Let A be a square matrix of order 3 × 3, then | kA | is equal

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 Multiple Choice QuestionsShort Answer Type

121.

Solve the equation
open vertical bar table row cell straight x plus straight a end cell cell space straight x end cell cell space straight x end cell row straight x cell space straight x plus straight a end cell cell space straight x end cell row straight x straight x cell space straight x plus straight a end cell end table close vertical bar space equals space 0 comma space space space straight a not equal to space 0.

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 Multiple Choice QuestionsLong Answer Type

122.

Solve:
open vertical bar table row cell straight x plus 1 end cell 2 3 row 3 cell straight x plus 2 end cell 1 row 1 2 cell straight x plus 3 end cell end table close vertical bar space equals 0

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123.

Prove that:
open vertical bar table row cell straight a squared plus 1 end cell cell straight a space straight b end cell cell straight a space straight c end cell row cell straight a space straight b end cell cell straight b squared plus 1 end cell cell straight b space straight c end cell row cell straight a space straight c end cell cell straight b space straight c end cell cell straight c squared plus 1 end cell end table close vertical bar space equals space 1 plus straight a squared plus straight b squared plus straight c squared.

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 Multiple Choice QuestionsShort Answer Type

124.

If x, y, z are different and
increment space equals space open vertical bar table row straight x cell space space space space straight x squared end cell cell space 1 plus straight x cubed end cell row straight y cell space space space straight y squared end cell cell space 1 plus straight y cubed end cell row straight z cell space space straight z squared end cell cell space 1 plus straight z cubed end cell end table close vertical bar space equals space 0 comma space space then space show space that space 1 plus xyz space equals 0

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 Multiple Choice QuestionsLong Answer Type

125.

Prove that:
open vertical bar table row straight x cell space space space straight x squared end cell cell space space 1 plus px cubed end cell row straight y cell space space straight y squared end cell cell space space 1 plus py cubed end cell row straight z cell space space straight z squared end cell cell space space 1 plus pz cubed end cell end table close vertical bar space equals space left parenthesis 1 plus pxyz right parenthesis thin space left parenthesis straight x minus straight y right parenthesis thin space left parenthesis straight y minus straight z right parenthesis thin space left parenthesis straight z minus straight x right parenthesis
where p is any scalar.

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 Multiple Choice QuestionsShort Answer Type

126.

If space open vertical bar table row straight a cell space space straight a cubed end cell cell space space straight a to the power of 4 minus 1 end cell row straight b cell space straight b cubed end cell cell space space straight b to the power of 4 minus 1 end cell row straight c cell space straight c cubed end cell cell space space straight c to the power of 4 minus 1 end cell end table close vertical bar space equals space 0 comma
show that
a + b + c = abc(ab + bc + ca) where a , b , c are all different.

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127.

Let the three digit numbers A28,3B9 and 62C where A, B, and C are any integers between 0 and 9, be divisible by a fixed integer k. Show that the determinant
open vertical bar table row straight A cell space space 3 end cell cell space space space 6 end cell row 8 cell space 9 end cell cell space space straight C end cell row 2 cell space straight B end cell cell space space 2 end cell end table close vertical bar
is divisible by k.

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 Multiple Choice QuestionsLong Answer Type

128.

Prove that:
open vertical bar table row cell left parenthesis straight b plus straight c right parenthesis squared end cell cell straight a squared end cell cell straight a squared end cell row cell straight b squared end cell cell left parenthesis straight c plus straight a right parenthesis squared end cell cell straight b squared end cell row cell straight c squared end cell cell straight c squared end cell cell left parenthesis straight a plus straight b right parenthesis squared end cell end table close vertical bar space equals space 2 space abc space left parenthesis straight a plus straight b plus straight c right parenthesis cubed.

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129.

Prove that:
open vertical bar table row cell negative 2 straight a end cell cell space space space space straight a plus straight b end cell cell space space space straight a plus straight c end cell row cell straight b plus straight a end cell cell space space minus 2 straight b end cell cell space space space straight b plus straight c end cell row cell straight c plus straight a end cell cell space space straight c plus straight b end cell cell space space minus 2 straight c end cell end table close vertical bar space equals space 4 left parenthesis straight a plus straight b right parenthesis thin space left parenthesis straight b plus straight c right parenthesis thin space left parenthesis straight c plus straight a right parenthesis.


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 Multiple Choice QuestionsMultiple Choice Questions

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130.

Let A be a square matrix of order 3 × 3, then | kA | is equal to 


  • k | A |
  • k2 | A |
  • 3| A |
  • 3| A |


C.

3| A |
Let space space space space straight A space equals space open square brackets table row cell straight a subscript 1 end cell cell space space space space straight b subscript 1 end cell cell space space space straight c subscript 1 end cell row cell straight a subscript 2 end cell cell space space space space straight b subscript 2 end cell cell space space space space straight c subscript 2 end cell row cell straight a subscript 3 end cell cell space space space straight b subscript 3 end cell cell space space space space straight c subscript 3 end cell end table close square brackets

therefore space space space space space space space space space space kA space equals space straight k open square brackets table row cell straight a subscript 1 end cell cell space space straight b subscript 1 end cell cell space space straight c subscript 1 end cell row cell straight a subscript 2 end cell cell space straight b subscript 2 end cell cell space space straight c subscript 2 end cell row cell straight a subscript 3 end cell cell space space straight b subscript 3 end cell cell space space straight c subscript 3 end cell end table close square brackets space equals space open square brackets table row cell ka subscript 1 end cell cell space space kb subscript 1 end cell cell space space kc subscript 1 end cell row cell ka subscript 2 end cell cell space space kb subscript 2 end cell cell space space kc subscript 2 end cell row cell ka subscript 3 end cell cell space kb subscript 3 end cell cell space space kc subscript 3 end cell end table close square brackets
therefore space space space space open vertical bar kA close vertical bar space equals space open vertical bar table row cell ka subscript 1 end cell cell space space kb subscript 1 end cell cell space kc subscript 1 end cell row cell ka subscript 2 end cell cell space kb subscript 2 end cell cell space kc subscript 2 end cell row cell ka subscript 3 end cell cell space kb subscript 3 end cell cell space kc subscript 3 end cell end table close vertical bar space equals space straight k cubed open vertical bar table row cell straight a subscript 1 end cell cell space straight b subscript 1 end cell cell space straight c subscript 1 end cell row cell straight a subscript 2 end cell cell space straight b subscript 2 end cell cell space straight c subscript 2 end cell row cell straight a subscript 3 end cell cell space straight b subscript 3 end cell cell space straight c subscript 3 end cell end table close vertical bar
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 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 space space space space space space space space space space space space space space left square bracket Taking space out space straight k space common space from space each space row right square bracket 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 space space space space space space space space space equals space straight k cubed open vertical bar straight A close vertical bar
therefore space space space left parenthesis straight C right parenthesis space is space the space correct space answer.
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