An Inquiry-Based Introduction to Linear Algebra and Applications

Subsection Application: Area and Volume

Consider the problem of finding the area of a parallelogram determined by two vectors \(\vu\) and \(\vv\text{,}\) as illustrated at left in Figure 17.1.

We could calculate this area, for example, by breaking up the parallelogram into two triangles and a rectangle and finding the area of each. Now consider the problem of calculating the volume of the three-dimensional analog (called a parallelepiped) determined by three vectors \(\vu\text{,}\) \(\vv\text{,}\) and \(\vw\) as illustrated at right in Figure 17.1.

It is quite a bit more difficult to break this parallelepiped into subregions whose volumes are easy to compute. However, all of these computations can be made quickly by using determinants. The details are later in this section.

Subsection Introduction

We know that a non-zero vector \(\vx\) is an eigenvector of an \(n \times n\) matrix \(A\) if \(A \vx = \lambda \vx\) for some scalar \(\lambda\text{.}\) Note that this equation can be written as \((A-\lambda I_n)\vx=\vzero\text{.}\) Until now, we were given eigenvalues of matrices and have used the eigenvalues to find the eigenvectors. In this section we will learn an algebraic technique to find the eigenvalues ourselves. We will also be able to justify why an \(n\times n\) matrix has at most \(n\) eigenvalues.

A scalar \(\lambda\) is an eigenvalue of \(A\) if \((A - \lambda I_n)\vx=\vzero\) has a non-trivial solution \(\vx\text{,}\) which happens if and only if \(A-\lambda I_n\) is not invertible. In this section we will find a scalar whose value will tell us when a matrix is invertible and when it is not, and use this scalar to find the eigenvalues of a matrix.

Subsection The Determinant of a Square Matrix

Around 1900 or so determinants were deemed much more important than they are today. In fact, determinants were used even before matrices. According to Tucker³³ determinants (not matrices) developed out of the study of coefficients of systems of linear equations and were used by Leibniz 150 years before the term matrix was coined by J. J. Sylvester in 1848. Even though determinants are not as important as they once were, the determinant of a matrix is still a useful quantity. We saw in Preview Activity 17.1 that the determinant of a matrix tells us if the matrix is invertible and how it can help us find eigenvalues. In this section, we will see how to find the determinant of any size matrix and how to use this determinant to find the eigenvalues.

The determinant of a \(2 \times 2\) matrix \(A = \left[ \begin{array}{cc} a\amp b \\ c\amp d \end{array} \right]\) is \(\det(A)=ad-bc\text{.}\) The matrix \(A\) is invertible if and only if \(\det(A) \neq 0\text{.}\) We will use a recursive approach to find the determinants of larger size matrices building from the \(2\times 2\) determinants. We present the result in the \(3 \times 3\) case here — a more detailed analysis can be found at the end of this section.

To find the determinant of a \(3 \times 3\) matrix \(A = \left[ \begin{array}{ccc} a_{11}\amp a_{12}\amp a_{13} \\ a_{21}\amp a_{22}\amp a_{23}\\ a_{31}\amp a_{32}\amp a_{33} \end{array} \right]\text{,}\) we will use the determinants of three \(2\times 2\) matrices. More specifically, the determinant of \(A\text{,}\) denoted \(\det(A)\) is the quantity

This sum is called a cofactor expansion of the determinant of \(A\text{.}\) The smaller matrices in this expansion are obtained by deleting certain rows and columns of the matrix \(A\text{.}\) In general, when finding the determinant of an \(n\times n\) matrix, we find determinants of \((n-1)\times (n-1)\) matrices, which we can again reduce to smaller matrices to calculate.

We will use the specific matrix

as an example in illustrating the cofactor expansion method in general.

• We first pick a row or column of \(A\text{.}\) We will pick the first row of \(A\) for this example.

