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When looking for the determinant of some square matrix, one can row reduce until the matrix is in upper triangular form.

My question is: we are allowed to add or subtract multiples of one row to another row, but when we want to multiply/divide a singular row by a constant, we need to multiply the entire determinant by this constant. Why is this?

Thanks!

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  • $\begingroup$ This goes back to the definition of the determinant and its basic properties; I'm sure it must be written in your textbook, whatever book on linear algebra it is.. $\endgroup$ – Peter Franek Feb 27 '15 at 20:32
  • $\begingroup$ I don't have one, actually! I just remembered this property while tutoring somebody and forgot the why it works. Do you know? $\endgroup$ – Bliebervik Feb 27 '15 at 20:34
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    $\begingroup$ The formula that expresses the determinant of a matrix in terms of its minors might be helpful to "easily" justify this. en.wikipedia.org/wiki/… $\endgroup$ – megas Feb 27 '15 at 20:37
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The determinant is linear in each of its $n$ arguments, e.g.: $$ \mbox{det} (c a_1, \ldots, a_n) = c \,\mbox{det} (a_1, \ldots, a_n) $$ This is called multilinear. It is a form, because it is a map fron $V^n$ to its field $K$. Then it switches sign, if you switch two arguments, this is a consequence of it being alternating, i.e. vanishing if two arguments are the same.

This gives: alternating multilinear form

This can be verified with its definition: $$ \mbox{det} A = \sum_{\pi\in S_n} \mbox{sgn}(\pi) \,a_{1\,\pi(1)}\cdots a_{n\,\pi(n)} $$ where $S_n$ is the symmetric group consisting of the permutations of the set $\{1,\ldots,n\}$.

Or you think of the determinant as signed volume of the paralel epiped spawned by the argument vectors. Scaling a single vector will scale the volume.

And $$ \mbox{det}(a_1 + c a_j, \cdots, a_n) = \mbox{det}(a_1, \cdots, a_n) + c \, \mbox{det}(a_j, \cdots, a_n) $$ where the second term vanishes because $a_j$ shows up twice, which is the alternating property. This cancels the multiplier $c$.

In the volume picture this would result in a deformation which is not volume changing, which is less obvious for higher dimensions to me.

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