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I found this exercise in the linear algebra script of my university.

For $A, B, C, D \in \mathbb{F}^{n \times n}$, show that

$$\det{\begin{pmatrix} A & B \\ C & D\end{pmatrix}} = \det{A}\det{D} - \det{B}\det{C}$$

While I know this would be true for $C = 0$, in that case we can bring $A$ and $D$ in row echelon form. But how about the case where $C \neq 0$?

As an example I tried the matrix $\begin{pmatrix}2 & 3 & 1 & 0 \\ 3 & 2 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{pmatrix}$, which should have a determinant of $-6$ if the lemma was true. But the determinant of the matrix is $-8$.

Can I conclude that the exercise has a typo?

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    $\begingroup$ have you tried doing examples? $\endgroup$
    – Asinomás
    Sep 21, 2021 at 7:46
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    $\begingroup$ where does this exercise come from? $\endgroup$ Sep 21, 2021 at 7:50
  • $\begingroup$ It's known that the LHS is $\det(AD-BC)$, which is not $\det(AD)-\det(BC)$ in general. $\endgroup$
    – J.G.
    Sep 21, 2021 at 7:51
  • $\begingroup$ @SiongThyeGoh I found the exercise in a script of my university while preparing for the exam $\endgroup$ Sep 21, 2021 at 7:52
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    $\begingroup$ Looks like a good counterexample ! I considered the permutation matrix of any permutation that sends an element from one half to the other, but as well also sends an element of the first half to the same half, you'll get that every submatrix has a $0$ column, thus each small determinant is $0$, yet the permutation matrix is invertible so the big determinant is not $0$, the simplest matrix would be $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$. $\endgroup$
    – Asinomás
    Sep 21, 2021 at 7:56

2 Answers 2

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Here is a route you could take.

You have rightly asserted that if we have $$\det\begin{pmatrix}A&B\\0&D\end{pmatrix}=\det(A)\det(D)$$.

Similarly one can trivially write $$\det\begin{pmatrix}A&0\\C&D\end{pmatrix}=\det(A)\det(D)$$

Now observe that $$\begin{pmatrix}I&0\\-CA^{-1}&I\end{pmatrix}\begin{pmatrix}A&B\\C&D\end{pmatrix}=\begin{pmatrix}A&B\\0&D-CA^{-1}B\end{pmatrix}$$

Thus one can write that $$\det(I)\det(I)\det\begin{pmatrix}A&B\\C&D\end{pmatrix}=\det(A)\det(D-CA^{-1}B)$$

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The correct result is $$\det\begin{bmatrix} A & B \\ C & D\end{bmatrix}= \det(AD-BC)$$

https://en.wikipedia.org/wiki/Block_matrix

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