Determinants

If

$$\begin{equation*} A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \end{equation*}$$

is a $2 \times 2$ matrix with entries from a field $F$, then we define the determinant of $A$, denoted $\det(A)$ or $|A|$, to be the scalar

$$\begin{equation*} ad - bc. \end{equation*}$$

The function $\det : M_{2\times2}(F) \to F$ is a linear function of each row of a $2\times2$ matrix when the other row is held fixed. That is, if $u, v,$ and $w$ are in $F^2$ and $k$ is a scalar, then

$$\begin{equation*} \det \begin{pmatrix} u + kv \\ w \end{pmatrix} = \det \begin{pmatrix} u \\ w \end{pmatrix} + k \det \begin{pmatrix} v \\ w \end{pmatrix}. \end{equation*}$$

and

$$\begin{equation*} \det \begin{pmatrix} w \\ u + kv \end{pmatrix} = \det \begin{pmatrix} w \\ u \end{pmatrix} + k \det \begin{pmatrix} w \\ v \end{pmatrix}. \end{equation*}$$

For this definition, it is convenient to introduce the following notation: Given $A \in M_{n\times n}(F)$, for $n \ge 2$, denote the $(n-1)\times(n-1)$ matrix obtained from $A$ by deleting row $i$ and column $j$ by $\tilde{A}_{ij}$.

Let $A \in M_{n\times n}(F)$. If $n = 1$, so that $A = (A_{11})$, we define

$$\begin{equation*} \det(A) = A_{11}. \end{equation*}$$

For $n \ge 2$, we define $\det(A)$ recursively as

$$\begin{equation*} \det(A) = \sum_{j=1}^{n} (-1)^{1+j} A_{1j} \cdot \det(\tilde{A}_{1j}). \end{equation*}$$

The scalar $\det(A)$ is called the determinant of $A$ and is also denoted by $|A|$. The scalar

$$\begin{equation*} (-1)^{i+j} \det(\tilde{A}_{ij}) \end{equation*}$$

is called the cofactor of the entry of $A$ in row $i$, column $j$.

Letting

$$\begin{equation*} c_{ij} = (-1)^{i+j} \det(\tilde{A}_{ij}) \end{equation*}$$

denote the cofactor of the row $i$, column $j$ entry of $A$, we can express the formula for the determinant of $A$ as

$$\begin{equation*} \det(A) = A_{11} c_{11} + A_{12} c_{12} + \cdots + A_{1n} c_{1n}. \end{equation*}$$

The determinant of an $n \times n$ matrix is a linear function of each row when the remaining rows are held fixed. That is, for $1 \le r \le n$, we have

$$\begin{equation*} \det \begin{pmatrix} a_1 \\ \vdots \\ a_{r-1} \\ u + kv \\ a_{r+1} \\ \vdots \\ a_n \end{pmatrix} = \det \begin{pmatrix} a_1 \\ \vdots \\ a_{r-1} \\ u \\ a_{r+1} \\ \vdots \\ a_n \end{pmatrix} + k \det \begin{pmatrix} a_1 \\ \vdots \\ a_{r-1} \\ v \\ a_{r+1} \\ \vdots \\ a_n \end{pmatrix}. \end{equation*}$$

whenever $k$ is a scalar and $u$, $v$, and each $a_i$ are row vectors in $F^n$.

Theorems

1. If $A \in M_{n \times n}(F)$ has a row consisting entirely of zeros, then

$$\begin{equation*} \det(A) = 0. \end{equation*}$$

2. The determinant of a square matrix can be evaluated by cofactor expansion along any row. That is, if $A \in M_{n \times n}(F)$, then for any integer $i$ ($1 \le i \le n$),

$$\begin{equation*} \det(A) = \sum_{j=1}^{n} (-1)^{i+j} A_{ij} \cdot \det(\tilde{A}_{ij}). \end{equation*}$$

3. If $A \in M_{n \times n}(F)$ and $B$ is a matrix obtained from $A$ by interchanging any two rows of $A$, then

   $$\begin{equation*} \det(B) = -\det(A). \end{equation*}$$

4. Let $A \in M_{n \times n}(F)$, and let $B$ be a matrix obtained by adding a multiple of one row of $A$ to another row of $A$. Then

   $$\begin{equation*} \det(B) = \det(A). \end{equation*}$$

5. If $A \in M_{n \times n}(F)$ has rank less than $n$, then

$$\begin{equation*} \det(A) = 0. \end{equation*}$$

6. For any $A, B \in M_{n \times n}(F)$, we have

   $$\begin{equation*} \det(AB) = \det(A)\,\det(B). \end{equation*}$$

7. For any $A \in M_{n \times n}(F)$, we have

   $$\begin{equation*} \det(A^{t}) = \det(A). \end{equation*}$$

8. Let $Ax = b$ be the matrix form of a system of $n$ linear equations in $n$ unknowns, where $x = (x_1, x_2, \ldots, x_n)^t.$If $\det(A) \neq 0$, then the system has a unique solution, and for each $k$ ($k = 1,2,\ldots,n$),

   $$\begin{equation*} x_k = \frac{\det(M_k)}{\det(A)}, \end{equation*}$$

   where $M_k$ is the $n \times n$ matrix obtained from $A$ by replacing column $k$ of $A$ with $b$.