• For each entry in the row (or column) we choose, in this case the first row, we will calculate the determinant of a smaller matrix obtained by removing the row and the column the entry is in. Let \(A_{ij}\) be the smaller matrix found by deleting the \(i\)th row and \(j\)th column of \(A\text{.}\) For entry \(a_{11}\text{,}\) we find the matrix \(A_{11}\) obtained by removing first row and first column:

  \begin{equation*} A_{11} = \left[ \begin{array}{cc} 4 \amp 3 \\ 2 \amp 1 \end{array} \right]\,\text{.} \end{equation*}

  For entry \(a_{12}\text{,}\) we find

  \begin{equation*} A_{12} = \left[ \begin{array}{cc} 1 \amp 3 \\ 2 \amp 1 \end{array} \right]\,\text{.} \end{equation*}

  Finally, for entry \(a_{13}\text{,}\) we find

  \begin{equation*} A_{13} = \left[ \begin{array}{cc} 1 \amp 4 \\ 2 \amp 2 \end{array} \right] \,\text{.} \end{equation*}

• Notice that in the \(3 \times 3\) determinant formula in (17.1) above, the middle term had a (-) sign. The signs of the terms in the cofactor expansion alternate within each row and each column. More specifically, the sign of a term in the \(i\)th row and \(j\)th column is \((-1)^{i+j}\text{.}\) We then obtain the following pattern of the signs within each row and column:

  \begin{equation*} \left[ \begin{array}{cccc} + \amp - \amp + \amp \cdots \\ - \amp + \amp - \amp \cdots \\ + \amp - \amp + \amp \cdots \\ \vdots \amp \amp \amp \end{array} \right] \end{equation*}

  In particular, the sign factor for \(a_{11}\) is \((-1)^{1+1}=1\text{,}\) for \(a_{12}\) is \((-1)^{1+2}=-1\text{,}\) and for \(a_{13}\) is \((-1)^{1+3}=1\text{.}\)

• For each entry \(a_{ij}\) in the row (or column) of \(A\) we chose, we multiply the entry \(a_{ij}\) by the determinant of \(A_{ij}\) and the sign \((-1)^{i+j}\text{.}\) In this case, we obtain the following numbers

  \begin{equation*} a_{11} (-1)^{1+1} \det(A_{11}) = 1 \det \left[ \begin{array}{cc} 4 \amp 3 \\ 2 \amp 1 \end{array} \right] = 1(4-6)=-2 \end{equation*}
  \begin{equation*} a_{12} (-1)^{1+2} \det(A_{12}) = -2 \det \left[ \begin{array}{cc} 1 \amp 3 \\ 2 \amp 1 \end{array} \right] = -2(1-6)=10 \end{equation*}
  \begin{equation*} a_{13} (-1)^{1+3} \det(A_{13}) = 0 \end{equation*}

  Note that in the last calculation, since \(a_{13}=0\text{,}\) we did not have to evaluate the rest of the terms.

• Finally, we find the determinant by adding all these values:

  \begin{align*} \det(A) \amp = a_{11} (-1)^{1+1} \det(A_{11}) + a_{12} (-1)^{1+2} \det(A_{12})\\ \amp \qquad + a_{13} (-1)^{1+3} \det(A_{13})\\ \amp = 8\text{.} \end{align*}

Subsection Cofactors

We will now define the determinant of a general \(n \times n\) matrix \(A\) in terms of a cofactor expansion as we did in the \(3 \times 3\) case. To do so, we need some notation and terminology.

• We let \(A_{ij}\) be the submatrix of \(A = [a_{ij}]\) found by deleting the \(i\)th row and \(j\)th column of \(A\text{.}\) The determinant of \(A_{ij}\) is called the \(ij\)th minor of \(A\) or the minor corresponding to the entry \(a_{ij}\text{.}\)

• Notice that in the \(3 \times 3\) case, we used the opposite of the 1,2 minor in the sum. It will be the case that the terms in the cofactor expansion will alternate in sign. We can make the signs in the sum alternate by taking \(-1\) to an appropriate power. As a result, we define the \(ij\)th cofactor \(C_{ij}\) of \(A\) as

  \begin{equation*} C_{ij} = (-1)^{i+j} \det\left(A_{ij}\right)\text{.} \end{equation*}

• Finally, we define the determinant of \(A\text{.}\)

This method for computing determinants is called the cofactor expansion or Laplace expansion of \(A\) along the 1st row. The cofactor expansion reduces the computation of the determinant of an \(n \times n\) matrix to \(n\) computations of determinants of \((n-1) \times (n-1)\) matrices. These smaller matrices can be reduced again using cofactor expansions, so it can be a long and grueling process for large matrices. It turns out that we can actually take this expansion along any row or column of the matrix (a proof of this fact is given in Section 22). For example, the cofactor expansion along the 2nd row is

and along the 3rd column the formula is

Note that when finding a cofactor expansion, choosing a row or column with many zeros makes calculations easier.

