So, the problem is this. I can infer that the determinant of a matrix with two identical rows is equal to $0$ because exchanging two rows negates the determinant of the matrix (which is relatively straightforward to prove). Since there is no net-change in the values of the matrix, the only conclusion to draw is that the determinant of the matrix is indeed $0$.

From this, I was curious how to prove that multiplying a row $r$ with a scalar $a$ and adding it to row $k$ (where $r \not= k$, for the sake of argument) will not change the determinant. So, it could be said that a matrix $G$ contains the presumed change to the matrix.

$$\det{G} = \sum_{j=0}^{n}{G_{kj}C_{kj}(G)}$$ $$\det{G} = \sum_{j=0}^{n}{(F_{kj} +aF_{rj})C_{kj}(F)}$$ $$\det{G} = \sum_{j=0}^{n}{(F_{kj} +aF_{rj})C_{kj}(F)}$$ $$\det{G} = \sum_{j=0}^{n}{F_{kj}C_{kj}(F)} +a\sum_{j=0}^{n}{F_{rj}C_{kj}(F)}$$ $$\det{G} = \sum_{j=0}^{n}{F_{kj}C_{kj}(F)} +a\sum_{j=0}^{n}{F_{rj}C_{kj}(F)}$$

Clearly, the lhs of the addition accounts for $\det{F}$, so it's almost there. Somehow, I must prove to myself that the rhs is $0$ to validate this notion in my head.

$$\det{G} = \det{F} +a\sum_{j=0}^{n}{F_{rj}C_{kj}(F)}$$

All the books I've sampled from thus far didn't really feel like validating the rhs as $0$, they just stated that the rhs is simply the determinant of a matrix $F$ where the contents of row $k$ have been replaced with the values of $r$. I can almost visualize that, but can someone provide a little bit more why everyone seems to take it "for granted". Perhaps I don't see it, but the rhs doesn't seem like anything has been exchanged.

How I see it: The fact that $r$ now stands where $k$ stood seems to imply that $r$ is actually taking $k$'s place (or as if it is taking its place) and the matrix now seems to have two rows $r$, therefore the determinant is evaluating to $0$ and yielding:

$$\det{G} = \det{F}$$

Am I close, too far off? If someone could explain a bit further, I'd be most grateful.

  • $\begingroup$ You're right! =p This is exactly the principle! Many books define determinant as an "alternating multilinear operator in the rows" (that is, the two properties you've shown, and that the determinant of the identity is 1), so that property holds for definition. But, in the end, you can show (easy, but much working) that any definition of determinant coincides with that you used (Laplace expansion, right?). And you just shown that this is in fact an alternating multilinear function. =p $\endgroup$
    – Yuki
    Jun 4, 2012 at 23:13

1 Answer 1


I think it's easier to use the multiplicative property of the determinant and turn row operations into matrix multiplication. To over-simplify: What's the determinant of the matrix

$E=\begin{bmatrix} 1 & 0 \\ a & 1 \\ \end{bmatrix}$

$E \cdot A$ for 2x2 $A$ is the same as performing the row operation $ar_1+r_2\to r_2$

Turns out all elementary row ops are nearly diagonal like that.


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