I have to calculate the determinant of the following matrix:

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

Using following hint:

Calculate determinant of matrix $AA^{T}$ and use the theorem that $\det(AB) = \det A \cdot \det B$

I simply don't see how can I use this hint in this calculation.

  • 3
    $\begingroup$ Have you got the product $AA^T$? If yes, please include it in the text of the quesiton. Also, do you know how $\det A^T$ and $\det A$ are related? $\endgroup$ – Berci Jun 1 '13 at 11:35

We have $$AA^T=(a^2+b^2+c^2+d^2)I_4$$ then $$\det(A)^2=\det (AA^T)=(a^2+b^2+c^2+d^2)^4$$ hence $$\det(A)=-(a^2+b^2+c^2+d^2)^2$$ and we choose the negative sign since the $\det$ function is continuous with its components and $\det(A)=-1$ in the case $a=1$ and $b=c=d=0$

  • 1
    $\begingroup$ Thanks for the tip about sign of the result : ) $\endgroup$ – Świstak35 Jun 1 '13 at 11:51
  • $\begingroup$ @Świstak35 you're welcome. $\endgroup$ – user63181 Jun 1 '13 at 11:52

Here is a little hint:

$$\det(A^T) = \det(A).$$

This makes the questions a lot shorter. Can you think how?

  • 1
    $\begingroup$ Answer here: $AA^{T}$ just looks like big, boring to calculate matrix. In fact, almost all fields in matrix are reducing to 0's, and in result we have a matrix which has nonzero values only on it's main diagonal. Thus, our determinent is product of values on main diagonal. Then, using that $det A^{T} = det A$ the result is easy to calculate. Thanks! $\endgroup$ – Świstak35 Jun 1 '13 at 11:49

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