# 4.3: Determinants and Volumes

- Page ID
- 70204

- Understand the relationship between the determinant of a matrix and the volume of a parallelepiped.
- Learn to use determinants to compute volumes of parallelograms and triangles.
- Learn to use determinants to compute the volume of some curvy shapes like ellipses.
*Pictures:*parallelepiped, the image of a curvy shape under a linear transformation.*Theorem:*determinants and volumes.*Vocabulary word:***parallelepiped**.

In this section we give a geometric interpretation of determinants, in terms of *volumes.* This will shed light on the reason behind three of the four defining properties of the determinant, Definition 4.1.1 in Section 4.1. It is also a crucial ingredient in the change-of-variables formula in multivariable calculus.

## Parallelograms and Paralellepipeds

The determinant computes the volume of the following kind of geometric object.

The **paralellepiped **determined by \(n\) vectors \(v_1, v_2,\ldots,v_n\) in \(\mathbb{R}^n \) is the subset

\[ P = \bigl\{a_1x_1 + a_2x_2 + \cdots + a_nx_n \bigm| 0 \leq a_1,a_2,\ldots,a_n\leq 1\bigr\}. \nonumber \]

In other words, a parallelepiped is the set of all linear combinations of \(n\) vectors with coefficients in \([0,1]\). We can draw parallelepipeds using the parallelogram law for vector addition.

Example \(\PageIndex{1}\): The Unit Cubte

The parallelepiped determined by the standard coordinate vectors \(e_1,e_2,\ldots,e_n\) is the unit \(n\)-dimensional cube.

Figure \(\PageIndex{1}\)

Example \(\PageIndex{2}\): Parallelograms

When \(n = 2\text{,}\) a paralellepiped is just a paralellogram in \(\mathbb{R}^2 \). Note that the edges come in parallel pairs.

Figure \(\PageIndex{2}\)

Example \(\PageIndex{3}\)

When \(n=3\text{,}\) a parallelepiped is a kind of a skewed cube. Note that the faces come in parallel pairs.

Figure \(\PageIndex{3}\)

When does a parallelepiped have zero volume? This can happen only if the parallelepiped is flat, i.e., it is squashed into a lower dimension.

Figure \(\PageIndex{4}\)

This means exactly that \(\{v_1,v_2,\ldots,v_n\}\) is *linearly dependent,* which by Corollary 4.1.1 in Section 4.1 means that the matrix with rows \(v_1,v_2,\ldots,v_n\) has determinant zero. To summarize:

The parallelepiped defined by \(v_1,v_2,\ldots,v_n\) has zero volume if and only if the matrix with rows \(v_1,v_2,\ldots,v_n\) has zero determinant.

## Determinants and Volumes

The key observation above is only the beginning of the story: the volume of a parallelepiped is *always* a determinant.

Theorem \(\PageIndex{1}\): Determinants and Volumes

Let \(v_1,v_2,\ldots,v_n\) be vectors in \(\mathbb{R}^n \text{,}\) let \(P\) be the parallelepiped determined by these vectors, and let \(A\) be the matrix with rows \(v_1,v_2,\ldots,v_n\). Then the absolute value of the determinant of \(A\) is the volume of \(P\text{:}\)

\[ |\det(A)| = \text{vol}(P). \nonumber \]

**Proof**-
Since the four defining properties, Definition 4.1.1 in Section 4.1, characterize the determinant, they also characterize the absolute value of the determinant. Explicitly, \(|\det|\) is a function on square matrices which satisfies these properties:

- Doing a row replacement on \(A\) does not change \(|\det(A)|\).
- Scaling a row of \(A\) by a scalar \(c\) multiplies \(|\det(A)|\) by \(|c|\).
- Swapping two rows of a matrix does not change \(|\det(A)|\).
- The determinant of the identity matrix \(I_n\) is equal to \(1\).

