I'm a bit confused with something I read and I hope you can help me. I'm studying determinants and right now how matrix row operations change the determinants.

I read (and in fact quote):

The effect of multiplying a row of $A$ by a scalar $k$ is to multiply $|A|$ by $k$.

I saw the proof and I think I understand it (which if you ask me, just makes things worst).

However, lets see this example: $A= \begin{bmatrix}3& 3\\ 2& 1\end{bmatrix}$, $|A|$ is clearly $3-2(3) = -3$

Now, if I divide the first row of $A$ by $3$ I get $\begin{bmatrix}1& 1\\ 2& 1\end{bmatrix}$ and the determinant should be $3\cdot (1\cdot 1 - 2\cdot 1) = -3$ This starts to be fishy... I divided the first row by 3, this is the same as multiplying by $\frac{1}{3}$.By the definition above the determinant should be $\frac{1}{3}(1\cdot 1 - 2\cdot 1) = -\frac{1}{3}$

Do you see my point?

Now to make things even more confusing: if $A = \begin{bmatrix}1/3& 1/3 \\ 2& 1\end{bmatrix}, |A| = -1/3$. If I multiply the first row by 3 I get: $\begin{bmatrix}1& 1 \\ 2& 1\end{bmatrix}$. Now I am multiplying by 3 so the determinant should be $3\cdot \left|\begin{matrix}1& 1 \\ 2& 1\end{matrix}\right|$ but it isn't! So it is in fact $\frac{1}{3}|[1,1;2,1]|$

I guess my definition is wrong (or incomplete) not sure... what am I missing???

Thanks in advance!

  • 2
    $\begingroup$ If you multiply a row of A by a scalar k, the resulting determinant will be $k\cdot |A|$. If you divide by a constant $k$, (equivalent to multiplying a row by $\frac 1k)$, then the determinant of the resulting matrix is $\frac 1k\cdot |A|$. $\endgroup$ – amWhy Mar 12 '15 at 21:42
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    $\begingroup$ You made mistakes, your first determinant is -3, so by multiplying by 1/3 you get -1. This is also what you get by multiplying the first column by 1/3 : $\begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}$ has a determinant of -1. $\endgroup$ – krirkrirk Mar 12 '15 at 21:44

When you see $|A|$, you have to remember it's talking about the original determinate. Take your original example, where A is $\begin{pmatrix} 3 & 3 \\ 2 & 1 \end{pmatrix}$. You calculate the determinate correctly. Now perform a row operation, multiplying the first row by $\frac{1}{3}$. You get a new matrix $B= \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}$. What your theorem tells you is that since B was gotten from a row operation on A, $|B|=\frac{1}{3}|A|$. If you calculate things out directly, you get $\frac{1}{3}|A|=\frac{-3}{3}=-1$, which is the correct value.

It's the same situation for your second example. Your original matrix A has a row multiplied by 3 to give a matrix B. If we want to find the determinate of B, we need to compute $3\cdot|A|$. You found $|B|=-1$ and $|A|=\frac{-1}{3}$, and these values satisfy the equation.

You have to think of performing a row operation as creating a new matrix. To find the determinate of this new matrix, you can use the determinate of the original matrix prior to performing the row operation. It is this old determinate that you multiply by a constant.


See the answer here by @Ellya (I believe it's the same in your textbook) to understand the cause of the effect Explanation:

Let $A\in M_n(\mathbb F)$.

The proof of the proposition comes from the determinant definition itself:

In the formula: $$\operatorname{det}(A)=\sum_{\sigma\in S_n}\operatorname{sign}(\sigma)\prod_{i=1}^na_{i,\sigma_i}$$

A general sumand $$\operatorname{sign}(\sigma)\prod_{i=1}^na_{i,\sigma_i}$$ is a product of $n$ matrix entries neither 2 of which are in the same row nor column, i.e., a general sumand consists of one and only one entry from the same row (column). That is where the term algebraic complement or cofactor comes from.

Therefore, if we multiply a row(column) of a matrix by, let's say, $\lambda^m\in\mathbb F,m\in\mathbb N$,

$m$ factors (matrix entries) are multiplied by $\lambda$.


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