As with any new idea, like the determinant, we must ask what properties are satisfied. We state the following theorem without proof for the time being. For the interested reader, the proof of many of these properties is given in Section 22 and others in the exercises.

Note that if we were to find the determinant of a \(4\times 4\) matrix using the cofactor method, we will calculate determinants of 4 matrices of size \(3\times 3\text{,}\) each of which will require 3 determinant calculations again. So, we will need a total of 12 calculations of determinants of \(2\times 2\) matrices. That is a lot of calculations. There are other, more efficient, methods for calculating determinants. For example, we can row reduce the matrix, keeping track of the effect that each row operation has on the determinant.

Subsection The Determinant of a \(3 \times 3\) Matrix

Earlier we defined the determinant of a \(3 \times 3\) matrix. In this section we endeavor to understand the motivation behind that definition.

We will repeat the process we went through in the \(2 \times 2\) case to see how to define the determinant of a \(3 \times 3\) matrix. Let

To find the inverse of \(A\) we augment \(A\) by the \(3 \times 3\) identity matrix

and row reduce the matrix (using appropriate technology) to obtain

where

In this case, we can see that the inverse of the \(3 \times 3\) matrix \(A\) will be defined if and only if \(d \neq 0\text{.}\) So, in the \(3 \times 3\) case the determinant of \(A\) will be given by the value of \(d\) in Equation (17.2). What remains is for us to see how this is related to determinants of \(2 \times 2\) sub-matrices of \(A\text{.}\)

To start, we collect all terms involving \(a_{11}\) in \(d\text{.}\) A little algebra shows that

Now let's collect the remaining terms involving \(a_{12}\text{:}\)

Finally, we collect the terms involving \(a_{13}\text{:}\)

Now we can connect the determinant of \(A\) to determinants of \(2 \times 2\) sub-matrices of \(A\text{.}\)

• Notice that

  \begin{equation*} a_{33} a_{22} - a_{32} a_{23} \end{equation*}

  is the determinant of the \(2 \times 2\) matrix \(\left[ \begin{array}{cc} a_{22}\amp a_{23} \\ a_{32}\amp a_{33} \end{array} \right]\) obtained from \(A\) by deleting the first row and first column.

• Similarly, the expression

  \begin{equation*} a_{33} a_{21} - a_{31} a_{23} \end{equation*}

  is the determinant of the \(2 \times 2\) matrix \(\left[ \begin{array}{cc} a_{21}\amp a_{23} \\ a_{31}\amp a_{33} \end{array} \right]\) obtained from \(A\) by deleting the first row and second column.

• Finally, the expression

  \begin{equation*} a_{32} a_{21} - a_{31} a_{22} \end{equation*}

  is the determinant of the \(2 \times 2\) matrix \(\left[ \begin{array}{cc} a_{21}\amp a_{22} \\ a_{31}\amp a_{32} \end{array} \right]\) obtained from \(A\) by deleting the first row and third column.

Putting this all together gives us formula (17.1) for the determinant of a \(3 \times 3\) matrix as we defined earlier.

Subsection Two Devices for Remembering Determinants

There are useful ways to remember how to calculate the formulas for determinants of \(2 \times 2\) and \(3 \times 3\) matrices. In the \(2 \times 2\) case of \(A = \left[ \begin{array}{cc} a_{11}\amp a_{12} \\ a_{21}\amp a_{22} \end{array} \right]\text{,}\) we saw that

This makes \(|A|\) the product of the diagonal elements \(a_{11}\) and \(a_{22}\) minus the product of the off-diagonal elements \(a_{12}\) and \(a_{21}\text{.}\) We can visualize this in an array by drawing arrows across the diagonal and off-diagonal, with a plus sign on the diagonal arrow indicting that we add the product of the diagonal elements and a minus sign on the off-diagonal arrow indicating that we subtract the product of the off-diagonal elements as shown in Figure 17.4.

