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In the below determinant, without expanding the determinant we have to prove that the determinant is zero.

enter image description here

I know how to solve it by expanding, but without that I have no idea.

a25bedc5-3d09-41b8-82fb-ea6c353d75ae
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  • Look at point $1)$ here: http://math.stackexchange.com/questions/2068032/determinant-of-skew-symmetric-matrices – Maffred Dec 24 '16 at 03:45
  • Any time there exists symmetry in the matrix, you can guarantee that the determinant is 0. To see this, perform a cofactor expansion and see for yourself! – Anthony P Dec 24 '16 at 04:47

2 Answers2

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Hint. The determinant of a skew-symmetric matrix $A$ of odd order is always $0$. Check this by $A=-A^t$ and $\det A^t=\det A$.

Cave Johnson
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1

The matrix has a null space containing $\begin{pmatrix}a\\ c\\b\end{pmatrix}$ at least. As long as one of $a,b,c$ are nonzero then the nullspace is nontrivial, but if they're all $0$ then the matrix is the $0$ matrix to begin with. Either way, the determinant is $0$ because it has a nontrivial nullspace.

Bob Jones
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