 cell space space space 2 end cell cell space space 5 end cell row 2 cell space space 3 end cell cell space space 1 space end cell row cell negative 1 end cell cell space space 1 end cell cell space 1 end cell end table close square brackets space open square brackets table row 2 cell space space space 3 end cell cell space space space minus 13 end cell row cell negative 3 end cell cell space space 6 end cell cell space space space space 9 end cell row 5 cell negative 3 end cell cell space space minus 1 end cell end table close square brackets

                  equals space 1 over 21 open square brackets table row cell 2 minus 6 plus 25 end cell cell space space space 4 plus 12 minus 15 end cell cell space space space 13 plus 18 minus 5 end cell row cell 4 minus 9 plus 25 end cell cell space 6 plus 18 minus 3 end cell cell space minus 26 plus 27 minus 1 end cell row cell negative 2 minus 3 plus 5 end cell cell space minus 3 plus 6 minus 3 end cell cell space 13 plus 9 minus 1 end cell end table close square brackets
equals space 1 over 21 space open square brackets table row 21 cell space space space 0 end cell cell space space 0 end cell row 0 cell space 21 end cell cell space 0 end cell row 0 0 cell space 21 end cell end table close square brackets space equals space open square brackets table row 1 cell space space 0 end cell cell space 0 end cell row 0 cell space 1 end cell cell space 0 end cell row 0 cell space 0 end cell cell space 1 end cell end table close square brackets space equals space straight I
therefore space space we space have space straight A to the power of negative 1 end exponent straight A space equals space straight I space equals space straight A space straight A to the power of negative 1 end exponent.

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