Theorem: For $n \times n$ matrices $A$ and $B$, $\det(AB) = \det(A)\det(B)$.

Proof: We prove this using properties of elementary matrices and the relationship between row operations and determinants.

Step 1: Prove the result when $A$ is an elementary matrix.

There are three types of elementary matrices:

Type I (row swap): If $E$ swaps two rows, then $\det(E) = -1$, and multiplying by $E$ on the left swaps two rows of $B$, so: $\det(EB) = -\det(B) = \det(E)\det(B)$

Type II (row scaling): If $E$ multiplies a row by scalar $c$, then $\det(E) = c$, and multiplying by $E$ scales a row of $B$ by $c$, so: $\det(EB) = c\det(B) = \det(E)\det(B)$

Type III (row addition): If $E$ adds a multiple of one row to another, then $\det(E) = 1$, and this operation doesn't change the determinant of $B$: $\det(EB) = \det(B) = 1 \cdot \det(B) = \det(E)\det(B)$

Thus the theorem holds when $A$ is any elementary matrix.

Step 2: Extend to invertible matrices using factorization.

If $A$ is invertible, it can be written as a product of elementary matrices: $A = E_1E_2 \cdots E_k$

Then by repeated application of Step 1: $\det(AB) = \det(E_1E_2 \cdots E_kB) = \det(E_1)\det(E_2 \cdots E_kB)$ $= \det(E_1)\det(E_2)\det(E_3 \cdots E_kB)$ $= \det(E_1)\det(E_2) \cdots \det(E_k)\det(B)$ $= \det(E_1E_2 \cdots E_k)\det(B) = \det(A)\det(B)$

Step 3: Extend to all matrices.

If $A$ is not invertible, then $\det(A) = 0$. We claim that $AB$ is also not invertible (so $\det(AB) = 0$).

Suppose, for contradiction, that $AB$ were invertible. Then there exists $(AB)^{-1}$ such that: $(AB)(AB)^{-1} = I$

Multiplying on the left by $B^{-1}A^{-1}$ (if these exist) or rearranging doesn't work directly. Instead, we use rank arguments:

If $A$ is not invertible, then $\ker(A) \neq \{\mathbf{0}\}$. There exists $\mathbf{v} \neq \mathbf{0}$ with $A\mathbf{v} = \mathbf{0}$.

Then $(AB)\mathbf{v} = A(B\mathbf{v}) = \mathbf{0}$ as well, so $AB$ has nontrivial kernel and cannot be invertible.

Therefore $\det(AB) = 0 = \det(A)\det(B)$.

More rigorously: $\text{rank}(AB) \leq \text{rank}(A) < n$, so the columns of $AB$ are linearly dependent, giving $\det(AB) = 0$.

Alternative approach for Step 3: View determinant as a function. Define $f(A) = \det(AB)$ for fixed $B$. This function:

• Is multilinear in the rows of $A$
• Is alternating (swapping rows changes sign)
• Satisfies $f(I) = \det(B)$

By uniqueness of the determinant function with these properties, $f(A) = \det(A) \cdot \det(B)$.

Conclusion: In all cases, $\det(AB) = \det(A)\det(B)$. ∎