5.1. Determinants as Areas or Volumes

5.1.1. Introduction

The word “determinant” already appeared in the section about invertibility: a \(2\times2\) matrix \(A = \left[\begin{array}{cc} a & b \\ c & d\end{array} \right]\) is invertible if and only if

(5.1.1)\[ad-bc \neq 0.\]

We called the expression \(ad-bc\) the determinant of the matrix \(A\). Formula (5.1.1) is also equivalent to the statement that the columns of the matrix \(A\) are linearly independent.

Likewise, by row reducing a general \(3 \times 3\) matrix

\[\begin{split} A = \left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{array} \right] \end{split}\]

we might end up with an expression containing all the entries \(a_{ij}\) that tells us whether \(A\) is invertible or not. In this section we will use a geometric approach to derive such an expression, and will again call this the determinant of the matrix \(A\). Its formula (Equation (5.1.2)), when looked at from the right perspective, shows an opportunity to generalize the concept to higher dimensions. We will follow that route in Section 5.2.

We will start by introducing determinants as a way to compute areas (in the plane) and volumes (in the space \(\R^3\)).

5.1.2. Area, orientation, determinant

We start with two vectors \(\vect{u}\) and \(\vect{v}\) (as always, starting from the origin) in \(\R^2\). They ‘span’ a parallelogram \(OACB\), where \(A\) and \(B\) are the end points of \(\vect{u}\) and \(\vect{v}\), and \(C\) corresponds to the vector \(\vect{u}+\vect{v}\). See Figure 5.1.1

Fig. 5.1.1 The parallelogram OACB

Proposition 5.1.1

The area of the parallelogram \(OACB\), spanned by the vectors \( \vect{u} =\left[\begin{array}{c} a \\ b \end{array}\right]\) and \(\vect{v}=\left[\begin{array}{c} c \\ d \end{array}\right]\) is given by \(|ad-bc|\), i.e., the absolute value of   \(ad-bc\).

Proof. The quickest way to prove this is to translate it to the cross product (Section 1.3). To make use of the cross product we introduce the vectors.

\[\begin{split} \tilde{\vect{u}} = \left[\begin{array}{c} a \\ b \\ 0 \end{array}\right] \quad \text{and} \quad \tilde{\vect{v}} = \left[\begin{array}{c} c \\ d \\ 0 \end{array}\right] \end{split}\]

Thus we embed the plane into \(\R^3\) as the \(x_1\)-\(x_2\)-plane. See Figure 5.1.2.

Fig. 5.1.2 Oriented area

So we embed the plane \(\R^2\) as the \(x\)-\(y\)-plane into \(\R^3\).

We then have

\[\begin{split} \tilde{\vect{u}} \times \tilde{\vect{v}} = \left[\begin{array}{c} 0 \\ 0 \\ ad-bc \end{array}\right]. \end{split}\]

The length of this cross product is equal to \(|ad-bc|\). This gives the area of the parallelogram spanned by the vectors \(\tilde{\vect{u}}\) and \(\tilde{\vect{v}}\), and this is an identical copy of the original parallelogram spanned by \(\vect{u}\) and \(\vect{v}\).

Now what about the sign here?

Proposition 5.1.2

\(ad - bc = \norm{\vect{u}}\norm{\vect{v}}\sin(\varphi)\), where \(\varphi\) is the angle from \(\vect{u}\) counterclockwise to \(\vect{v}\). We will call this the directed angle from \(\vect{u}\) to \(\vect{v}\).

Proof. Again we can resort to properties of the cross product, but in fact it is not necessary to go up one dimension. By a small twist we can turn the determinant into an inner product:

\[\begin{split} ad-bc = \left[\begin{array}{c} -b \\ a \end{array}\right] \ip \left[\begin{array}{c} c \\ d \end{array}\right] = \vect{u}^{\perp} \ip \vect{v}, \end{split}\]

where \(\vect{u}^{\perp}\) is the vector that is perpendicular to \(\vect{u}\), points ‘to the left’ of \(\vect{u}\), and has the same length as \(\vect{u}\). See Figure 5.1.3.

