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What a determinant measures: the area or volume spanned by the columns, the factor by which a map scales every region, and the sign that records whether space was turned over.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to read a determinant as the area of the parallelogram its columns span and as the factor by which the map scales every region at once, use its sign to say whether the map reverses orientation, explain why a zero determinant means the plane has been collapsed onto a line and therefore why the matrix has no inverse, see the multiplicative rule and the free row operation as statements about scalings and shears, and compute the area of a triangle or test three points for collinearity with a single determinant.
You can find a determinant two ways, and you know the rules it obeys: multiplicative over products, unchanged by a transpose, multiplied by $c^n$ when the whole matrix is scaled, zero exactly when the matrix has no inverse. Every one of those is a fact you have been told. This lesson makes them one fact you can see.
The parallelogram spanned by two vectors $u$ and $v$ is the set of all $su + tv$ with $s$ and $t$ between $0$ and $1$; in three dimensions three vectors span a parallelepiped. A basis of the plane is positively oriented when the second vector lies anticlockwise from the first, and negatively oriented otherwise; a map that exchanges the two is said to reverse orientation. A matrix that collapses space onto something smaller is singular, and its image is then a proper subspace.
Let $A$ be $2 \times 2$. The unit square has area $1$, and $A$ sends it to the parallelogram spanned by the two columns of $A$ — because the columns are exactly where $A$ sends $(1, 0)$ and $(0, 1)$.
$$\text{area of the image} = |\det A|$$
That is not a special fact about the square. Every region of the plane has its area multiplied by the same factor $|\det A|$: cut any shape into small squares, each is scaled by $|\det A|$, and the total is too. In three dimensions the same statement holds with volume and a $3 \times 3$ determinant.
The sign is orientation. Area is never negative, so the sign is carrying something else: whether the plane came out the right way round. A positive determinant means the columns are in the same rotational order as $(1,0)$ and $(0,1)$; a negative one means the map turned the plane over, so a reflection is part of what it does.
Three things you already knew now have one-line pictures:
Another way: picture
Draw the unit square with corners at $(0,0)$, $(1,0)$, $(1,1)$ and $(0,1)$ and watch where $A$ sends it. The corner $(1,0)$ goes to the first column, $(0,1)$ goes to the second, and $(1,1)$ goes to their sum, so the square becomes the parallelogram on those two columns. Its area is $|\det A|$. Now imagine the whole plane tiled with tiny copies of that square: every one of them is sheared and stretched the same way, so every area in the plane is multiplied by the same number.
Another way: steps
Suppose $\det A = 0$ for a $2 \times 2$ matrix. Then the parallelogram spanned by the columns has zero area, which for two vectors in a plane means exactly one thing: they are parallel, or one of them is zero. Either way both columns lie on a single line through the origin.
Every output of $A$ is a combination of its columns, so every output lies on that line. The whole plane — two dimensions of it — has been folded onto one. Points that were different now coincide, and no rule can decide which of them an image came from. That is what having no inverse looks like, and it is why no inverse, zero determinant, zero area and parallel columns are four descriptions of one situation rather than four facts.
The same argument runs in three dimensions with one extra case: a $3 \times 3$ matrix with determinant zero flattens space onto a plane, onto a line, or onto the origin, according to how many independent columns survive. The determinant reports that something was lost without saying how much, which is the question the rank answers instead.
Two constructions cover almost everything asked of this idea.
A triangle from three points. Vertices $A$, $B$, $C$ give edge vectors $B - A$ and $C - A$. The parallelogram on those has area $|\det[\,B-A \;\; C-A\,]|$, and the triangle is half of it:
$$\text{area} = \tfrac{1}{2}\left|\det \begin{pmatrix} x_B - x_A & x_C - x_A \\ y_B - y_A & y_C - y_A \end{pmatrix}\right|$$
This is one formula for every triangle in the plane, with no cases for obtuse or right-angled, no need for a height, and no trigonometry. The subtraction is the step that is forgotten: a determinant always measures from the origin, so the figure has to be brought there.
A volume from three vectors. In three dimensions, $|\det[\,u \;\; v \;\; w\,]|$ is the volume of the parallelepiped they span, and a zero value says the three vectors lie in a common plane. That test — are these three vectors coplanar? — is one determinant, and it is the three-dimensional form of are these two vectors parallel?
A useful consequence: three points in the plane are collinear exactly when the triangle they make has area zero, so one determinant settles collinearity too.
