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How can I prove that the determinant of the matrix is $0$ only when $a=b=c=d=0$? I tried with the Laplace Method of Expansion but I cannot solve the final equation.

\begin{bmatrix} a & b & c & \ d \\ b & -a & d & \ -c \\ c & -d & -a & \ b \\ d & c & -b & \ -a \\ \end{bmatrix}

Thanks in advance!

Gibbs
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jameslamatte
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    What is your final equation? – Gibbs May 08 '18 at 15:14
  • I came to this equation: a^4 -(ad)^2 -bda(-c)-bd(-a)c-b^2c^2+c^4 = 0 – jameslamatte May 08 '18 at 15:34
  • Alternatively, the determinant is just the signed volume of the parallepiped spanned by the columns. In this case, the four columns (regarded as vectors in $\mathbb R^4$) have the same length and are pairwise orthogonal, so that the parallelepiped they span is a hypercube with side length $s := \sqrt{a^2 + b^2 + c^2 + d^2}$ and hence (hyper)volume $s^4 = (a^2 + b^2 + c^2 + d^2)^2$, which is zero iff $a = b = c = d = 0$. – Travis Willse May 08 '18 at 15:46

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