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

k | A |
k² | A |
k ³| A |
k ³| A |

C.

k 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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