Determinant of a $5\times 5$ matrix using properties

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?

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

• You can change rows of the determinant to make it diagonal but keep in mind the change in sign. – Riju Apr 13 '17 at 18:09
• 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. – Wintermute Apr 13 '17 at 18:10
• So $\det(A)=\pm 5!$, without any $2\times 2$ manipulation. – Dietrich Burde Apr 13 '17 at 18:10
• 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. – Victor S. Apr 13 '17 at 18:24