Determinant of a symmetric $7\times 7$-matrix

Question

Is there any simple method to calculate the determinant of the following symmetric $7\times 7$-matrix $$M:=\begin{pmatrix} 1 & 0 & 0 & 0 & a_{2} & a_{3} & a_{4}\\ 0 & 1 & 0 & 0 & -a_{1} & -a_{4} & a_{3}\\ 0 & 0 & 1 & 0 & a_{4} & -a_{1} & -a_{2}\\ 0 & 0 & 0 & 1 & -a_{3} & a_{2} & -a_{1}\\ a_{2} & -a_{1} & a_{4} & -a_{3} & a_0 & 0 & 0\\ a_{3} & -a_{4} & -a_{1} & a_{2} & 0 & a_0 & 0\\ a_{4} & a_{3} & -a_{2} & -a_{1} & 0 & 0 & a_0\\ \end{pmatrix}$$ where $a_i$ are real numbers.

Answer 1

Here is a rather simple method.

Consider the block partition of matrix $M$:

$$M=\pmatrix{I_4&B^T\\B&a_0I_3} \ \ \text{of the form} \ \ \pmatrix{R&S\\T&U}.$$

Using a classical determinantal formula (https://en.wikipedia.org/wiki/Schur_complement) valid for any partition where the diagonal blocks are square and the upper left block is invertible:

$$det(A)=det(R)det(U-TR^{-1}S)=det(I_4)det(a_0I_3-BI_4^{-1}B^T)$$

where $U-TR^{-1}S$ is the classical Schur's complement. Thus:

$$\tag{1}det(A)=det(a_0I_3-BB^T)$$

But $BB^T=(a_1^2 + a_2^2 + a_3^2 + a_4^2)I_3$. Then:

$$\tag{2}det(A)=det((a_0-a_1^2 - a_2^2 - a_3^2 - a_4^2)I_3)=(a_0-a_1^2 - a_2^2 - a_3^2 - a_4^2)^3$$

Answer 2

The determinant of $M$ is given by $$ \det(M)=(a_0 - a_1^2 - a_2^2 + a_3^2 - a_4^2)(a_0 - a_1^2 - a_2^2 - a_3^2 - a_4^2)^2. $$ Note that the matrix is not symmetric.

For the symmetric variant we have $$ \det(M)=(a_0 - a_1^2 - a_2^2 - a_3^2 - a_4^2)^3, $$ where we could use rules of the determinant for block matrices (the upper left corner being the identity).