An important property that the determinant satisfies is the following: $\det(AB) = \det(A)\det(B)$ where $$A$$ and $$B$$ are $$n \times n$$ matrices. A immediate and useful consequence is $\det(A^k) = \det(A)^k$ for all natural numbers $$k$$.

### Examples

1. Let $$A, B\in \mathbb{R}^{3\times 3}$$ be such that $$\det(A) = 3$$ and $$\det(B) = -2$$. Then $$\det(AB) = \det(A)\det(B) = -6$$.

2. Let $$A = \begin{bmatrix} -2 & 1 \\ -1 & 1\end{bmatrix}$$. Then $$\det(A^{100}) = \det(A)^{100} = (-1)^{100} = 1.$$

## $$2 \times 2$$ case

We now prove that if $$A$$ and $$B$$ are $$2\times 2$$ matrices then $$\det(AB) = \det(A)\det(B)$$. Proving the result for the general case is somewhat challenging and is left as an exercise.

Suppose that $$A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$$ and $$B = \begin{bmatrix} s & t \\ u & v \end{bmatrix}$$. Then $$AB = \begin{bmatrix} as + bu & at + bv \\ cs + du & ct + dv \end{bmatrix}$$. Hence, \begin{eqnarray*} \det(AB) - \det(A)\det(B)& = & \begin{vmatrix} as + bu & at + bv \\ cs + du & ct + dv \end{vmatrix} - \begin{vmatrix} a & b \\ c & d \end{vmatrix} \begin{vmatrix} s & t \\ u & v \end{vmatrix} \\ & = & (as+bu)(ct+dv) - (cs + du)(at+bv) - (ad-bc)(sv - ut) \\ & = & (acst + adsv+bctu + bduv) - (acst + bcsv + adtu + bduv) - (adsv - adtu - bcsv + bctu) \\ & = & acst + adsv + bctu + bduv - acst - bcsv - adtu - bduv - adsv + adtu + bcsv - bctu \\ & = & 0. \end{eqnarray*} Hence, $$\det(AB) = \det(A)\det(B)$$.

## Exercises

1. Compute the determinants of each of the following products:

1. $$\begin{bmatrix} 2 & 3 \\ 1 & 2\end{bmatrix} \begin{bmatrix} 3 & -1 \\ 0 & 1\end{bmatrix}$$

2. $$\begin{bmatrix} 1 & 0 & i \\ 0 & 2 & 1 \\ i & 1 &-1\end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{bmatrix}$$

2. Let $$A = \begin{bmatrix} 1 & 2-i \\ 0 & 1+i\end{bmatrix}$$. Give the value of $$\det(A^{16})$$.

3. Let $$A$$ and $$B$$ be $$3\times 3$$ matrices. Prove that $$\det(AB) = \det(A)\det(B)$$.

4. Let $$A$$ and $$B$$ be $$n\times n$$ matrices such that $$B$$ is obtained from $$A$$ by adding a constant multiple of a row of $$A$$ to another row of $$A$$. Prove that $$\det(B) = \det(A)$$. (Hint: Express $$B$$ as $$MA$$ where $$M$$ is an elementary matrix representing the row operation performed to obtain $$B$$. What must $$\det(M)$$ be?)

5. Let $$A$$ and $$B$$ be $$n\times n$$ matrices such that $$B$$ is obtained from $$A$$ by interchanging two rows of $$A$$. Prove that $$\det(B) = -\det(A)$$.