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I understand that adding a multiple of one row to another in a matrix has no effect on the determinant, which seems to contradict something I learned earlier: if I understand correctly, for a $n\times n$ matrix with rows $[v_1, v_2, v_3,\ldots v_n]$,

$$\det[v_1+v_1', v_2, v_3,\ldots v_n]=\det[v_1, v_2, v_3,\ldots v_n]+\det[v_1', v_2, v_3,\ldots v_n].$$

Thanks for the clarification!

MattAllegro
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Gabriel
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    Well, if $v'_1$ is, say, $v_2$, you have on the r.h.s.: $,\operatorname{det}[v_2,v_2,\dots,v_n]$, which is $0$, since a determinant is antisymmetric. – Bernard Feb 14 '15 at 21:47
  • Have you understood the problem? If so, can you mark the right answer? – Eugene Zhang Feb 17 '15 at 08:20

2 Answers2

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Determinants are multilinear, and equal to zero when two components (columns) are equal. What you've written is correct. On the other hand since you're adding $v_1'$, a column of the determinant, to a different column, then upon doing your expansion, you get the original determinant, plus $det[v_1',v_2,v_3,...,v_1',...,v_n]$ which is $0$ because of the repeated column.

NashEw.
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Alex R.
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If you add a multiple of a column already in a matrix $A$ it will not affect the determinant at all. This is because, if you have a matrix where one column(row) is a multiple of another column(row), you immediately get that the column(row) vectors in your matrix aren't linearly independent. This means, by default, that the determinant of that matrix is zero.

Scounged
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