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Linear property of determinates

  1. Dec 16, 2007 #1
    So I am looking at the proof for this in a linear algebra book and I half way get it:


    If the all elements of the jth column of a determinate D are linear combinations of two columns of numbers, i.e., if
    [tex] D=\lambda b_{i}+uc{i}[/tex] where lambda and mu are fixed numbers, then D is equalto a linear comination of the two determinates:
    [tex] D=D_{1}\lambda+D{2}u [/tex]

    Here both determinates D1 and D2 have the same columns as the determinate D except for the jth column; the jth colum of D1 consists of the numbers [tex] b_{i}[/tex] wile the jth column of D2 consists of the numbers [tex] c_{i}[/tex]
  2. jcsd
  3. Dec 16, 2007 #2
    Actually I figured it out, sorry for the dumb question.
    I have never dealt with any proofs before but I get it.
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