The absolute value of the determinant is the

*only*such function: indeed, by Recipe: Computing Determinants by Row Reducing in Section 4.1, if you do some number of row operations on \(A\) to obtain a matrix \(B\) in row echelon form, then\[ |\det(A)| = \left|\frac{\text{(product of the diagonal entries of $B$)}} {\text{(product of scaling factors used)}}\right|. \nonumber \]

For a square matrix \(A\text{,}\) we abuse notation and let \(\text{vol}(A)\) denote the volume of the paralellepiped determined by the rows of \(A\). Then we can regard \(\text{vol}\) as a function from the set of square matrices to the real numbers. We will show that \(\text{vol}\) also satisfies the above four properties.

- For simplicity, we consider a row replacement of the form \(R_n = R_n + cR_i\). The volume of a paralellepiped is the volume of its base, times its height: here the “base” is the paralellepiped determined by \(v_1,v_2,\ldots,v_{n-1}\text{,}\) and the “height” is the perpendicular distance of \(v_n\) from the base.

Figure \(\PageIndex{5}\)

- Translating \(v_n\) by a multiple of \(v_i\) moves \(v_n\) in a direction parallel to the base. This changes neither the base nor the height! Thus, \(\text{vol}(A)\) is unchanged by row replacements.

Figure \(\PageIndex{6}\)

- For simplicity, we consider a row scale of the form \(R_n = cR_n.\) This scales the length of \(v_n\) by a factor of \(|c|\text{,}\) which also scales the perpendicular distance of \(v_n\) from the base by a factor of \(|c|\). Thus, \(\text{vol}(A)\) is scaled by \(|c|\).

Figure \(\PageIndex{7}\)

- Swapping two rows of \(A\) just reorders the vectors \(v_1,v_2,\ldots,v_n\text{,}\) hence has no effect on the parallelepiped determined by those vectors. Thus, \(\text{vol}(A)\) is unchanged by row swaps.

Figure \(\PageIndex{8}\)

- The rows of the identity matrix \(I_n\) are the standard coordinate vectors \(e_1,e_2,\ldots,e_n\). The associated paralellepiped is the unit cube, which has volume \(1\). Thus, \(\text{vol}(I_n) = 1\).

Since \(|\det|\) is the only function satisfying these properties, we have

\[ \text{vol}(P) = \text{vol}(A) = |\det(A)|. \nonumber \]

This completes the proof.

Since \(\det(A) = \det(A^T)\) by the transpose property, Proposition 4.1.4 in Section 4.1, the absolute value of \(\det(A)\) is also equal to the volume of the paralellepiped determined by the *columns* of \(A\) as well.

Example \(\PageIndex{4}\): Length

A \(1\times 1\) matrix \(A\) is just a number \(\left(\begin{array}{c}a\end{array}\right)\). In this case, the parallelepiped \(P\) determined by its one row is just the interval \([0,a]\) (or \([a,0]\) if \(a\lt0\)). The “volume” of a region in \(\mathbb{R}^1 = \mathbb{R}\) is just its length, so it is clear in this case that \(\text{vol}(P) = |a|\).

Figure \(\PageIndex{9}\)

Example \(\PageIndex{5}\): Area

When \(A\) is a \(2\times 2\) matrix, its rows determine a parallelogram in \(\mathbb{R}^2 \). The “volume” of a region in \(\mathbb{R}^2 \) is its area, so we obtain a formula for the area of a parallelogram: it is the determinant of the matrix whose rows are the vectors forming two adjacent sides of the parallelogram.

Figure \(\PageIndex{10}\)

It is perhaps surprising that it is possible to compute the area of a parallelogram without trigonometry. It is a fun geometry problem to prove this formula by hand. [Hint: first think about the case when the first row of \(A\) lies on the \(x\)-axis.]

Find the area of the parallelogram with sides \((1,3)\) and \((2,-3)\).