Area is the absolute value. $\det A = -6$ means an area scaling of $6$ with the plane turned over. Reporting an area of $-6$ is reporting something that does not exist.
A determinant measures from the origin. For three given points, $\det$ of the matrix of their coordinates is not the area of anything you want. Subtract one vertex from the other two first; the answer is then independent of where the triangle sits, as an area must be.
The triangle is a half. The determinant gives the parallelogram. Forgetting the $\tfrac{1}{2}$ doubles every answer, and the mistake is invisible unless you check against a simple case.
Scaling the matrix is not scaling one side. Doubling every entry of a $2 \times 2$ matrix doubles both columns, so it quadruples the area — $\det(2A) = 4\det A$. The picture makes the exponent obvious in a way the algebra does not.
A small determinant is not a small matrix. $\begin{pmatrix} 1000 & 999 \\ 1001 & 1000 \end{pmatrix}$ has determinant $-999$ while its entries are enormous; and a matrix of tiny entries can be invertible. The determinant reports the area spanned, not the size of anything.
$A = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}$, so $\det A = 6 - 1 = 5$.
One determinant, computed once.
The unit square becomes a parallelogram of area $5$. A circle of area $\pi$ becomes an ellipse of area $5\pi$. A triangle of area $0.2$ becomes one of area $1$.
The factor does not depend on the shape.
The determinant is positive, so nothing was turned over, and $\det(A^{-1}) = \tfrac{1}{5}$ — the inverse shrinks every area back by the same factor, as it must.
Scalings compose by multiplying.
Vertices $(1, 1)$, $(5, 2)$, $(3, 6)$. Edge vectors from the first: $(4, 1)$ and $(2, 5)$.
Subtract a vertex before anything else.
$\det \begin{pmatrix} 4 & 2 \\ 1 & 5 \end{pmatrix} = 20 - 2 = 18$, so the parallelogram has area $18$ and the triangle has area $9$.
Halve the parallelogram.
Now $(1,1)$, $(3,2)$, $(7,4)$: the edge vectors are $(2,1)$ and $(6,3)$, the determinant is $6 - 6 = 0$, and the three points are collinear — the triangle has no area because it is not a triangle.
Zero area, zero determinant, one line.
The determinant is $0 \times 0 - (-2)(2) = 4$.
Compute it before interpreting it.
Positive, so orientation is preserved: nothing has been turned over.
The sign first, then the size.
Every area is multiplied by $4$. The matrix is in fact a quarter turn combined with a scaling by $2$ in both directions, and two directions each doubled multiply area by four.
Write any $2 \times 2$ matrix of whole numbers whose two columns span a parallelogram of area $8$, taken with positive orientation.
This task has no paper form; do it on a device.
Find the area of the parallelogram spanned by the vectors $(9, 7)$ and $(1, 2)$.
Answer:
Match each matrix to what it does to areas in the plane.
| Leaves every area exactly as it was | Multiplies every area by $20$ | Collapses every area to zero | |
|---|---|---|---|
| $\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}$ | |||
| $\begin{pmatrix} 5 & 0 \\ 0 & 4 \end{pmatrix}$ | |||
| $\begin{pmatrix} 3 & 6 \\ 3 & 6 \end{pmatrix}$ |
Write a $2 \times 2$ matrix, with no entry equal to zero, that collapses the whole plane onto a single line.
This task has no paper form; do it on a device.
A $2 \times 2$ matrix has determinant $-9$. What does the sign tell you?
A triangle has vertices $A = (4, 1)$, $B = (10, 1)$ and $C = (10, 10)$. Fill in the table to find its area.
| Value | |
|---|---|
| First component of $B - A$ | |
| Second component of $B - A$ | |
| First component of $C - A$ | |
| Second component of $C - A$ | |
| Determinant of the matrix with those two as its columns | |
| Area of the triangle |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Write any $2 \times 2$ matrix of whole numbers whose two columns span a parallelogram of area $7$, taken with positive orientation.
This task has no paper form; do it on a device.
You can read an area, a volume and an orientation off a determinant, and say what a determinant of zero does to the plane. Say in your own words why $\det(AB) = \det A \det B$ is obvious once the determinant is a scaling factor. Next: vector spaces, where the word dimension stops being something you can only draw.
10. Your turn: what does $\begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}$ do to areas?, step 3
The picture agrees with the number.