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

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:
$$\begin{bmatrix} 1&2&3&4 \\5&6&7&8 \\2&6&4&8 \\3&1&1&2\end{bmatrix}= \begin{bmatrix}1&2&3&4\\0&-4&-8&-12\\0&2&-2&0\\0&-5&-8&-10\end{bmatrix}$$
"clearing" the first column. Now, seeing the "-4" in the pivot for the second column/row,
$$=-4\begin{bmatrix}1&2&3&4\\0&1&2&3\\0&0&-6&-6\\0&0&2&5\end{bmatrix}$$
$$=(-4)(-6)\begin{bmatrix}1&2&3&4\\0&1&2&3\\0&0&1&1\\0&0&0&3\end{bmatrix}$$
$$=(-4)(-6)(3)\begin{bmatrix}1&2&3&4\\0&1&2&3\\0&0&1&1\\0&0&0&1\end{bmatrix}$$
and now, since we have "1"s along the main diagonal, the determinant is (-4)(-6)(3)= 72.

 

[Brutus]

Great, thanks.