Fig. 5.1.3 The parallelogram \(OACB\) and the orthogonal vector \(\vect{u}^{\perp}\)

So

\[ ad-bc = \vect{u}^{\perp} \ip \vect{v} = \norm{\vect{u}^{\perp}} \norm{\vect{v}}\cos(\vartheta), \]

where \(\vartheta\) is the angle between \(\vect{u}^{\perp}\) and \(\vect{v}\). Here \(h = \norm{\vect{v}}\cos(\vartheta)\) is \((\pm)\) the length of the projection of \(\vect{v}\) onto the line perpendicular to \(\vect{u}\), which can be interpreted as the height of the parallelogram. So then

\[\begin{split} \begin{array}{rcl} \norm{\vect{u}^{\perp}} \norm{\vect{v}}\cos(\vartheta) &=& \norm{\vect{u}}\times \norm{\vect{v}}\cos(\vartheta) \\ &=& \pm \text{(base length)} \times \text{height} \\ &=& \pm \text{area of } OACB. \end{array} \end{split}\]

After some rewriting

\[\begin{split} \begin{array}{ccl} ad - bc &=& \norm{\vect{u}^{\perp}}\norm{\vect{v}}\cos(\vartheta) \\ &=& \norm{\vect{u}}\norm{\vect{v}}\cos(\tfrac12\pi-\varphi)\\ &=& \norm{\vect{u}}\norm{\vect{v}}\sin(\varphi), \end{array} \end{split}\]

where \(\varphi\) is the angle from \(\vect{u}\) to the left (= counterclockwise) to \(\vect{v}\).

We see that \(ad-bc\) is equal to the area of the parallelogram if the angle from \(\vect{u}\) to \(\vect{v}\) is less then \(\pi\), and minus this area if the angle is between \(\pi\) and \(2\pi\).

Definition 5.1.1

The determinant of the ordered set \((\vect{u},\vect{v})\) of two vectors \(\vect{u} =\left[\begin{array}{c} a \\ b \end{array}\right] \) and \(\vect{v}=\left[\begin{array}{c} c \\ d \end{array}\right] \) in \(\R^2\) is defined as

\[ \det{(\vect{u},\vect{v})} = ad - bc. \]

Alternatively, the determinant can be seen as an operator working on \(2 \times 2\) matrices, coming with its own notation:

\[\begin{split} \det{\big([\, \vect{u} \,\, \vect{v}\, ]\big)} = \det{\left(\left[\begin{array}{cc} a & c \\ b & d \end{array}\right]\right)} = \left|\begin{array}{cc} a & c \\ b & d \end{array}\right|= ad-bc. \end{split}\]

Apart from the area, the determinant also says something about the relative position of the two vectors \(\vect{u}\) and \(\vect{v}\). In fact, we can use the determinant to define the orientation of two vectors in the plane (and, later, also of \(n\) vectors in \(\R^n\)).

Definition 5.1.2

The ordered set \((\vect{u},\vect{v})\) of two linearly independent vectors \(\vect{u}\) and \(\vect{v}\) is said to be positively oriented if \(\det{(\vect{u},\vect{v})} > 0\), and negatively oriented if \(\det{(\vect{u},\vect{v})} < 0\).

Proposition 5.1.3

Two by two determinants obey the following rules:

1. \(\det{(\vect{v},\vect{u})} = - \det{(\vect{u},\vect{v})}\).

2. \(\det{(\vect{u},\vect{v}+\vect{w})} = \det{(\vect{u},\vect{v})} + \det{(\vect{u},\vect{w})}\).

3. \(\det{(\vect{u},k\vect{v})} = k \det{(\vect{u},\vect{v})}\), \(k \in \R\).

4. \(\det{(\vect{e_1},\vect{e_2})} = 1\).

These properties can also be expressed using matrices.

Two by two determinants obey the following rules:

1. \( \begin{vmatrix} c & a \\ d & b \end{vmatrix} = - \begin{vmatrix} a & c \\ b & d \end{vmatrix}\).

2. \(\begin{vmatrix} a_{1} & b_{1}+ c_1\\ a_{2} & b_{2}+ c_2 \end{vmatrix} = \begin{vmatrix} a_{1} & b_{1}\\ a_{2} & b_{2}\end{vmatrix}+ \begin{vmatrix} a_{1} & c_1\\ a_{2} & c_2 \end{vmatrix}\).

3. \(\begin{vmatrix} a_{1} & k b_{1}\\ a_{2} & k b_{2}\end{vmatrix} = k \begin{vmatrix} a_{1} & b_{1}\\ a_{2} & b_{2}\end{vmatrix}\).

4. \(\begin{vmatrix} 1 & 0 \\ 0 & 1\end{vmatrix} = 1\).

These properties are verified by applying the definition

\[\begin{split} \left|\begin{array}{cc} a & c \\ b & d \end{array}\right| = ad-bc. \end{split}\]

Exercise 5.1.1

Verify the four properties of Proposition 5.1.3

The properties have a clear geometric interpretation using the notion of ‘signed area’. The following alternative proof uses this geometric viewpoint.

Proof. Three of the four properties are quickly settled.

1. Interchanging \(\vect{u}\) and \(\vect{v}\) does not alter the parallelogram. However, it changes the orientation of the two vectors, and thus the signed area changes sign.

3. \(\det{(\vect{u},k\vect{v})} = k\times\det{(\vect{u},\vect{v})}\), \(k \in \R\).

