Block matrices determinant - What am I missing?

Question

A property of determinants

If $M$ is a block matrix with square matrices $A,D$ then $\det\begin{pmatrix} A & B \\ 0 & D \end{pmatrix}=\det(A)\det(D)$

An example of using that property

$\det\begin{pmatrix} 2&2&6&3 \\ 5&4&8&5 \\ 8&2&0&6 \\ 0&5&1&7 \end{pmatrix}=-414$

$\det\begin{pmatrix} 2&2 \\ 5&4 \end{pmatrix}=-2$

$\det\begin{pmatrix} 0&6 \\ 1&7 \end{pmatrix}=-6$

We were told this property exists and It should work, but somehow in my example it does not work as $-6 \cdot -2$ does not equal $-414$. What am I missing?

Edit: added math formating

Answer 1

For block matrices $\begin{bmatrix}A & B \\ C & D \end{bmatrix}$, if $A$ is non-singular, then we have

$$\det\begin{bmatrix}A & B \\ C & D \end{bmatrix}=\det(A)\det(D-CA^{-1}B)$$

This is due to

$$\begin{bmatrix}A & B \\ C & D \end{bmatrix}= \begin{bmatrix}A & 0 \\ C & I \end{bmatrix}\begin{bmatrix}I & A^{-1}B \\ 0 & D-CA^{-1}B \end{bmatrix}$$

Similarly, if $D$ is non-singular, we have

$$\det\begin{bmatrix}A & B \\ C & D \end{bmatrix}=\det(D)\det(A-BD^{-1}C)$$

If $B=0$ or $C=0$, we do have $\det\begin{bmatrix}A & B \\ C & D \end{bmatrix}=\det(A)\det(D)$. This condition doesn't hold in your example so you can't apply it.

You can read about it at the wikipedia page.

Answer 2

What you are missing is that $C$ should be $\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right)$ in order that your formula works.