# Trying to explain to first year students that $\det$Diag$(A,B)$ = $\det A \cdot \det B$

Let $A$ be an $m$ by $m$ matrix and $B$ an $r$ by $r$ matrix and let Diag$(A,B)$ be a block diagonal matrix. It is an $(m+r)$ by $(m+r)$ matrix where the top left $m$ by $m$ corner is $A$, the bottom right $r$ by $r$ corner is $B$ and $0$ else where.

I am trying to explain to first year students that $\det$Diag$(A,B)$ = $\det A \cdot \det B$. I was wondering if someone could assist me with how I can explain to them in a clear manner (other than just giving an example or two)? Thank you!

Maybe you can use:

$$\begin{bmatrix} A & 0\\ 0 & B \end{bmatrix} = \begin{bmatrix} A & 0\\ 0 & I_r \end{bmatrix} \begin{bmatrix} I_m & 0\\ 0 & B \end{bmatrix}$$

$$\det\begin{bmatrix} A & 0\\ 0 & B \end{bmatrix} = \det \begin{bmatrix} A & 0\\ 0 & I_r \end{bmatrix} \det \begin{bmatrix} I_m & 0\\ 0 & B \end{bmatrix} = \det A \det B$$.

It should be easy to demonstrate that

$$\det \begin{bmatrix} A & 0\\ 0 & I \end{bmatrix} = \det A$$

based on the normal technique involving minors.

• O this is a simple trick! Thank you! – Johnny T. Feb 28 '16 at 23:31

You could prove this directly from the definition of determinant. Or using the Laplace expansion. This should be quite straightforward as the upper and lower off-diagonal blocks will now have zero contribution to the determinant.