Matrix determinant

What relationship could exist between the systems of linear equations with unique solutions, the linear independence between vectors, and the area of a parallelogram? In this topic, you will have a first approach with one of the most well-known and useful concepts of linear algebra: the determinant

The 2x2 system of equations with a unique solution

Let's analyze the $2 \times 2$ system of linear equations:

$$\begin{align*} ax + by &= e \\ cx + dy &= f \end{align*}$$You can easily eliminate the $y$ variable simply by multiplying the first equation by $d$ and the second by $-b$:$$\begin{align*} adx + bdy &= de \\ -bcx - bdy &= -bf \end{align*}$$Now, add both equations:$$(ad-bc)x= de -bf$$But don't rush! Here comes the tricky point. Just in case, that $ad-bc \neq 0$, you get that $x= \frac{de -bf}{ad-bc}$. And from the first equation, you can determine that $y= \frac{af -ec}{ad-bc}$. Thus, only if $ad-bc \neq 0$ the system has a unique solution, and you can calculate it explicitly!

Let's introduce some notation. Remember that the system can be written in matrix form:

$$\begin{pmatrix}a&b\\c&d \end{pmatrix} \begin{pmatrix} x \\y \end{pmatrix} = \begin{pmatrix} e \\f \end{pmatrix}$$ The special number $ad-bc$ determines if the system has a unique solution. It's so important it deserves a suggestive name. Let's introduce some notation. If you denote the matrix of the system by $A=\begin{pmatrix}a&b\\c&d \end{pmatrix}$, then the determinant of $A$ is $ad-bc$. It's usually denoted as $|A|$ or $\det A$ or even just $\begin{vmatrix}a&b\\c&d \end{vmatrix}$.

Linear independence between two vectors

Let's take two vectors in the plane $v=\begin{pmatrix}a \\ c \end{pmatrix}$ and $w=\begin{pmatrix}b \\ d \end{pmatrix}$.

You already know that they're linearly independent when $x\ v + y\ w=0$ implies $x=y=0$. Let's take a closer look at the first equation. It can be written more explicitly as $x \begin{pmatrix}a \\ c \end{pmatrix} + y \begin{pmatrix} b \\ d \end{pmatrix}=0$. But wait a minute, this is a system of equations:

$$\begin{align*} ax + by &= 0 \\ cx + dy &= 0 \end{align*}$$You can use two vectors to build a matrix $A$. For this, you simply put the vectors as the column of the matrix:

$$A=[v \, | \, w]= \begin{pmatrix}a&b\\c&d \end{pmatrix}$$

Recognizing what you discovered in the previous section, if the determinant of $A$ is different than $0$, then the values of $x$ and $y$ are uniquely determined as:$$x= \frac{0d - 0b}{|A|} = 0 \qquad y= \frac{0a - 0c}{|A|}=0$$

Let's summarize what you just discovered!

The area between vectors

Again consider two vectors in the plane $v=\begin{pmatrix}a \\ c \end{pmatrix}$ and $w=\begin{pmatrix}b \\ d \end{pmatrix}$. Let's denote the origin by $O$.

As you know, the sum of two vectors is the parallelogram formed by both, so let's start by drawing it. Let's denote by $L$ the line connecting $v$ with $v+w$:

The strategy will be to convert the parallelogram into a rectangle. For this, you will cut two triangles from the parallelogram and place them appropriately. First, imagine a horizontal line of height $d$. Let us denote by $P$ the point where it intersects $L$. Also, the point where $L$ intersects the horizontal axis will be important, let's write it as $Q=(x,0)$:

Now, If you look closely, the triangle formed by $w$, $P$ and $v +w$ is congruent to the triangle formed by $O$, $Q$ and $v$. That is, they're the same triangle but in different positions. This means that the area of the original parallelogram is the same as the area of the parallelogram formed by $O$, $Q$, $P$ and $w$:

Similarly, consider the vertical line through $Q$. This line and the vertical axis are crossed by the line connecting $w$ to $P$ at points $(x, d)$ and $(0, d)$, respectively:

Well, the triangle formed by $P$, $Q$ and $(x, d)$ is congruent to the triangle formed by $O$, $w$ and $(0, d)$. Thus, the area of the original parallelogram is just the same as the area of the rectangle formed by $O$, $Q$, $(x, d)$ and $(0, d)$!

