# Determinant of $14 \times 14$ matrix

The determinant of the following $14 \times 14$ matrix

$$\begin{bmatrix} D_1 & D_2 \\ D_3 & D_4 \end{bmatrix}$$

where the $D_i$ blocks are $7 \times 7$ diagonal matrices given by

$$D_1 = \mbox{diag} (a_1, a_2, \ldots a_7)$$

$$D_2 = \mbox{diag} (a_8, a_9, \ldots a_{14})$$

$$D_3 = \mbox{diag} (a_8, a_{14}, a_{13},\ldots a_9)$$

$$D_4 = \mbox{diag} (a_1, a_7, a_6, \ldots a_2)$$

is

$$(a_1^2-a_8^2)(a_2a_7-a_9a_{14})^2(a_3a_6-a_{10}a_{13})^2(a_4a_5-a_{11}a_{12})^2$$

Am I correct?

• How did you get there? You will get more attention if you show some work. – Arnaud Mortier Mar 6 '18 at 10:36
• I don't think this is correct. I believe that it should be $(a_1\cdot ...\cdot a_7)^2-(a_8\cdot ...\cdot a_{14})^2$ – Levent Mar 6 '18 at 10:37
• @ Arnaud Mortier By finding it for smaller values of $n$. – Mittal G Mar 6 '18 at 10:40
• @ Levent Actually you will be right if $D_1 = D_4$ and $D_2 = D_3$ which is not so. – Mittal G Mar 6 '18 at 10:40
• The definition of $D_3$ and $D_4$ is beyond my capabilities. Could you spell out all diagonal matrix elements to remove remaining doubts? – Axel Kemper Mar 6 '18 at 10:47

A classical result is the following one.

Being given a $2 \times 2$ block matrix with square blocks:

$$\mathbf{S} := \begin{pmatrix} A & B\\ C & D \end{pmatrix},$$

if $DC=CD$ [which is the case here : any diagonal matrix commutes with any other], then :

$$\det(S)=det(AD-BC)$$

Can you take it from here ?

Reference: (Determinant of block matrix with commuting blocks)