Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Determinants

Long 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 $straight A to the power of negative 1 end exponent space space equals fraction numerator straight A squared minus 3 straight I over denominator 2 end fraction.$

straight A space equals space open square brackets table row 0 cell space space 1 end cell cell space 1 space end cell row 1 cell space space 0 end cell 1 row 1 cell space 1 end cell 0 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 0 cell space space space space 1 end cell cell space space space 1 end cell row 1 cell space space space 0 end cell cell space space space space 1 end cell row 1 cell space space space 1 end cell cell space space space 0 end cell end table close vertical bar space equals space 0 space open vertical bar table row 0 cell space space 1 end cell row 1 cell space space 0 end cell end table close vertical bar space minus space 1 space open vertical bar table row 1 cell space space 1 end cell row 1 cell space space 0 end cell end table close vertical bar space plus space 1 space open vertical bar table row 1 cell space space 0 end cell row 1 cell space space 1 end cell end table close vertical bar space space space space space space space space space space space space space space space equals space 0 left parenthesis 0 minus 1 right parenthesis space minus 1 space left parenthesis 0 minus 1 right parenthesis space plus space 1 left parenthesis 1 minus 0 right parenthesis space equals space 0 plus 1 plus 1 space equals space 2 space 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 first row of | A | are

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

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

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

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

open vertical bar table row 1 cell space space 1 end cell row 0 cell space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 0 cell space space space 1 end cell row 1 cell space space 1 end cell end table close vertical bar comma space space open vertical bar table row 0 cell space space 1 end cell row 1 cell space space 0 end cell end table close vertical bar space space space straight i. straight e. space space space space space 1 comma space 1 comma space minus 1 space respectively.

therefore space space space space adj space straight A space equals space open square brackets table row cell negative 1 end cell 1 cell space space 1 end cell row cell space space 1 end cell cell negative 1 end cell cell space space 1 end cell row cell space space 1 end cell 1 cell negative 1 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 1 end cell cell space space space space 1 end cell cell space space space space space 1 end cell row 1 cell negative 1 end cell cell space space space space space space 1 end cell row 1 cell space space 1 end cell cell space space minus 1 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 1 half open square brackets table row cell negative 1 end cell cell space space space 1 end cell cell space space space 1 end cell row 1 cell negative 1 end cell cell space space space space 1 space end cell row 1 cell space 1 end cell cell negative 1 end cell end table close square brackets space space space space space space space space space space space space straight A squared space equals space open square brackets table row 0 cell space space 1 end cell cell space space space 1 end cell row 1 cell space space 0 end cell cell space space space 1 end cell row 1 cell space space 1 end cell cell space space space 0 end cell end table close square brackets space space open square brackets table row 0 cell space space 1 end cell cell space space 1 end cell row 1 cell space space 0 end cell cell space space 1 end cell row 1 cell space 1 end cell cell space space 0 end cell end table close square brackets space equals space open square brackets table row 2 cell space space 1 end cell cell space space space 1 end cell row 1 cell space space 2 end cell cell space space 1 end cell row 1 cell space space 1 end cell cell space space 2 end cell end table close square brackets therefore space space space straight A squared minus 3 straight I space equals space open square brackets table row 2 cell space space space 1 end cell cell space space 1 end cell row 1 cell space space 2 end cell cell space space 1 end cell row 1 cell space space 1 end cell cell space space 2 end cell end table close square brackets space minus space space 3 open square brackets table row 1 cell space 0 end cell cell space 0 end cell row 0 cell space 1 space 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 open square brackets table row cell negative 1 end cell cell space 1 end cell cell space space space space 1 end cell row 1 cell negative 1 end cell cell space space space space 1 end cell row 1 cell space space 1 end cell cell space space minus 1 end cell end table close square brackets therefore space space space space fraction numerator straight A squared minus 3 straight I over denominator 2 end fraction space equals space 1 half open square brackets table row cell negative 1 end cell cell space space space 1 end cell cell space space space space 1 end cell row cell space space 1 end cell cell space minus 1 end cell cell space space space space 1 end cell row cell space space 1 end cell cell space space 1 end cell cell space space space space 1 end cell end table close square brackets therefore space space space space space straight A to the power of negative 1 end exponent space equals space fraction numerator straight A squared space minus space 3 straight I over denominator 2 end fraction

77 Views

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 .

86 Views

Short Answer Type

203.

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

76 Views

Long Answer Type

204.

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

73 Views

205.

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

74 Views

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

75 Views

Previous Year Papers

Test Series

Test Yourself

Long Answer Type

201. Find if Also show that

Short Answer Type

Long Answer Type

Short Answer Type

Long Answer Type