Therefore, the area you're looking for is simply $x \, d$. All that is needed is to calculate the value of $x$!

For this, it'll be convenient for you to remember the formula for the slope of a line. If the points $(x_1, y_1)$ and $(x_2, y_2)$ lie on the same line, then the slope of the line is $\frac{y_2-y_1}{x_2 - x_1}$. Then, the slope $m$ of $L$, which contains $v +w$ and $v$, is:

$$m = \frac{(c + d) - c}{(a+b) - a} = \frac{d}{b}$$But $L$ contains the points $Q$ and $v$, so you can also calculate $m = \frac{c-0}{a-x} = \frac{c}{a-x}$. Equating the two values of $m$, it is immediate that $\frac{d}{b} = \frac{c}{a-x}$, thus:$$x=\frac{ad-bc}{d}$$

Finally, the area is $x \, d = \frac{ad-bc}{d} \, d = ad-bc$. This is the determinant of the matrix of $v$ and $w$!

What if the two vectors are linearly dependent? Geometrically it means that they are on the same line, so the parallelogram formed by both is degenerate. Connecting with the result of the previous section, when the vectors are linearly dependent, their determinant is 0, so the area of their parallelogram is 0.

The importance of the determinant

Up to this point, you have only dealt with $2 \times 2$ matrices, but the determinant can be generalized. Actually, it can be defined for any square matrix regardless of its size. In the following topic, you'll delve into its definition and the techniques to calculate it. In particular, $1 \times 1$ matrices can be thought of as simple numbers, so define their determinant to be that number. So, if $A =(a)$ is a $1 \times 1$ matrix, then $\det(A) = a$.

Virtually all of the results you constructed in this topic are true for larger dimensions: square systems of any size $n$, the linear independence of $n$ vectors in $\mathbb{R}^n$, and the "hypervolume" formed by $n$ vectors in $\mathbb{R}^n$.

But the determinant is not limited to just that. From a theoretical point of view, it is a very special function called multilinear, and it has unique properties that you'll learn about in future topics. In addition, its applications are very diverse, help to simplify, and have geometric interpretations. Some of the best-known are:

• Invertibility: The determinant can quickly tell you if a matrix is invertible or not. This is a crucial property when solving systems of linear equations.

• Eigenvalues and Eigenvectors: Determinants play a vital role in finding the eigenvalues of a matrix. You will learn about them in the future. They're the solutions of an equation that is derived from the determinant.

• Change of Basis: Determinants are used in the change of basis in linear algebra. When changing the basis of a vector space, the determinant of the change of the basis matrix can provide information about the "volume" change between the old basis and the new one.

• Matrix Diagonalization: This process consists of expressing a complicated matrix as the product of simpler ones with a simple geometric interpretation.

• Linear transformations. They are functions that get along well with the structure of the vector space. The determinant gives you a measure of how much they distort space.

• Calculation of areas of arbitrary regions. The integral is a widely used tool to calculate areas of sets that are much more general than simple rectangles. What is remarkable here is the calculation of certain integrals radically involves the determinant.

Conclusion

Let's consider the following system of linear equations:

$$\begin{align*} ax + by &= e \\ cx + dy &= f \end{align*}$$Its matrix is $A=\begin{pmatrix}a&b\\c&d \end{pmatrix}$. Let also $v=\begin{pmatrix}a \\ c \end{pmatrix}$ and $w=\begin{pmatrix}b \\ d \end{pmatrix}$ two vectors in the plane. Notice that $A=[v \, | \, w]$

• If $∣A∣ \neq 0$, then the system of equations has a unique solution given by

$$x= \frac{de -bf}{|A|} \qquad y= \frac{af -ce}{|A|}$$

• If $∣A∣ \neq 0$, then $v$ and $w$ are linearly independent.
• The area of the parallelogram formed by $v$ and $w$ is $∣A∣$.