# Linear Algebra: row reducing in order to find the determinant

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!

• 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.. – Peter Franek Feb 27 '15 at 20:32
• I don't have one, actually! I just remembered this property while tutoring somebody and forgot the why it works. Do you know? – Bliebervik Feb 27 '15 at 20:34
• 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/… – megas Feb 27 '15 at 20:37

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 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 $\{1,\ldots,n\}$.
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 $c$.