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

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

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  • $\begingroup$ Look at point $1)$ here: math.stackexchange.com/questions/2068032/… $\endgroup$ – Maffred Dec 24 '16 at 3:45
  • $\begingroup$ 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! $\endgroup$ – Anthony P Dec 24 '16 at 4:47
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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 $\det A^t=\det A$.

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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.

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