   Giving one of the vectors a factor \(k\) changes the area with a factor \(|k|\). If \(k > 0\), the orientation of the two vector does not change, so the determinant gets a factor \(|k|\), which in this case is equal to \(k\). If however \(k < 0\), then the orientation does change, so the determinant gets a factor \(-|k|\), which in this case is again equal to \(k\).

4. \(\det{(\vect{e_1},\vect{e_2})} \) is the area of the unit square.

The remaining property,

2. \(\det{(\vect{u},\vect{v}+\vect{w})} = \det{(\vect{u},\vect{v})} + \det{(\vect{u},\vect{w})}\),

is the most interesting one. The two pictures of Figure 5.1.4 tell the story.

Fig. 5.1.4 The sum rule in a picture. Note, this is a 2D picture.

In the picture on the left, both \((\vect{u},\vect{v})\) and \((\vect{u},\vect{w})\) are positively oriented. So there

\[\begin{split} \begin{array}{lcl} \det{(\vect{u},\vect{v})} + \det{(\vect{u},\vect{w})} &=& \text{area}(OABC) + \text{area}(CBDE) \\ &=& \text{area}(OADE) = \det{(\vect{u},\vect{v}+\vect{w})}, \end{array} \end{split}\]

since the two triangles \(OCE\) and \(ABD\) are congruent, so have equal areas.

In the picture on the right, the orientation of \((\vect{u},\vect{v})\) is positive, the orientation of \((\vect{u},\vect{w})\) is negative, and the orientation of \((\vect{u},\vect{v}+\vect{w})\) is positive again. So there

\[\begin{split} \begin{array}{lcl} \det{(\vect{u},\vect{v})} + \det{(\vect{u},\vect{w})} &=& \text{area}(OAFC) - \text{area}(OBDA) \\ &=& \text{area}(OAFC) - \text{area}(CGEF) \\ &=& \text{area}(OAEG) = \det{(\vect{u},\vect{v}+\vect{w})} \end{array} \end{split}\]

since now the areas of \(OAFC\) and \(CGEB\) add up to the area of \(OABC\), owing to the congruence of the triangles \(OGC\) and \(AEF\).

There are more pairwise orientations to consider, but the idea is hopefully clear to you.

5.1.3. \(\, 3 \times 3\) determinants: volume and orientation

Suppose \(\vect{a}, \vect{b}, \vect{c}\) are three vectors in \(\R^3\). For the moment, suppose they are linearly independent. So \(\vect{a}, \vect{b}\) are not multiples of each other, and \(\vect{c}\) is not in the plane spanned by \(\vect{a}, \vect{b}\). Then the three vectors can be interpreted as three edges of a parallelepiped. See Figure 5.1.5.

Fig. 5.1.5 Volume equals base area time height

Proof. Just as the area of a parallelogram can be computed as ‘base length times height’, the volume of a parallelepiped can be computed as ‘base area times height’. As base region we can take the parallelogram spanned by \(\vect{a}\) and \(\vect{b}\), and then the base area becomes

\[ A = \norm{\vect{a} \times \vect{b}}. \]

The height is found by projecting \(\vect{c}\) onto the line through the origin that is perpendicular to the plane spanned by \(\vect{a}\) and \(\vect{b}\). A direction vector of this line is precisely given by \(\vect{a} \times \vect{b}\). So

\[ h = \vect{c}\ip\vect{n} \]

where \(\vect{n}\) is the unit vector perpendicular to the ‘base plane’ that points to the same side as the vector \(\vect{c}\). This unit vector is given by

\[ \vect{n} = \pm \frac{\vect{a} \times \vect{b}}{\norm{\vect{a} \times \vect{b}}} \]

And then the formula ‘height times base area’ gives

\[ V = \vect{c}\ip\vect{n} = \pm \vect{c}\ip\frac{\vect{a} \times \vect{b}}{\norm{\vect{a} \times \vect{b}}} \cdot \norm{\vect{a} \times \vect{b}} = (\pm) \vect{c}\ip(\vect{a} \times \vect{b}). \]

The statement about the sign follows immediately from the third defining property of the cross product.

Note that the expression \(\vect{c}\ip(\vect{a} \times \vect{b})\) must be invariant under cyclic permutations of the three vectors.

Proposition 5.1.4

\[ \vect{c}\ip(\vect{a} \times \vect{b}) = \vect{a}\ip(\vect{b} \times \vect{c})= \vect{b}\ip(\vect{c} \times \vect{a}) \]

and swapping two vectors gives a minus sign:

\[ \vect{c}\ip(\vect{a} \times \vect{b})= - \vect{c}\ip(\vect{b} \times \vect{a}). \]

Proof. The parallelepiped spanned by the three vectors does not change under any permutation, and the orientation remains the same under a cyclic permutation.

