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Given the A matrix as follow: $$A = \begin{pmatrix}B & D\\\ 0 & C\end{pmatrix}$$ Where B and C are square matrices. Matrix A is said to be in block (upper) triangular form with the formula for determinant as follow: $$det A = (det B)(det C)$$

Prove this formula in the case when A is a 4 x 4 matrix. Note that there are three cases for B and C to consider. Hint: The cases are about the possible sizes of B and C, not about what their entries are.

So far, what I have done is proving 1 easy case with B and C as a 2 x 2 square matrix by expanding down the first column. I'm really confuse by how to determine the other two cases.

So my question is how can I find the other two cases and how can I prove them?

Any help is appreciate. Thanks a lot.

Peter
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  • @DietrichBurde I don't think it answer my question as what I'm trying is to prove the formula of determinant of block triangular matrix, but topic you mentioned is proving the 3 different results. Thanks for your help – Peter Jun 05 '21 at 11:10
  • The first result $(1)$ is what you need. It gives that $\det(A)=\det(B)\det(C)$, and it is proved there. – Dietrich Burde Jun 05 '21 at 11:52

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Let us denote $\operatorname{size}(A)=n$ if $A\in M_{n\times n}(\mathbb{K})$. The only restriction here is that $\operatorname{size}(B)+\operatorname{size}(C)=4$ (and, for the sake of it being nontrivial, both sizes being positive). Therefore, the other two cases are $$\operatorname{size}(B)=1,\ \operatorname{size}(C)=3$$ and $$\operatorname{size}(B)=3,\ \operatorname{size}(C)=1.$$ These cases are also easily proven by expanding.

Samuel M. A. Luque
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