📂Matrix Algebra

Determinants

Definition

Let $A$ be the following $2 \times 2$ matrix.

$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

The determinant of $A$ is defined as follows and is denoted by $\det(A)$.

$$\det(A) := ad - bc$$

Explanation

To discuss determinants, we cannot ignore the very purpose of linear algebra. Most problems in mathematics can be summarized as ‘Can we solve the equation?’ A simple equation like

$$ax = b$$

demonstrates that unless $a = 0$, this equation clearly has a solution. Quadratic equations such as

$$a x^2 + b x + c = 0$$

can also be easily solved using the quadratic formula. Consequently, mathematicians increased the degree of $x$, aiming to tackle more challenging problems. However, it was proven by the unfortunate genius Abel that ‘algebraic equations of degree 5 or higher have no general solution.’

Meanwhile, exploring paths by increasing variables or the number of equations remained open. And thus came determinants. Although it might seem from its Korean name that determinants appeared after matrices, historically, determinants emerged before matrices¹, and in fact, the English words “determinant” and “matrix” are not particularly related. The term “determinant” refers to a formula used to determine whether a system of linear equations with two variables has a solution.

$$\left\{ \begin{align*} ax + by &= 0 \\ cx + dy &= 0 \end{align*} \right.$$

When provided with a system of equations like above, if $ad-bc = 0$ then only the trivial solution $x=y=0$ exists. If $ad-bc \ne 0$, it has a unique non-trivial solution. Therefore, $ad-bc$ becomes the formula that determines the existence of solutions to the given system of equations, hence the name determinant.

As you know, systems of equations can be expressed in the form of matrices. A ‘simple’ system of equations can be represented as follows:

$$A \mathbf{x} = \mathbf{b}$$

Recall that the solution of $ax = b$ was $x = \dfrac{b}{a}$. Since $\dfrac{1}{a}$ is the inverse of $a$, multiplying both sides could leave only $x$. Regarding the conditions for the existence of solutions, $a= 0$ does not have an inverse, thus a solution for $ax = b$ does not exist. Similarly, the question of the existence of $A \mathbf{x} = \mathbf{b}$ also boils down to whether the inverse of $A$ can be found. The existence of the inverse of $A$ itself determines whether the linear system represented by $A$ has a solution, and finding this inverse is equivalent to finding the solution. It is evident that the condition for the existence of the inverse of $A$ coincides with the condition for the linear system represented by $A$ to have a unique solution.

The inverse of $A = \begin{bmatrix} a & b \\ c & d\end{bmatrix}$ is as follows:

$$A^{-1} = \dfrac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

The method to prove this is simply to multiply $A$ with $A^{-1}$. If $\det (A) = ad - bc = 0$ holds, regardless of the shape of the matrix, the constant in front of $A^{-1}$ is $\dfrac{1}{0}$, meaning no inverse can exist. This is the reason invertibility is often called nonsingularity. The term “singular” translates to “peculiar,” which in mathematical terms, suggests a notion akin to “dividing by zero.”

On the other hand, from the perspective of viewing a determinant as a function mapping $n\times n$ real numbers to $1$ real numbers, it can be defined as follows.

General Definition

Definition by Properties

A function $\det : \mathbb{R}^{n \times n } \to \mathbb{R}$ is defined as a determinant if it satisfies the following conditions:

• For the identity matrix $I_{n}$, $\det(I_{n}) = 1$
• For $1 \le i,j \le n$, $\det \begin{bmatrix} \mathbb{r_{1}} \\ \vdots \\ \mathbb{r_{i}} \\ \vdots \\ \mathbb{r_{j}} \\ \vdots \\ \mathbb{r_{n}} \end{bmatrix} = - \det \begin{bmatrix} \mathbb{r_{1}} \\ \vdots \\ \mathbb{r_{j}} \\ \vdots \\ \mathbb{r_{i}} \\ \vdots \\ \mathbb{r_{n}} \end{bmatrix}$
• $\det \begin{bmatrix} k \mathbb{r_{1}} + l \mathbb{r_{1}}^{\prime} \\ \vdots \\ \mathbb{r_{n}} \end{bmatrix} = k \det \begin{bmatrix} \mathbb{r_{1}} \\ \vdots \\ \mathbb{r_{n}} \end{bmatrix} + l \det \begin{bmatrix} \mathbb{r_{1}}^{\prime} \\ \vdots \\ \mathbb{r_{n}} \end{bmatrix}$

Definition by Permutations

Let the permutation of the set $\left\{ 1, 2, \dots, n \right\}$ be $\sigma$. Let $A = [a_{ij}]$ be a $n \times n$ matrix. Then, the determinant of $A$ is defined as follows:

$$\begin{align*} \det (A) &= \sum_{\sigma \in S_{n}} \sgn(\sigma) a_{1\sigma(1)} a_{2\sigma(2)} \cdots a_{n\sigma(n)} \\ &= \sum_{\sigma \in S_{n}} \sgn(\sigma) \prod_{i=1}^{n} a_{i\sigma(i)} \\ \end{align*}$$

Here, $S_{n}$ is the symmetric group, and $\sgn$ is the sign of the permutation.

Explanation

By broadening our understanding of determinants, it becomes significantly easier to discuss whether solutions for a system of equations exist. The culmination of such discussions is encapsulated in the theorem below.

$$\forall A \in \mathbb{C}^{n \times n},\quad \exists A^{-1} \iff \det{A} \ne 0$$

Though regarded almost as a definition, this is a fact so obvious that it can be accepted as a theorem. However, if one cannot properly explain why such a theorem exists, or why it is truly obvious, then it implies a lack of understanding of determinants. Especially in the case of determinants, concepts often precede definitions, so dedicate ample time to ensuring comprehension if necessary.

See Also

• Properties of Determinants
• Laplace Expansion $\det A = \sum\limits_{i,j} a_{ij} [\operatorname{adj} A]_{ij}$
• Jacobi’s Formula $\mathrm{d}(\det A) = \Tr \Big( (\operatorname{adj}A) \mathrm{d}A \Big) = \det A \cdot \Tr(A^{-1} \mathrm{d}A)$