# Choosing the sign of determinant when taking a square root

Calculate the determinant $$\det(A)=\begin{vmatrix}a&b&c&d\\ \:\:\:-b&a&d&-c\\ \:\:\:-c&-d&a&b\\ \:\:\:-d&c&-b&a\end{vmatrix}$$

I found that $$\det(A)\det(A^T)=\det(A)^2=(a^2+b^2+c^2+d^2)^4$$ From this we get $$\det(A) = \pm (a^2+b^2+c^2+d^2)^2$$ Now, how to choose the sign? Any help is appreciated.

• en.wikipedia.org/wiki/Quaternion#Matrix_representations your matrix is pretty similar to the first one, you just need to negate your $b,c,d$ – Will Jagy Jul 6 at 18:06
• @WillJagy - Yes, that is the transpose of this matrix. Yet, I didn't notice if they provided the way of calculation... Thanks anyway :) – VIVID Jul 7 at 11:27

Here is one quick way: Use the standard cofactor formula for the determinant. Expand only what you need. What is the sign of $$a^4$$?
Evaluate it at $$A=I$$ gives you the sign.
• @VIVID In the case $a=1$, $b,c,d=0$ we have $A=I_4$ and hence it's determinant is just $1$ which gives you the sign. – Peter Foreman Jul 6 at 7:41
• @PeterForeman - I'm pretty sure the question was "why should every matrix $A$ have the same sign choice?", not "how do you apply the formula to $A = I$ to get the sign?". – Paul Sinclair Jul 6 at 17:10