Figure \(\PageIndex{11}\)

**Solution**

The area is

\[ \left|\det\left(\begin{array}{cc}1&3\\2&-3\end{array}\right)\right| = |-3-6| = 9. \nonumber \]

Find the area of the parallelogram in the picture.

Figure \(\PageIndex{12}\)

**Solution**

We choose two adjacent sides to be the rows of a matrix. We choose the top two:

Figure \(\PageIndex{13}\)

Note that we do not need to know where the origin is in the picture: vectors are determined by their length and direction, not where they start. The area is

\[ \left|\det\left(\begin{array}{cc}-1&-4\\2&-1\end{array}\right)\right| = |1+8| = 9. \nonumber \]

Find the area of the triangle with vertices \((-1,-2), \,(2,-1),\,(1,3).\)

Figure \(\PageIndex{14}\)

**Solution**

Doubling a triangle makes a paralellogram. We choose two of its sides to be the rows of a matrix.

Figure \(\PageIndex{15}\)

The area of the parallelogram is

\[ \left|\det\left(\begin{array}{cc}2&5\\3&1\end{array}\right)\right| = |2-15| = 13, \nonumber \]

so the area of the triangle is \(13/2\).

You might be wondering: if the absolute value of the determinant is a volume, what is the geometric meaning of the determinant without the absolute value? The next remark explains that we can think of the determinant as a *signed* volume. If you have taken an integral calculus course, you probably computed negative areas under curves; the idea here is similar.

Theorem \(\PageIndex{1}\) on determinants and volumes tells us that the *absolute value* of the determinant is the volume of a paralellepiped. This raises the question of whether the sign of the determinant has any geometric meaning.

A \(1\times 1\) matrix \(A\) is just a number \(\left(\begin{array}{c}a\end{array}\right)\). In this case, the parallelepiped \(P\) determined by its one row is just the interval \([0,a]\) if \(a \geq 0\text{,}\) and it is \([a,0]\) if \(a\lt0\). In this case, the sign of the determinant determines whether the interval is to the left or the right of the origin.

For a \(2\times 2\) matrix with rows \(v_1,v_2\text{,}\) the sign of the determinant determines whether \(v_2\) is counterclockwise or clockwise from \(v_1\). That is, if the counterclockwise angle from \(v_1\) to \(v_2\) is less than \(180^\circ\text{,}\) then the determinant is positive; otherwise it is negative (or zero).

Figure \(\PageIndex{16}\)

For example, if \(v_1 = {a\choose b}\text{,}\) then the counterclockwise rotation of \(v_1\) by \(90^\circ\) is \(v_2 = {-b\choose a}\) by Example 3.3.8 in Section 3.3, and

\[ \det\left(\begin{array}{cc}a&b\\-b&a\end{array}\right) = a^2 + b^2 > 0. \nonumber \]

On the other hand, the *clockwise* rotation of \(v_1\) by \(90^\circ\) is \(b\choose -a\text{,}\) and

\[ \det\left(\begin{array}{cc}a&b\\b&-a\end{array}\right) = -a^2 - b^2 \lt 0. \nonumber \]

For a \(3\times 3\) matrix with rows \(v_1,v_2,v_3\text{,}\) the *right-hand rule* determines the sign of the determinant. If you point the index finger of your right hand in the direction of \(v_1\) and your middle finger in the direction of \(v_2\text{,}\) then the determinant is positive if your thumb points roughly in the direction of \(v_3\text{,}\) and it is negative otherwise.

Figure \(\PageIndex{17}\)

In higher dimensions, the notion of signed volume is still important, but it is usually *defined* in terms of the sign of a determinant.

