Why Reusing Rows & Cols To Find Determinants Is Not Allowed

[Brutus]

Why am I not allowed to reuse rows to find the determinant via elementary operations?

Hi,

I am learning about matrices and determinants and there is something I can't figure out, straight to the point with an example:

Evaluating the determinant...
<br /> \begin{bmatrix}<br /> 1&amp;2&amp;3&amp;4 \\<br /> 5&amp;6&amp;7&amp;8 \\<br /> 2&amp;6&amp;4&amp;8 \\<br /> 3&amp;1&amp;1&amp;2<br /> \end{bmatrix}=2<br /> \begin{bmatrix}<br /> 1&amp;2&amp;3&amp;4 \\<br /> 5&amp;6&amp;7&amp;8 \\<br /> 1&amp;3&amp;2&amp;4 \\<br /> 3&amp;1&amp;1&amp;2<br /> \end{bmatrix}=2<br /> \begin{bmatrix}<br /> 1&amp;2&amp;3&amp;4 \\<br /> 0&amp;-4&amp;-8&amp;-12 \\<br /> 0&amp;1&amp;-1&amp;0 \\<br /> 0&amp;-5&amp;-8&amp;-10<br /> \end{bmatrix}=8<br /> \begin{bmatrix}<br /> 1&amp;2&amp;3&amp;4 \\<br /> 0&amp;-1&amp;-2&amp;-3 \\<br /> 0&amp;1&amp;-1&amp;0 \\<br /> 0&amp;-5&amp;-8&amp;-10<br /> \end{bmatrix}=8<br /> \begin{bmatrix}<br /> 1&amp;2&amp;3&amp;4 \\<br /> 0&amp;-1&amp;-2&amp;-3 \\<br /> 0&amp;0&amp;-\frac{1}{2}&amp;-2 \\<br /> 0&amp;0&amp;-\frac{1}{2}&amp;0<br /> \end{bmatrix}=8<br /> \begin{bmatrix}<br /> 1&amp;2&amp;3&amp;4 \\<br /> 0&amp;-1&amp;-2&amp;-3 \\<br /> 0&amp;0&amp;-\frac{1}{2}&amp;-2 \\<br /> 0&amp;0&amp;0&amp;2<br /> \end{bmatrix}=8<br />

Obviously, it's wrong, the right answer is 72(see [1]), I didn't work on the main diagonal on purpose, I wanted to see if I was allowed to reuse rows(or cols), so why am I not allowed? what am I really doing to the matrix each time I reuse a row(or col) ?

I am not asking for a proof or anything, just a simple explanation for human beings ;).

Also, working with both rows and columns is not allowed either, I guess this is a particular case of the above since I am reusing a cell.

Thank you.

[1] http://www.sosmath.com/matrix/determ1/determ1.html

 

[Dick]

You can reuse rows. You can also mix row and column operations. But how did you get from the fourth matrix to the fifth? I'm baffled.

 

[Brutus]

Nevermind, I made a stupid mistake, in step 4 I did: row 3 - row 1 * 1/2 and forgot to write the -1/2 in A_{3 1} ...
I also forgot to include the 5/2 in A_{4 1} when doing row 4 - row 1 * (-5/2)
damn...

I guess my mind was expecting a failure and it tricked itself into it...

Thank you.

 

[HallsofIvy]

You can, as Dick said, use a row more than once. But it is more orderly and less error-prone if you work from upper left to lower right working with each row and column in turn. I would have done this as follows:
\begin{bmatrix} 1&amp;2&amp;3&amp;4 \\5&amp;6&amp;7&amp;8 \\2&amp;6&amp;4&amp;8 \\3&amp;1&amp;1&amp;2\end{bmatrix}= \begin{bmatrix}1&amp;2&amp;3&amp;4\\0&amp;-4&amp;-8&amp;-12\\0&amp;2&amp;-2&amp;0\\0&amp;-5&amp;-8&amp;-10\end{bmatrix}
"clearing" the first column. Now, seeing the "-4" in the pivot for the second column/row,
=-4\begin{bmatrix}1&amp;2&amp;3&amp;4\\0&amp;1&amp;2&amp;3\\0&amp;0&amp;-6&amp;-6\\0&amp;0&amp;2&amp;5\end{bmatrix}
=(-4)(-6)\begin{bmatrix}1&amp;2&amp;3&amp;4\\0&amp;1&amp;2&amp;3\\0&amp;0&amp;1&amp;1\\0&amp;0&amp;0&amp;3\end{bmatrix}
=(-4)(-6)(3)\begin{bmatrix}1&amp;2&amp;3&amp;4\\0&amp;1&amp;2&amp;3\\0&amp;0&amp;1&amp;1\\0&amp;0&amp;0&amp;1\end{bmatrix}
and now, since we have "1"s along the main diagonal, the determinant is (-4)(-6)(3)= 72.

 

[Brutus]

Great, thanks.