Determinant of Block Matrices: How Do Non-Zero Blocks Affect the Determinant?

[Dethrone]

View attachment 4098

I don't quite follow this, can anyone explain?

 

[Jameson]

Hi Rido12. I haven't actually used this fact but it appears what is going on it that we are blocking the matrix into this form:

\(\displaystyle \left(\begin{array}{cc}A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right)\)

More particularly we are choosing $A_{12}=0$ so we end up with a lower triangular block matrix that looks like this:

\(\displaystyle \left(\begin{array}{cc}A_{11} &0 \\ A_{21} & A_{22} \end{array}\right)\)

So, $\text{det}\left(\begin{array}{cc}A_{11} &0 \\ A_{21} & A_{22} \end{array}\right)=\text{det}(A_{11}A_{22}-0A_{21})=\text{det}(A_{11}A_{22})=\text{det}(A_{11})\text{det}(A_{22})$

I think that's the basic argument. Here is a link with some useful info on it as well. :)

 

[Dethrone]

Hi Jameson! :D

Thanks for the reply - so it appears that the blocks don't have to be of the same size? By that, I mean $A_{11}$ appears to be of size $2$ by $2$ while $A_{21}$ seems to be of $3$ by $2$. If so, this seems to be a pretty useful tool! (Cool)

 

[I like Serena]

Hi Jameson! :D

Thanks for the reply - so it appears that the blocks don't have to be of the same size? By that, I mean $A_{11}$ appears to be of size $2$ by $2$ while $A_{21}$ seems to be of $3$ by $2$. If so, this seems to be a pretty useful tool! (Cool)

Hey Rido! ;)

See here.

So the block matrices can be of any size, but either right top, or left bottom has to be zero.