Finding the determinant of a block diagonal matrix

Question

We have the following square block matrix $X= \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}$, where $A$ and $B$ are square. I want to get the determinant of this matrix (I know it equals $\det A \times \det B$).

I've looked through the question catalog and found a lot of questions similar to this one, but the answers they received use techniques I am not familiar with - Leibniz' formula, techniques beyond the scope of an introductory linear algebra course, etc.

So I want to know how to do this with Laplace expansion. I know how Laplace expansion works for regular matrices but I don't know how to use it here. Let's say we want to use Laplace expansion along the first row of $A$, then we get that $ \det A = \displaystyle \sum_{j=1}^n (-1)^{j+1}a_{1j}\det A_{1j} $, but how do we use this to obtain the result for the entire block matrix?

Can we say that because the elements of $A$ are always in different rows and columns as compared to the elements of $B$, the entire matrix $B$ is always present in the minors of $A$? If so, the result seems a bit more intuitive to me, but I still don't know how to computationally obtain it.

Answer 1

You should already know that $\det(MN) = \det(M)\det(N)$.

With that in mind, we can write $$ \pmatrix{A&0\\0&B} = \pmatrix{A&0\\0&I} \pmatrix{I&0\\0&B} $$ where $I$ denotes an identity matrix. So now, it's sufficient to show that $$ \det\pmatrix{A&0\\0&I} = \det \pmatrix{I&0\\0&A} = \det(A) $$ you'll find that it's fairly easy to show that this one is true using Laplace expansion along the top "$I$-row".

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Computation: let's take that last matrix. Suppose that $$ X = \pmatrix{I_{k \times k} \\&A} = \\ \pmatrix{1\\&1\\&&\ddots\\&&&1\\ &&&&a_{11} & \cdots & a_{1n}\\ &&&& \vdots & \ddots & \vdots\\ &&&& a_{n1} & \cdots & a_{nn}} $$ Taking the Laplace expansion along the first row gives us $$ \det(X) = 1 \det\pmatrix{I_{(k-1)\times (k-1)} \\&A} $$ and this will work whenever $k \geq 1$. When $k = 1$, we have $$ \det(X) = \pmatrix{1&0 & \cdots & 0\\ 0&a_{11} & \cdots & a_{1n}\\ 0& \vdots & \ddots & \vdots\\ 0& a_{n1} & \cdots & a_{nn}} = 1 \det(A) $$ So, by induction, we will have $\det(X) = \det(A)$ for any integer $k \geq 1$.