Using properties of determinants, prove that from Mathematics D

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 Multiple Choice QuestionsShort Answer Type

311.

For what values of k, the system of linear equations

x + y + z = 2
2x + y - z = 3
3x + 2y + kz = 4

has a unique solution?

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 Multiple Choice QuestionsLong Answer Type

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312.

Using properties of determinants, prove that

open vertical bar table row cell left parenthesis straight x plus straight y right parenthesis squared end cell zx zy row zx cell left parenthesis straight z plus straight y right parenthesis squared end cell xy row zy xy cell left parenthesis straight z plus straight x right parenthesis squared end cell end table close vertical bar space space equals space 2 xyz left parenthesis straight x plus straight y plus straight z right parenthesis cubed


open vertical bar table row cell left parenthesis straight x plus straight y right parenthesis squared end cell zx zy row zx cell left parenthesis straight z plus straight y right parenthesis squared end cell xy row zy xy cell left parenthesis straight z plus straight x right parenthesis squared end cell end table close vertical bar space equals space 2 xyz left parenthesis straight x plus straight y plus straight z right parenthesis cubed
straight L. straight H. straight S comma space Multiplying space straight R subscript 1 comma space straight R subscript 2 space and space straight R subscript 3 space by space straight z comma space straight x space comma space straight y space respectively

equals space 1 over xyz space open vertical bar table row cell straight z left parenthesis straight x plus straight y right parenthesis squared end cell zx cell straight z squared straight y end cell row cell zx squared end cell cell straight x left parenthesis straight z plus straight y right parenthesis squared end cell cell straight x squared straight y end cell row cell zy squared end cell cell xy squared end cell cell straight y left parenthesis straight z plus straight x right parenthesis squared end cell end table close vertical bar
taking space common space straight z comma space straight x comma space straight y comma space from space straight C subscript 1 comma end subscript straight C subscript 2 comma end subscript and space straight C subscript 3

equals space xyz over xyz space open vertical bar table row cell left parenthesis straight x plus straight y right parenthesis squared end cell straight z cell straight z squared end cell row cell straight x squared end cell cell left parenthesis straight z plus straight y right parenthesis squared end cell cell straight x squared end cell row cell straight y squared end cell cell straight y squared end cell cell left parenthesis straight z plus straight x right parenthesis squared end cell end table close vertical bar
straight C subscript 1 rightwards arrow straight C subscript 2 minus straight C subscript 3 space and space straight C subscript 2 rightwards arrow straight C subscript 2 minus straight C subscript 3

taking space common space straight x plus straight y plus straight z space from space straight C subscript 1 space and space straight C subscript 2
left parenthesis straight x plus straight y plus straight z right parenthesis squared space open vertical bar table row cell straight x plus straight y minus straight z end cell 0 cell straight z squared end cell row 0 cell left parenthesis straight z plus straight y right parenthesis squared end cell cell straight x squared end cell row cell straight y minus straight z minus straight x end cell cell straight y minus straight z minus straight x end cell cell left parenthesis straight z plus straight x right parenthesis squared end cell end table close vertical bar
straight R subscript 3 rightwards arrow straight R subscript 3 rightwards arrow left parenthesis straight R subscript 1 plus straight R subscript 2 right parenthesis

equals space left parenthesis straight x plus straight y plus straight z right parenthesis squared space open vertical bar table row cell straight x plus straight y minus straight z end cell 0 cell straight z squared end cell row 0 cell straight z plus straight y minus straight x end cell cell straight x squared end cell row cell negative 2 straight x end cell cell negative 2 zx end cell cell 2 xz end cell end table close vertical bar
straight C subscript 1 rightwards arrow zC subscript 1 comma space straight C subscript 2 space rightwards arrow xC subscript 3

equals space fraction numerator left parenthesis straight x plus straight y plus straight z right parenthesis squared over denominator xz end fraction space open vertical bar table row cell straight z left parenthesis straight x plus straight y minus straight z right parenthesis end cell 0 cell straight z squared end cell row 0 cell straight x left parenthesis straight z plus straight y minus straight x right parenthesis end cell cell straight x squared end cell row cell negative 2 xz end cell cell negative 2 zx end cell cell 2 xz end cell end table close vertical bar
straight C subscript 1 rightwards arrow straight C subscript 1 plus straight C subscript 3 comma space space straight C subscript 2 space rightwards arrow straight C subscript 2 plus straight C subscript 3