## Volumes of Regions

Let \(A\) be an \(n\times n\) matrix with columns \(v_1,v_2,\ldots,v_n\text{,}\) and let \(T\colon\mathbb{R}^n \to\mathbb{R}^n \) be the associated matrix transformation, Definition 3.1.3 in Section 3.1, \(T(x)=Ax\). Then \(T(e_1)=v_1\) and \(T(e_2)=v_2\text{,}\) so \(T\) takes the unit cube \(C\) to the parallelepiped \(P\) determined by \(v_1,v_2,\ldots,v_n\text{:}\)

Figure \(\PageIndex{18}\)

Since the unit cube has volume \(1\) and its image has volume \(|\det(A)|\text{,}\) the transformation \(T\) scaled the volume of the cube by a factor of \(|\det(A)|\). To rephrase:

Note \(\PageIndex{1}\)

If \(A\) is an \(n\times n\) matrix with corresponding matrix transformation \(T\colon\mathbb{R}^n \to\mathbb{R}^n \text{,}\) and if \(C\) is the unit cube in \(\mathbb{R}^n \text{,}\) then the volume of \(T(C)\) is \(|\det(A)|.\)

The notation \(T(S)\) means the image of the region \(S\) under the transformation \(T\). In set builder notation, Note 2.2.3 in Section 2.2, this is the subset

\[ T(S) = \bigl\{T(x)\mid x \text{ in } S\bigr\}. \nonumber \]

In fact, \(T\) scales the volume of *any* region in \(\mathbb{R}^n \) by the same factor, even for curvy regions.

Let \(A\) be an \(n\times n\) matrix, and let \(T\colon\mathbb{R}^n \to\mathbb{R}^n \) be the associated matrix transformation \(T(x)=Ax\). If \(S\) is any region in \(\mathbb{R}^n \text{,}\) then

\[ \text{vol}(T(S))=|\det(A)|\cdot\text{vol}(S). \nonumber \]

**Proof**-
Let \(C\) be the unit cube, let \(v_1,v_2,\ldots,v_n\) be the columns of \(A\text{,}\) and let \(P\) be the paralellepiped determined by these vectors, so \(T(C) = P\) and \(\text{vol}(P) = |\det(A)|\). For \(\epsilon > 0\) we let \(\epsilon C\) be the cube with side lengths \(\epsilon\text{,}\) i.e., the paralellepiped determined by the vectors \(\epsilon e_1,\epsilon e_2,\ldots,\epsilon e_n\text{,}\) and we define \(\epsilon P\) similarly. By the second defining property, \(T\) takes \(\epsilon C\) to \(\epsilon P\). The volume of \(\epsilon C\) is \(\epsilon^n\) (we scaled each of the \(n\) standard vectors by a factor of \(\epsilon\)) and the volume of \(\epsilon P\) is \(\epsilon^n|\det(A)|\) (for the same reason), so we have shown that \(T\) scales the volume of \(\epsilon C\) by \(|\det(A)|\).

Figure \(\PageIndex{19}\)

By the first defining property, Definition 3.3.1 in Section 3.3, the image of a translate of \(\epsilon C\) is a translate of \(\epsilon P\text{:}\)

\[ T(x + \epsilon C) = T(x) + \epsilon T(C) = T(x) + \epsilon P. \nonumber \]

Since a translation does not change volumes, this proves that \(T\) scales the volume of a translate of \(\epsilon C\) by \(|\det(A)|\).

At this point, we need to use techniques from multivariable calculus, so we only give an idea of the rest of the proof. Any region \(S\) can be approximated by a collection of very small cubes of the form \(x + \epsilon C\). The image \(T(S)\) is then approximated by the image of this collection of cubes, which is a collection of very small paralellepipeds of the form \(T(x) + \epsilon P\).

