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Apparently the determinant of the following matrix can be transformed into the determinant of a $2\times 2$ matrix multiplied by a scalar. Would anyone show me how?

$$ \left( \begin{array}{ccccc} 0 & 0 & 1 & 0 & 0 \\ 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 4 \\ 0 & 5 & 0 & 0 & 0\end{array} \right)$$

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    $\begingroup$ You can change rows of the determinant to make it diagonal but keep in mind the change in sign. $\endgroup$ – Riju Apr 13 '17 at 18:09
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    $\begingroup$ Why no just permute the columns until you have a diagonal matrix? Each permutation changes the sign of the determinant and the determinant of a diagonal matrix is the product of the diagonals. $\endgroup$ – Wintermute Apr 13 '17 at 18:10
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    $\begingroup$ So $\det(A)=\pm 5!$, without any $2\times 2$ manipulation. $\endgroup$ – Dietrich Burde Apr 13 '17 at 18:10
  • $\begingroup$ Clearly I'm not up to date with my introduction to linear algebra course since I didn't know about the laplacian expansion or that I could permute and multiply the diagonal... thank you everyone. $\endgroup$ – Victor S. Apr 13 '17 at 18:24
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I agree with the rest that permutations of rows and columns is sufficient.

Another way is by doing Laplace expansion.

\begin{align} \begin{vmatrix} 0 & 0 & 1 & 0 & 0 \\ 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 4 \\ 0 & 5 & 0 & 0 & 0\end{vmatrix} &= \begin{vmatrix} 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 4 \\ 0 & 5 & 0 & 0\end{vmatrix} \\ &= 2\begin{vmatrix} 0 & 3 & 0 \\ 0 & 0 & 4 \\ 5 & 0 & 0\end{vmatrix} \\ &= 2(5)\begin{vmatrix} 3 & 0 \\ 0 & 4 \\ \end{vmatrix} \end{align}

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