Product rule

Recall that the row operations (a)-(c) correspond to elementary matrices, say $ E_a$, $ E_b$, and $ E_c$ (and recall how we constructed $ E_a$, $ E_b$, and $ E_c$ in elementary matrices). Since $ \det I_n = 1$, we obtain $ \det E_a = \det I = 1$, $ \det E_b = -\det I = -1$, and $ \det E_c = c\cdot\det I = c$. Then the properties of determinant over row operations can be summarized in

$\displaystyle \det E A = (\det E)(\det A)$   $\displaystyle \mbox{ for every elementary matrix $E$. }$ (6)

If $ A$ is invertible, there is a series of elementary matrices $ E_1, E_2, \ldots, E_p$ so that $ E_p \cdots E_2 E_1 A = I_n$. By applying the equation above repeatedly, we obtain $ (\det E_p)\cdots (\det E_2)(\det E_1)(\det A) = 1$, and therefore $ {\det A \neq 0}$. Similarly if $ A$ is not invertible, we obtain $ {\det A = 0}$ (why?). Together we conclude that

$\displaystyle \det A \neq 0$    if and only if $ A$ is invertible.

If $ A$ is invertible, we can express $ A = E_1^{-1} \cdots E_p^{-1}$ (why?). Since $ E_i^{-1}$'s are also elementary matrices, we obtain

$\displaystyle \det AB = \det E_1^{-1} \cdots E_p^{-1} B
= (\det E_1^{-1} \cdots E_p^{-1}) (\det B)
= (\det A) (\det B)
$

Thus, we have shown that $ \det AB = (\det A) (\det B)$. [If $ A$ is not invertible, neither is $ A B$ (why?); thus, $ {\det AB = (\det A) (\det B) = 0}$.]



Department of Mathematics
Last modified: 2005-10-21