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A set of rules in my textbook is as follows:

a. If $A$ has a zero row (column), then $\det A = 0$

b. If $B$ is obtained by interchanging two rows (columns) of $A$, then $\det B = -\det A$

c. If $A$ has two identical rows (columns), then $\det A = 0$

d. If $B$ is obtained by multiplying a row (column) of $A$ by $k$, then $\det B = k\cdot\det A$

e. If $A$, $B$, and $C$ are identical except that the $i$-th row (column) of $C$ is the sum of the $i$-th rows (columns) of $A$ and $B$, then $\det C = \det B + \det A$

f. If $B$ is obtained by adding a multiple of one row (column) of $A$ to another row (column), then $\det B = \det A$

I don't understand why "column" is in parentheses after every instance of row. Is that the same as saying, for example with clause a: "if $A$ has a zero row or a zero column, then $\det A = 0$"? As in, it's saying that column and row can be used interchangably in the statement, as the statement holds true either way?

If the above interpretation is correct, then how? Why would these statements that apply to rows also apply to columns. The only situation I could see it applying is if the matrix is symmetrical, but the question doesn't specify that, it only says that the matrix is square.

Any help is appreciated.

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  • $\begingroup$ Yes, the same rules apply to columns too. This startled me as well when I was taught it. See J. G.'s answer. And welcome to the site. $\endgroup$
    – timtfj
    Commented Jan 1, 2019 at 22:40
  • $\begingroup$ $\det(M^T)=\det(M)$. $\endgroup$
    – amd
    Commented Jan 2, 2019 at 4:41

2 Answers 2

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The trick is that $\det A^T=\det A$. Therefore, if there's a zero column in $A$ there's a zero row in $A^T$, implying $\det A^T=0$. You can check all the other copy-paste-to-column results in that way.

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    $\begingroup$ And, if you define determinants by the formula using signs of permutations, $\det A^T=\det A$ just amounts to the fact that the sign of a permutation is the same as the sign of its inverse. $\endgroup$
    – Eric Wofsey
    Commented Jan 1, 2019 at 20:59
  • $\begingroup$ @EricWolsey Which, since signs multiply on composition of permutations, amounts to the primary-school fact that $(\pm 1)^2=1$. (Or, if we work with parities, $1+1=2$ does it.) $\endgroup$
    – J.G.
    Commented Jan 1, 2019 at 21:08
  • $\begingroup$ @EricWofsey: I definitely prefer the definition by permutations, because it immediately gives all the properties of determinants. I don't know why so many people even bother with the cofactor definition, which is really quite useless in my opinion. $\endgroup$
    – user21820
    Commented Jan 2, 2019 at 5:34
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That's because your textbook isn't really explaining determinants at all to give you any intuition. It's just listing a bunch of their properties.

The key thing to remember is that $\det A$ is the volumetric scaling factor of a matrix $A$ (i.e., $A$ scales the unit cube's volume by $\det A$).
If you've seen eigenvalues, this means that $\det A$ is the product of $A$'s eigenvalues. (This can be a little mindblowing the first time you realize it, if you've already learned linear algebra for some time.)

If you start from this, then it's far easier to see why rows and columns don't make a difference:

  • The unit cube is represented by its orthogonal edges via $I$, the identity matrix.

  • Right-multiplying by $I$ (which is an operation linearly combining $A$'s columns) gives the same result ($A$) as as left-multiplying by $I$ (which is an operation linearly combining $A$'s rows)

The fact that we have the same output solid regardless of whether we operate on rows or columns of $A$ means that the scaling factor is the same, and hence the determinant is the same.

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