The last proposition sets the way to take the determinant one dimension higher.

Definition 5.1.3

The determinant of the ordered set of three vectors \(\vect{a}, \vect{b}\) and \( \vect{c}\) in \(\R^3\) is defined by the expression

\[ \det{(\vect{a}, \vect{b},\vect{c})} = \vect{a}\ip(\vect{b} \times \vect{c}). \]

Note that the determinant is a real number.

Alternatively, we can define the determinant as a function working on \(3 \times 3\) matrices. And that is the sole interpretation we will use from now on.

If we put

\[\begin{split} A = [ \,\vect{a}\,\,\vect{b}\,\,\vect{c}\,] = \left[\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{array} \right] , \end{split}\]

then we define

\[\begin{split} \det{A} = \left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{array} \right|= \vect{a}\ip(\vect{b} \times \vect{c}). \end{split}\]

Proposition 5.1.5

(5.1.2)\[\begin{split}\left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{array} \right|= a_1\left|\begin{array}{cc} b_2 & c_2\\ b_3 & c_3 \end{array} \right|- a_2 \left|\begin{array}{cc} b_1 & c_1 \\ b_3 & c_3 \end{array} \right|+ a_3 \left|\begin{array}{cc} b_1 & c_1 \\ b_2 & c_2\end{array} \right|.\end{split}\]

The last expression can be further evaluated as

\[a_1b_2c_3 - a_1b_3c_2 - a_2b_1c_3 +a_2b_3c_1 + a_3b_1c_2 - a_3b_2c_1. \]

The expression involving the three \(2 \times 2\) determinants will be the stepping stone to define the determinant of a general \(n \times n\) matrix.

Proof. The identities are verified by evaluating the triple product:

\[\begin{split} \left[\begin{array}{c} a_1 \\ a_2\\ a_3 \end{array} \right] \ip \left( \left[\begin{array}{c} b_1 \\ b_2\\ b_3 \end{array} \right] \times \left[\begin{array}{c} c_1 \\ c_2\\ c_3 \end{array} \right] \right) = \left[\begin{array}{c} a_1 \\ a_2\\ a_3 \end{array} \right] \ip \left[\begin{array}{c} b_2c_3-b_3c_2 \\ b_3c_1-b_1c_3\\ b_1c_2-b_2c_1 \end{array} \right] = \end{split}\]

\[\begin{split} \begin{array}{cl} =& a_1(b_2c_3 - b_3c_2) - a_2(b_1c_3 -b_3c_1) + a_3(b_1c_2 - b_2c_1)\\ =& a_1b_2c_3 - a_1b_3c_2 - a_2b_1c_3 + a_2b_3c_1) + a_3b_1c_2 - a_3b_2c_1). \end{array} \end{split}\]

The next proposition summarizes the relevant properties of \(3 \times 3\) determinants.

Proposition 5.1.6

For the determinant

\[\begin{split} D = \det{A} = \det{\left[\,\vect{a}\,\, \vect{b}\,\,\vect{c}\, \right] } = \left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{array} \right| \end{split}\]

the following properties hold

1. \(|D|\), the absolute value of the determinant, is equal to the volume of the parallelepiped with the edges \(\vect{a},\vect{b}\) and \(\vect{c}\).

2. \(D=0\) if and only if the matrix \(A\) is singular.

   Equivalently,

   \(D\neq 0\) if and only if the matrix \(A\) is invertible, thus if
   the vectors \(\{\vect{a}, \vect{b},\vect{c}\}\) are linearly independent.

3. \(D > 0\) if and only if the ordered set \((\vect{a},\vect{b},\vect{c})\) is positively oriented. That is, oriented in the same way as the basis \((\vect{e}_1,\vect{e}_2,\vect{e}_3)\).

4. \(\det{I} = \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right| = 1\).

5.1.4. Grasple Exercises

Grasple Exercise 5.1.1

Area of paralellogram in the plane (with vertex at (0,0)).

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Grasple Exercise 5.1.2

Area of paralellogram in the plane (no vertex at (0,0)).

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Grasple Exercise 5.1.3

Area of triangle in the plane.

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Grasple Exercise 5.1.4

Volume of paralellepiped (with vertex at (0,0,0)).

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Grasple Exercise 5.1.5

Volume of paralellepiped (with vertex at (0,0,0)).

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Grasple Exercise 5.1.6

Considerations about cross product versus determinant.

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Grasple Exercise 5.1.7

Considerations about cross product/determinant/volume.

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Grasple Exercise 5.1.8

Finding a 3x3 determinant using orthogonality.

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Grasple Exercise 5.1.9

Using a determinant to check whether four points lie in the same plane.

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