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Determinants from any row or column

  1. Jul 6, 2011 #1
    I'm having a problem with this rule in general. Apparently one can calculate the determinant by multiplying the cofactors and entries of any row or any column of a matrix. I have a negative that pops up. I'll take a 3X3 matrix for simplicity.

    |a b c|
    |d e f|
    |g h i|

    calculating from the top row we have:
    det(A)=a(ei-fh)-b(di-fg)+c(dh-eg) = aei-afh-bdi+bfg+cdh-ceg
    calculating from the middle row we have
    det(A)=d(bi-ch)-e(ai-cg)+f(ah-bg) = -aei+afh+bdi-bfg-cdh+ceg

    The determinant seems to change by a factor of negative 1. This also seems to make sense from row operations. If for example, I swap the first and the second row to yield a matrix B, Then det(B)=-det(A). Yet if I calculate using entries and cofactors corresponding to the second row of B and the first row of A, the cofactor expansion of the two determinants will be identical. Though I'm told that we can start from any row or column, what did I miss? Thanks all.
  2. jcsd
  3. Jul 6, 2011 #2


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    Science Advisor

    That's because you are doing the second one incorrectly. The rule is that if you have [itex]a_{ij}[/itex] times its cofactor, you must multiply by [itex](-1)^{i+j}[/itex]. The simplest way to get that factor is to start at the top left and move by horizontally or vertically, alternating "+" and "-".

    So for your first calculation, "expanding on the first row", you should have "+, -, +" ([itex]a_{11}[/itex]: [itex](-1)^{1+1}= 1[/itex], [itex]a_{12}[/itex]: [itex](-1)^{1+2}= -1[/itex], [itex]a_{13}[/itex]: [itex](-1)^{1+ 3}= 1[/itex]) as you have. But "expanding on the second row", it should be "-, + , -" ([itex]a_{21}[/itex]: [itex](-1)^{2+1}= -1[/itex], [itex]a_{22}[/itex]: [itex](-1)^{2+2}= 1[/itex], [itex]a_{23}[/itex]: [itex](-1)^{2+3}= -1[/itex]) which changes the sign.
    Last edited by a moderator: Jul 6, 2011
  4. Jul 6, 2011 #3
    Thank you, perfect, so I was using the minor of each entry which doesn't include the (-1)^(i+j), which is necessary for the cofactor because Cij=(-1)(i+j)Mij. Thank you.
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