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Is there a way to write the following expression in terms of a determinant of one big matrix?

$\begin{align} A&=(A_1A_2-A_3^2)(A_1A_4-A_5^2)-(A_1A_6-A_7^2)^2\\ &=det(\mathbf{B})det(\mathbf{C})-det(\mathbf{D})^2\\ &=det(\mathbf{E}), \end{align}$

where $A_i$ are scalars, and

$\begin{align} \mathbf{B}&=\begin{bmatrix}A_1&A_3\\A_3&A_2\end{bmatrix}\\ \mathbf{C}&=\begin{bmatrix}A_1&A_5\\A_5&A_4\end{bmatrix}\\ \mathbf{D}&=\begin{bmatrix}A_1&A_7\\A_7&A_6\end{bmatrix}\\ \mathbf{E}&=\begin{bmatrix}det(\mathbf{B})&det(\mathbf{D})\\det(\mathbf{D})&det(\mathbf{C})\end{bmatrix} \end{align}$

In other words, I'm trying to write $A=det(\mathbf{X})$, where $\mathbf{X}$ is a big matrix with entries $A_i$

Cheers

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  • $\begingroup$ What are $A_i$ and $A$? $\endgroup$ – Wolfram Feb 2 '17 at 14:59
  • $\begingroup$ $A_i$ are scalars (updated above) $\endgroup$ – Pioneer83 Feb 2 '17 at 15:03
  • $\begingroup$ The question isn't clear. Can you explain it more explicit manner? $\endgroup$ – Domates Feb 2 '17 at 15:10
  • $\begingroup$ Then use the fact that $ab-c^2=\left|\begin{matrix}a &c\\c &b\end{matrix}\right|$ four times. $\endgroup$ – Wolfram Feb 2 '17 at 15:10
  • $\begingroup$ @H.Ergül: I have updated the question :) $\endgroup$ – Pioneer83 Feb 2 '17 at 15:35
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The best I could do was

$$ \begin{pmatrix} 1 & 0 & A_1 & 0\\ 0 &1 & 0 & A_1\\ A_2 & A_6 & A_3^2 & A_7^2\\ A_6 & A_4 & A_7^2 & A_5^2 \end{pmatrix} $$ that has the required determinant. I'm starting to think there doesn't exists a matrix as asked.

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