equals space fraction numerator left parenthesis straight x plus straight y plus straight z right parenthesis squared over denominator xz end fraction space open vertical bar table row cell straight z left parenthesis straight x plus straight y right parenthesis end cell cell straight z squared end cell cell straight z squared end cell row cell straight x squared end cell cell straight x left parenthesis straight z plus straight y right parenthesis end cell cell straight x squared end cell row 0 0 cell 2 xz end cell end table close vertical bar
taking space straight z space and space straight x space common space from space straight R subscript 1 space and space straight R subscript 2

expansion space along space space straight R subscript 3
equals space left parenthesis straight x plus straight y plus straight z right parenthesis squared space straight x space 2 xz space left parenthesis left parenthesis straight x plus straight y right parenthesis left parenthesis straight z plus straight y right parenthesis minus xz right parenthesis

equals space left parenthesis straight x plus straight y plus straight z right parenthesis squared space straight x space 2 xz left parenthesis xz plus xy plus yz plus straight y squared minus xz right parenthesis

equals space left parenthesis straight x plus straight y plus straight z right parenthesis squared space straight x space 2 xz left parenthesis xy space plus yz space plus straight y squared right parenthesis

equals space 2 xyz space left parenthesis straight x plus straight y plus straight z right parenthesis cubed
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 Multiple Choice QuestionsShort Answer Type

313.

Using the properties of determinants, solve the following for x:

open vertical bar table row cell straight x plus 2 end cell cell space space straight x plus 6 end cell cell space straight x minus 1 end cell row cell straight x plus 6 end cell cell straight x minus 1 end cell cell straight x plus 2 end cell row cell straight x minus 1 end cell cell straight x plus 2 end cell cell straight x plus 6 end cell end table close vertical bar space equals space 0

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314.

If open vertical bar table row cell 3 straight x end cell cell space 7 end cell row cell negative 2 end cell cell space 4 end cell end table close vertical bar space equals open vertical bar table row 8 cell space 7 end cell row 6 cell space 4 end cell end table close vertical bar comma find the value of x.

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315.

Use Properties of determinants, prove that:
open vertical bar table row cell 1 plus straight a end cell cell space 1 end cell cell space 1 end cell row 1 cell 1 plus straight b end cell 1 row 1 1 cell 1 plus straight c end cell end table close vertical bar space equals space abc plus bc plus ca plus ab

389 Views

316.

If open vertical bar table row cell straight x plus 1 end cell cell space space straight x minus 1 end cell row cell straight x minus 3 end cell cell space straight x plus 2 end cell end table close vertical bar space equals space open vertical bar table row 4 cell space space minus 1 end cell row 1 cell space space space space 3 end cell end table close vertical bar comma then write the value of x. 

193 Views

317.

Using properties of determinants prove the following:
open vertical bar table row 1 cell space space straight x end cell cell space space straight x squared end cell row cell straight x squared end cell cell space 1 end cell straight x row straight x cell space straight x end cell 1 end table close vertical bar space equals space left parenthesis 1 minus straight x cubed right parenthesis squared

247 Views

318.

Using properties of determinants, prove that 

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 Multiple Choice QuestionsLong Answer Type

319.

Use product open square brackets table row 1 cell negative 1 end cell 2 row 0 2 cell negative 3 end cell row 3 cell negative 2 end cell 4 end table close square brackets space space open square brackets table row cell negative 2 end cell 0 1 row 9 2 cell negative 3 end cell row 6 1 cell negative 2 end cell end table close square bracketsto solve the system of equations x + 3z = 9,
–x + 2y – 2z = 4, 2x – 3y + 4z = –3

733 Views

 Multiple Choice QuestionsShort Answer Type

320.

Using properties of determinants, prove that

111+3x1+ 3y1111+3z1 = 9 (3xyz + xy + yz + zx)


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