We can do a similar thing for the determinant of a \(3 \times 3\) matrix. In this case, we extend the \(3 \times 3\) array to a \(3 \times 5\) array by adjoining the first two columns onto the matrix. We then add the products along the diagonals going from left to right and subtract the products along the diagonals going from right to left as indicated in Figure 17.5.

Subsection Examples

What follows are worked examples that use the concepts from this section.

Subsection Summary

• The determinant of an \(n \times n\) matrix \(A = [a_{ij}]\) is found by taking the cofactor expansion of \(A\) along the first row. That is

  \begin{equation*} \det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} + \cdots + a_{1n}C_{1n}\text{,} \end{equation*}

  where

  • \(A_{ij}\) is the sub-matrix of \(A\) found by deleting the \(i\)th row and \(j\)th column of \(A\text{.}\)

  • \(C_{ij} = (-1)^{i+j} \det\left(A_{ij}\right)\) is the \(ij\)th cofactor of \(A\text{.}\)

• The matrix \(A\) is invertible if and only if \(\det(A) \neq 0\text{.}\)

Subsection Project: Area and Volume Using Determinants

The approach we will take to connecting area (volume) to the determinant will help shed light on properties of the determinant that we will discuss from an algebraic perspective in a later section. First, we mention some basic properties of area (we focus on area for now, but these same properties are valid for volumes as well). volume). As a shorthand, we denote the area of a region \(R\) by \(\Area(R)\text{.}\)

• Area cannot be negative.

• If two regions \(R_1\) and \(R_2\) don't overlap, then the area of the union of the regions is equal to the sum of the areas of the regions. That is, if \(R_1 \cap R_2 = \emptyset\text{,}\) then \(\Area(R_1 \cup R_2) = \Area(R_1) + \Area(R_2)\text{.}\)

• Area is invariant under translation. That is, if we move a geometric region by the same amount uniformly in a given direction, the area of the original region and the area of the transformed region are the same. A translation of a region is done by just adding a fixed vector to each vector in the region. That is, a translation by a vector \(\vv\) is a function \(T_{\vv}\) such that the image \(T_{\vv}(R)\) of a region \(R\) is defined as

  \begin{equation*} T_{\vv}(R) = \{\vr+\vv : \vr \in R\}\text{.} \end{equation*}

  Since area is translation invariant, \(\Area(T_{\vv}(R)) = \Area(R)\text{.}\)

• The area of a one-dimensional object like a line segment is \(0\text{.}\)

Now we turn our attention to areas of parallelograms. Let \(\vu\) and \(\vv\) be vectors in \(\R^2\text{.}\) The parallelogram \(P(\vu,\vv)\) defined by \(\vu\) and \(\vv\) with point \(Q\) as basepoint is the set

An illustration of such a parallelogram is shown at left in Figure 17.8.

If \(\vu = [u_1 \ u_2]^{\tr}\) and \(\vv = [v_1 \ v_2]^{\tr}\text{,}\) then we will also represent \(P(\vu,\vv)\) as \(P\left( \left[ \begin{array}{cc}u_1\amp u_2 \\ v_1\amp v_2 \end{array} \right] \right)\text{.}\)

Since area is translation and rotation invariant, we can translate our parallelogram by \(-\overrightarrow{OQ}\) to place its basepoint at the origin, then rotate by an angle \(\theta\) (as shown at left in Figure 17.8. This transforms the vector \(\vv\) to a vector \(\vv'\) and the vector \(\vu\) to a vector \(\vu'\) as shown at right in Figure 17.8. With this in mind we can always assume that our parallelograms have one vertex at the origin, with \(\vu\) along the \(x\)-axis, and \(\vv\) in standard position. Now we can investigate how to calculate the area of a parallelogram.

