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\begin{vmatrix} -2 & 2 & -1 \\ 1/2 & 0 & -1/2\\ 3/2 & -1 & 1/2 \end{vmatrix}

My book evaluates this determinant by using row-reduction to get the top row to \begin{vmatrix} 1 & 0 & 0\end{vmatrix} and then expands by cofactors in the first row.

I'm just wondering why they chose to do it this way. Wouldn't it be easier to take advantage of the fact that there's already a 0 in the middle; that way you just have to change one other entry to zero before expanding by cofactors? And why would they pick the first row and change the -2 to a 1? It would make more sense to me if they chose to do Gaussian elimination or something, but they just randomly expanded in the first row.

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  • $\begingroup$ We want to maximize the number of zeros, so as to reduce the complexity during the expansion. $\endgroup$ – Mc Cheng Oct 19 '16 at 15:06
  • $\begingroup$ The probably didn't change $-2$ to $1$, but swapped the first and third columns, changed the sign of the first (new) column and began elimination in the first row.. $\endgroup$ – Bernard Oct 19 '16 at 15:11
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I agree. I think it's arbitrary and we all have our preferred ways. Personally, I would add the first column to the third one, to obtain $$\begin{vmatrix} -2 & 2 & -3 \\ 1/2 & 0 & 0\\ 3/2 & -1 & 2 \end{vmatrix} =-\frac12\,\begin{vmatrix}2&-3\\ -1&2\end{vmatrix} =-\frac12\,\times1=-\frac12. $$

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A way to achieve this: add twice the third row to the first one, to get

$$\begin{vmatrix} 1 & 0 & 0 \\ 1/2 & 0 & -1/2\\ 3/2 & -1 & 1/2 \end{vmatrix}$$

Now, the determinant is trivially $-1/2$.

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