Find the inverse of the matrix:  from Mathematics Determinants

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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

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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

 


straight A space equals 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 space 2 end cell cell space space space space 3 end cell row 3 cell space space minus 2 end cell cell space space space 2 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 2 cell space space minus 3 end cell cell space space space 3 end cell row 3 cell space space space space 2 end cell cell space space 3 end cell row 3 cell negative 2 end cell cell space space 2 end cell end table close vertical bar space equals space 2 open vertical bar table row cell space space 2 end cell cell space space space 3 end cell row cell negative 2 end cell cell space space space 2 end cell end table close vertical bar minus left parenthesis negative 3 right parenthesis space open vertical bar table row 2 cell space space space 3 end cell row 3 cell space space space 2 end cell end table close vertical bar space plus space 3 open vertical bar table row 2 cell 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 space space space space space space space space space equals space 2 left parenthesis 4 plus 6 right parenthesis plus 3 left parenthesis 4 minus 9 right parenthesis plus 3 left parenthesis negative 4 minus 6 right parenthesis space equals space 20 minus 15 minus 30 space equals space minus 25 not equal to 0.
therefore space space space straight A to the power of negative 1 end exponent space exists.
Co-factors of the elements of the first row of | A | are
open vertical bar table row cell space space 2 end cell cell space space 3 end cell row cell negative 2 end cell cell space space 2 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space 3 end cell row 3 cell space space space space 2 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space space space space 2 end cell row 3 cell space space minus 2 end cell end table close vertical bar
i.e. 10, 5,  – 10 respectively
Co-factors of the elements of the second row of | A | are
negative open vertical bar table row cell negative 3 end cell cell space space space 3 end cell row cell negative 2 end cell cell space space 2 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space 3 end cell row 3 cell space space 2 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space minus 3 end cell row 3 cell space space minus 2 end cell end table close vertical bar
i.e., 0, – 5, – 5 respectively
Co-factors of the elements of the third row of | A | are
open vertical bar table row cell negative 3 end cell cell space space space 3 end cell row 2 cell space space space 3 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space 3 end cell row 2 cell space space 3 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space minus 3 end cell row 2 cell space space space space 2 end cell end table close vertical bar
i.e. – 15, 0, 10 respectively.
therefore space space space adj. space straight A space equals space open square brackets table row 10 cell space space 5 end cell cell space space space minus 10 end cell row 5 cell negative 5 end cell cell space space minus 5 end cell row cell negative 15 end cell cell space space 0 end cell cell space space space space space 10 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 10 cell space space 0 end cell cell space space minus 15 end cell row 5 cell space minus 5 end cell cell space space space space space space 0 end cell row cell negative 10 end cell cell space minus 5 end cell cell space space space space 10 end cell end table close square brackets
Now comma 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 negative 1 over 25 open square brackets table row 10 cell space space 0 end cell cell space space space minus 15 end cell row 5 cell negative 5 end cell cell space space space space space 0 end cell row cell negative 10 end cell cell space minus 5 end cell cell space space space space space 10 end cell end table close square brackets space equals 1 fifth open square brackets table row cell negative 2 end cell cell space space 0 end cell cell space space space space space space 3 end cell row cell negative 1 end cell cell space space 1 end cell cell space space space space space 0 end cell row 2 cell space space 1 end cell cell space space minus 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

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