Figure \(\PageIndex{20}\)

The volume of \(S\) is closely approximated by the sum of the volumes of the cubes; in fact, as \(\epsilon\) goes to zero, the limit of this sum is precisely \(\text{vol}(S)\). Likewise, the volume of \(T(S)\) is equal to the sum of the volumes of the paralellepipeds, take in the limit as \(\epsilon\to 0\). The key point is that

*the volume of each cube is scaled by \(|\det(A)|\).*Therefore, the sum of the volumes of the paralellepipeds is \(|\det(A)|\) times the sum of the volumes of the cubes. This proves that \(\text{vol}(T(S)) = |\det(A)|\text{vol}(S).\)

Let \(S\) be a half-circle of radius \(1\text{,}\) let

\[ A = \left(\begin{array}{cc}1&2\\2&1\end{array}\right), \nonumber \]

and define \(T\colon\mathbb{R}^2 \to\mathbb{R}^2 \) by \(T(x)=Ax\). What is the area of \(T(S)\text{?}\)

Figure \(\PageIndex{21}\)

**Solution**

The area of the unit circle is \(\pi\text{,}\) so the area of \(S\) is \(\pi/2\). The transformation \(T\) scales areas by a factor of \(|\det(A)| = |1 - 4| = 3\text{,}\) so

\[ \text{vol}(T(S)) = 3\text{vol}(S) = \frac{3\pi}2. \nonumber \]

Find the area of the interior \(E\) of the ellipse defined by the equation

\[ \left(\frac{2x-y}{2}\right)^2 + \left(\frac{y+3x}{3}\right)^2 = 1. \nonumber \]

**Solution**

This ellipse is obtained from the unit circle \(X^2+Y^2=1\) by the linear change of coordinates

\[ \begin{split} X \amp= \frac{2x-y}2 \\ Y \amp= \frac{y+3x}3. \end{split} \nonumber \]

In other words, if we define a linear transformation \(T\colon\mathbb{R}^2 \to\mathbb{R}^2 \) by

\[ T\left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c}(2x-y)/2 \\ (y+3x)/3\end{array}\right), \nonumber \]

then \(T{x\choose y}\) lies on the unit circle \(C\) whenever \(x\choose y\) lies on \(E\).

Figure \(\PageIndex{22}\)

We compute the standard matrix \(A\) for \(T\) by evaluating on the standard coordinate vectors:

\[ T\left(\begin{array}{c}1\\0\end{array}\right) = \left(\begin{array}{c}1\\1\end{array}\right) \qquad T\left(\begin{array}{c}0\\1\end{array}\right) = \left(\begin{array}{c}-1/2 \\ 1/3\end{array}\right) \quad\implies\quad A = \left(\begin{array}{cc}1&-1/2 \\ 1&1/3\end{array}\right). \nonumber \]

Therefore, \(T\) scales areas by a factor of \(|\det(A)| = |\frac13+\frac12| = \frac56.\) The area of the unit circle is \(\pi\text{,}\) so

\[ \pi = \text{vol}(C) = \text{vol}(T(E)) = |\det(A)|\cdot\text{vol}(E) = \frac56\text{vol}(E), \nonumber \]

and thus the area of \(E\) is \(6\pi/5\).

The above Theorem \(\PageIndex{2}\) also gives a geometric reason for multiplicativity of the (absolute value of the) determinant. Indeed, let \(A\) and \(B\) be \(n\times n\) matrices, and let \(T,U\colon\mathbb{R}^n \to\mathbb{R}^n \) be the corresponding matrix transformations. If \(C\) is the unit cube, then

\[ \begin{split} \text{vol}\bigl(T\circ U(C)\bigr) = \text{vol}\bigl(T(U(C))\bigr) \amp= |\det(A)|\text{vol}(U(C)) \\ \amp= |\det(A)|\cdot|\det(B)|\text{vol}(C) \\ \amp= |\det(A)|\cdot|\det(B)|. \end{split} \nonumber \]

On the other hand, the matrix for the composition \(T\circ U\) is the product \(AB\text{,}\) so

\[ \text{vol}\bigl(T\circ U(C)\bigr) = |\det(AB)|\text{vol}(C) = |\det(AB)|. \nonumber \]

Thus \(|\det(AB)| = |\det(A)|\cdot|\det(B)|\).