Finding Inverse of a 4*4 matrix Find the inverse of the following matrix
$$
\begin{array}{c}
{\left[\begin{array}{cccc}
c_{0} & c_{1} & c_{2} & c_{3} \\
c_{2} & c_{3} & c_{0} & c_{1} \\
c_{3} & -c_{2} & c_{1} & -c_{0} \\
c_{1} & -c_{0} & c_{3} & -c_{2}
\end{array}\right]} \\
\text { where } c_{0}=\frac{1+\sqrt{3}}{4 \sqrt{2}}, c_{1}=\frac{3+\sqrt{3}}{4 \sqrt{2}}, c_{2}=\frac{3-\sqrt{3}}{4 \sqrt{2}} \text { and } c_{3}=\frac{1-\sqrt{3}}{4 \sqrt{2}}
\end{array}
$$
I had thought of replacing the values with trigonometric values like Sin 15 , cos 15 .... But it seems very lengthy. Any shorter approach?
 A: Posting as an answer as it the MathJax for the matrix will not fit in a comment. Mathematica gives a rather nasty expression for the inverse of the matrix:
$$
\left(
\begin{array}{cccc}
 \frac{1}{560} \left(8 \sqrt{6}-29\right) & \frac{1}{560} \left(99-8 \sqrt{6}\right) & \frac{1}{560} \left(-8 \sqrt{6}-29\right) & \frac{1}{560} \left(8 \sqrt{6}+99\right) \\
 \frac{1}{560} \left(8 \sqrt{6}+99\right) & \frac{1}{560} \left(-8 \sqrt{6}-29\right) & \frac{1}{560} \left(8 \sqrt{6}-99\right) & \frac{1}{560} \left(29-8 \sqrt{6}\right) \\
 \frac{1}{560} \left(99-8 \sqrt{6}\right) & \frac{1}{560} \left(8 \sqrt{6}-29\right) & \frac{1}{560} \left(8 \sqrt{6}+99\right) & \frac{1}{560} \left(-8 \sqrt{6}-29\right) \\
 \frac{1}{560} \left(-8 \sqrt{6}-29\right) & \frac{1}{560} \left(8 \sqrt{6}+99\right) & \frac{1}{560} \left(29-8 \sqrt{6}\right) & \frac{1}{560} \left(8 \sqrt{6}-99\right) \\
\end{array}
\right)
$$
Showing this result analytically seems a daunting task.
A: The columns are pairwise orthogonal to each other.
So, with
$$A=
{\left[\begin{array}{cccc}
c_{0} & c_{1} & c_{2} & c_{3} \\
c_{2} & c_{3} & c_{0} & c_{1} \\
c_{3} & -c_{2} & c_{1} & -c_{0} \\
c_{1} & -c_{0} & c_{3} & -c_{2}
\end{array}\right]} \\
$$
You have
$$A^TA = qI_{4\times 4} \text{ with } q= \sum_{k=0}^3c_k^2 =1$$
Hence,
$$A^{-1} = A^T$$
So, the columns of matrix $A$ form an orthonormal basis.
