# Properties of Determinants

We will now look at some very important properties of determinants

Theorem 1: If $A$ is an $n \times n$ matrix and $k$ is any scalar, then $\det(kA) = k^n \det(A)$. |

**Proof:**Consider the determinant of an $n \times n$ matrix $A$ multiplied through by the scalar $k$, that is $\det(kA) = \begin{vmatrix} ka_{11} & ka_{12} & \cdots & ka_{1n} \\ ka_{21} & ka_{22} & \cdots & ka_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ ka_{n1} & ka_{n2} & \cdots & ka_{nn} \end{vmatrix}$. Now recall that if we take a matrix $A$ and multiply any row or column by a scalar $k$, the new determinant of that matrix will be $k$-times the original since cofactor expansion along that row would clearly yield a determinant $k$-times greater. In this case, all $n$-rows are multiplied by $k$, so our determinant $\det(kA)$ will be $\underbrace{k \cdot k \cdot … k}_{\mathrm{n-times}} = k^n$ greater than $\det(A)$. Hence, $\det(kA) = k^n \det (A)$. $\blacksquare$

Before we look at the next property concerning matrix products, we will first establish the following lemma (mini theorem) that we will need to prove the next couple of properties

Lemma 1: If $A$ and $E$ are two $n \times n$ matrices where $E$ is an elementary matrix, then $\det(EA) = \det(E) \det(A)$. |

**Proof:**Suppose that $E$ results by multiplying a row of $I$ by some scalar $k$. We thus know that the determinant of this matrix is $\det(E) = k$. Furthermore, we note that $EA$ results from multiplying a row in $A$ by $k$, so we have $\det(EA) = k \det (A)$. Making the substitution that $\det(E) = k$, we get that $\det(EA) = \det(E) \det(A)$.

- Now suppose $E$ results by interchanging rows. It thus follows that $\det(E) = -1$. Now if $EA$ is the result from interchanging two rows, then $\det(EA) = -\det(A)$. Making the substitution that $\det(E) = -1$, we have $\det(EA) = \det(E) \det(A)$.

- Lastly, suppose that $E$ results by adding a multiple of one row to another. We know that $\det(E) = 1$. Now if $EA$ is the result from adding a multiple of one row to another, then $\det(EA) = \det(A)$. Making the substitution that $\det(E) = 1$, we have the same result in that $\det(EA) = \det(E) \det(A)$. $\blacksquare$

We will now look at two more theorems regarding determinants.

Theorem 2: If $A$ is an $n \times n$ matrix and if $\det(A) ≠ 0$, then $A$ is invertible. If $\det(A) = 0$, then $A$ is not invertible. |

Theorem 3: If $A$ and $B$ are $n \times n$ matrices, then $\det(AB) = \det(A) \det(B)$. |