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!
Best Answer
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.