Using properties of determinants prove the following: from Math

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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

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

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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

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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. 

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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


increment space equals space open vertical bar table row 1 cell space space straight x end cell cell 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 squared end cell 1 end table close vertical bar
Applying straight R subscript 1 rightwards arrow straight R subscript 1 plus straight R subscript 2 plus straight R subscript 3 comma space we have
increment space equals space open vertical bar table row cell 1 plus straight x plus straight x squared end cell cell 1 plus straight x plus straight x squared end cell cell 1 plus straight x plus straight x squared end cell row cell straight x squared end cell 1 straight x row straight x cell straight x to the power of 21 end cell blank end table close vertical bar
equals space left parenthesis 1 plus straight x plus straight x squared right parenthesis space open vertical bar table row 1 cell space space 1 end cell cell space space space 1 end cell row cell straight x squared end cell cell space space 1 end cell cell space space straight x end cell row straight x cell space space straight x squared end cell cell space space space 1 end cell end table close vertical bar
  Applying space straight C subscript 2 rightwards arrow straight C subscript 2 minus straight C subscript 1 space and space straight C subscript 3 minus straight C subscript 1 comma space we space have colon
increment space equals space left parenthesis 1 plus straight x plus straight x squared right parenthesis space open vertical bar table row 1 cell space space space 0 end cell cell space space 0 end cell row cell straight x squared end cell cell space 1 minus straight x squared end cell cell space space straight x minus straight x squared end cell row straight x cell space 1 plus straight x squared end cell cell space 1 minus straight x end cell end table close vertical bar
equals left parenthesis 1 plus straight x plus straight x squared right parenthesis thin space left parenthesis 1 minus straight x right parenthesis left parenthesis 1 minus straight x right parenthesis space open vertical bar table row 1 cell space 0 end cell cell space 0 end cell row cell straight x squared end cell cell 1 plus straight x end cell cell space straight x end cell row straight x cell negative straight x end cell cell space 1 end cell end table close vertical bar
equals left parenthesis 1 minus straight x cubed right parenthesis left parenthesis 1 minus straight x right parenthesis open vertical bar table row 1 cell space 0 end cell cell space 0 end cell row cell straight x squared space end cell cell 1 plus straight x end cell cell space straight x end cell row straight x cell negative straight x end cell cell space 1 end cell end table close vertical bar
Expanding along R1, we have:
increment equals left parenthesis 1 minus straight x cubed right parenthesis left parenthesis 1 minus straight x right parenthesis thin space left parenthesis 1 right parenthesis space open vertical bar table row cell 1 plus straight x end cell cell space space space straight x end cell row cell negative straight x end cell cell space space 1 end cell end table close vertical bar
space space space equals left parenthesis 1 minus straight x cubed right parenthesis left parenthesis 1 minus straight x right parenthesis left parenthesis 1 plus straight x plus straight x squared right parenthesis
space space space equals left parenthesis 1 minus straight x cubed right parenthesis left parenthesis 1 minus straight x cubed right parenthesis
space space space equals left parenthesis 1 minus straight x cubed right parenthesis squared
Hence space proved.
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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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