Project Activity 17.3.

There are two situations to consider when we want to find the area of a parallelogram determined by vectors \(\vu\) and \(\vv\text{,}\) both shown in Figure 17.9. The parallelogram will be determined by the lengths of these vectors.

(a)

In the situation depicted at left in Figure 17.9, use geometry to explain why \(\Area(P(\vu,\vv)) = h |\vu|\text{.}\)

Hint.

What can we say about the triangles \(ODB\) and \(EAC\text{?}\)

(b)

In the situation depicted at right in Figure 17.9, use geometry to again explain why \(\Area(P(\vu,\vv)) = h |\vu|\text{.}\) (Hint: What can we say about \(\Area(AEC)\) and \(\Area(ODB)\text{?}\))

The result of Project Activity 17.3 is that the area of \(P(\vu,\vv)\) is given by \(h |\vu|\text{,}\) where \(h\) is the height of the parallelogram determined by dropping a perpendicular from the terminal point of \(\vv\) to the line determined by the vector \(\vu\text{.}\)

Now we turn to the question of how the determinant is related to area of a parallelogram. Our approach will use some properties of the area of \(P(\vu, \vv)\text{.}\)

Project Activity 17.4.

Let \(\vu\) and \(\vv\) be vectors that determine a parallelogram in \(\R^2\text{.}\)

(a)

Explain why

\begin{equation} \Area(P(\vu,\vv)) = \Area(P(\vv,\vu))\tag{17.3} \end{equation}

(b)

If \(k\) is any scalar, then \(k\vu\) either stretches or compresses \(\vu\text{.}\) Use this idea, and the result of Project Activity 17.3, to explain why

\begin{equation} \Area(P(k\vu,\vv)) = \Area(P(\vu,k\vv)) = |k| \Area(P(\vu,\vv))\tag{17.4} \end{equation}

for any real number \(k\text{.}\) A representative picture of this situation is shown at left in Figure 17.9 for a value of \(k > 1\text{.}\) You will also need to consider what happens when \(k \lt 0\text{.}\)

(c)

Finally, use the result of Project Activity 17.3 to explain why

\begin{equation} \Area(P(\vu+k\vv,\vv)) = \Area(P(\vu,\vv+k\vu)) = \Area(P(\vu,\vv))\tag{17.5} \end{equation}

for any real number \(k\text{.}\) A representative picture is shown at right in Figure 17.10.

Properties (17.4) and (17.5) will allow us to calculate the area of the parallelogram determined by vectors \(\vu\) and \(\vv\text{.}\)

We can apply the same arguments as above using rotations, translations, shearings, and scalings to show that the properties of area given above work in any dimension. Given vectors \(\vu_1\text{,}\) \(\vu_2\text{,}\) \(\ldots\text{,}\) \(\vu_n\) in \(\R^n\text{,}\) we let

If \(n = 2\text{,}\) then \(P(\vu_1,\vu_2)\) is the parallelogram determined by \(\vu_1\) and \(\vu_2\) with basepoint \(Q\text{.}\) If \(n = 3\text{,}\) then \(P(\vu_1, \vu_2, \vu_3)\) is the parallelepiped with basepoint \(Q\) determined by \(\vu_1\text{,}\) \(\vu_2\text{,}\) and \(\vu_3\text{.}\) In higher dimensions the sets \(P(\vu_1, \vu_2, \ldots,\vu_n)\) are called parallelotopes, and we use the notation \(\Vol(P(\vu_1, \vu_2, \ldots,\vu_n))\) for their volume. The \(n\)-dimensional volumes of these paralleotopes satisfy the following properties:

for any \(i\) and \(j\text{.}\)

for any real number \(k\) and any \(i\text{.}\)

for any real number \(k\) and any distinct \(i\) and \(j\text{.}\)

Properties (17.6), (17.7), and (17.8) involve the analogs of row operations on matrices, and we will prove algebraically that the determinant exhibits the same properties. In fact, the determinant can be uniquely defined by these properties. So in a sense, the determinant is an area or volume function.