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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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Short 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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Long Answer Type

204.
Find the inverse of the matrix $open square brackets table row 1 cell space space space space 0 end cell cell space space space space 0 end cell row 3 cell space space space 3 end cell cell space space space space space 0 end cell row 5 cell space space space space 2 end cell cell space space minus 1 end cell end table close square brackets$

straight A space equals space 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 2 end cell cell space space minus 1 end cell end table close square brackets therefore space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space space space 0 end cell cell space space space 0 end cell row 3 cell space space space space 3 end cell cell space space space space 0 end cell row 5 cell space space space space 2 end cell cell space minus 1 end cell end table close vertical bar space equals space left parenthesis 1 right parenthesis thin space left parenthesis 3 right parenthesis thin space left parenthesis negative 1 right parenthesis space equals space minus 3 space not equal to space 0 rightwards double arrow space space space space space straight A to the power of negative 1 end exponent space exists. 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 space space space left square bracket Product space of space diagonal space elements right square bracket

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

open vertical bar table row 3 cell space space space 0 end cell row 2 cell space space minus 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 3 cell space space space space space space 0 end cell row 5 cell space space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 3 cell space space space space space 3 end cell row 5 cell space space space 2 end cell end table close vertical bar

i.e. – 3, 3, – 9 respectively
Co-factors of the elements of the second row of | A | are

negative open vertical bar table row 0 cell space space space 0 end cell row 2 cell space space space 1 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space space space space 0 end cell row 5 cell space space minus 1 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space 0 end cell row 5 cell space space 2 end cell end table close vertical bar

i.e. 0, – 1, – 2 respectively
Co-factors of the elements of the third row of | A | are

open vertical bar table row 0 cell space space space 0 end cell row 3 cell space space space 0 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space 0 end cell row 3 cell space space 0 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space 0 end cell row 3 cell space space 3 end cell end table close vertical bar

i.e. 0, 0, 3 respectively

therefore space space space space adj space straight A space equals space open square brackets table row cell negative 3 end cell cell space space 3 end cell cell space space space minus 9 end cell row cell space 0 end cell cell space minus 1 end cell cell space space minus 2 end cell row cell space 0 end cell cell space 0 end cell cell space space space space 3 space end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 3 end cell cell space space space 0 end cell cell space space space 0 end cell row cell space space space 3 end cell cell space minus 1 end cell cell space space space 0 end cell row cell negative 9 end cell cell space minus 2 end cell cell space space space 3 end cell end table close square brackets Now 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 third open square brackets table row cell negative 3 end cell cell space space 0 end cell cell space space space 0 end cell row 3 cell negative 1 end cell cell space space 0 end cell row cell negative 9 end cell cell negative 2 end cell cell space space 3 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:
$open square brackets table row cell space space space 2 end cell cell space space space space minus 1 end cell cell space space space space space 1 end cell row cell negative 1 end cell cell space space space space space space space 2 end cell cell space space minus 1 end cell row cell space space 1 end cell cell space space space space minus 1 end cell cell space space space space space 2 end cell end table close square brackets$

75 Views

Short Answer Type

207.

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

72 Views

Long Answer Type

208.

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

82 Views

209.

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

74 Views

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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204. Find the inverse of the matrix

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