The Determinant

Section 17 The Determinant

Focus Questions

By the end of this section, you should be able to give precise and thorough answers to the questions listed below. You may want to keep these questions in mind to focus your thoughts as you complete the section.

How do we calculate the determinant of an $n \times n$ matrix?
What is one important fact the determinant tells us about a matrix?

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Subsection Application: Area and Volume

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Consider the problem of finding the area of a parallelogram determined by two vectors

u

and

v,

as illustrated at left in Figure 17.1.

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Figure 17.1. A parallelogram and a parallelepiped.

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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

u,

v,

and

w

as illustrated at right in Figure 17.1.

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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.

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Subsection Introduction

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We know that a non-zero vector

x

is an eigenvector of an

n \times n

matrix

A

A x = λ x

for some scalar

λ .

Note that this equation can be written as

(A - λ I_{n}) x = 0 .

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.

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A scalar

λ

is an eigenvalue of

A

(A - λ I_{n}) x = 0

has a non-trivial solution

x,

which happens if and only if

A - λ 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.

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Preview Activity 17.1.

In this activity, we will focus on $2 \times 2$ matrices. Let $A = [\begin{array}{cc} a & b \\ c & d \end{array}]$ be a $2 \times 2$ matrix. To see if $A$ is invertible, we row reduce $A$ by replacing row 2 with $a \cdot$ (row 2) $- c \cdot$ (row 1):

[\begin{array}{cc} a & b \\ 0 & a d - b c \end{array}] .

So the only way $A$ can be reduced $I_{2}$ is if $a d - b c \neq 0 .$ We call this quantity $a d - b c$ the determinant of $A,$ and denote the determinant of $A$ as $det (A)$ or $| A | .$ When $det (A) \neq 0,$ we know that

A^{- 1} = \frac{1}{a d - b c} [\begin{array}{rr} d & - b \\ - c & a \end{array}] .

We now consider how we can use the determinant to find eigenvalues and other information about the invertibility of a matrix.

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(a)

Let $A = [\begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array}] .$ Find $det (A)$ by hand. What does this mean about the matrix $A ?$ Can you confirm this with other methods?

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(b)

One of the eigenvalues of $A = [\begin{array}{cc} 1 & 3 \\ 2 & 2 \end{array}]$ is $λ = 4 .$ Recall that we can rewrite the matrix equation $A x = 4 x$ in the form $(A - 4 I_{2}) x = 0 .$ What must be true about $A - 4 I_{2}$ in order for 4 to be an eigenvalue of $A ?$ How does this relate to $det (A - 4 I_{2}) ?$

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(c)

Another eigenvalue of $A = [\begin{array}{cc} 1 & 3 \\ 2 & 2 \end{array}]$ is $λ = - 1 .$ What must be true about $A + I_{2}$ in order for $- 1$ to be an eigenvalue of $A ?$ How does this relate to $det (A + I_{2}) ?$

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(d)

To find the eigenvalues of the matrix $A = [\begin{array}{cc} 3 & 2 \\ 2 & 6 \end{array}],$ we rewrite the equation $A x = λ x$ as $(A - λ I_{2}) x = 0 .$ The coefficient matrix of this last system has the form $A - λ I_{2} = [\begin{array}{cc} 3 - λ & 2 \\ 2 & 6 - λ \end{array}] .$ The determinant of this matrix is a quadratic expression in $λ .$ Since the eigenvalues will occur when the determinant is 0, we need to solve a quadratic equation. Find the resulting eigenvalues. (Note: One of the eigenvalues is 2.)

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(e)

Can you explain why a $2 \times 2$ matrix can have at most two eigenvalues?

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Subsection The Determinant of a Square Matrix

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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.

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The determinant of a

2 \times 2

matrix

A = [\begin{array}{cc} a & b \\ c & d \end{array}]

det (A) = a d - b c .

The matrix

A

is invertible if and only if

det (A) \neq 0 .

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.

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To find the determinant of a

3 \times 3

matrix

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

we will use the determinants of three

2 \times 2

matrices. More specifically, the determinant of

A,

denoted

det (A)

is the quantity

\begin{matrix} (17.1) & a_{11} det ([\begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}]) - a_{12} det ([\begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array}]) + a_{13} det ([\begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}]) . \end{matrix}

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This sum is called a cofactor expansion of the determinant of

A .

The smaller matrices in this expansion are obtained by deleting certain rows and columns of the matrix

A .

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.

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We will use the specific matrix

A = [\begin{array}{ccc} 1 & 2 & 0 \\ 1 & 4 & 3 \\ 2 & 2 & 1 \end{array}]

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

We first pick a row or column of $A .$ 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 $A_{i j}$ be the smaller matrix found by deleting the $i$ th row and $j$ th column of $A .$ For entry $a_{11},$ we find the matrix $A_{11}$ obtained by removing first row and first column:

$A_{11} = [\begin{array}{cc} 4 & 3 \\ 2 & 1 \end{array}] .$

For entry $a_{12},$ we find

$A_{12} = [\begin{array}{cc} 1 & 3 \\ 2 & 1 \end{array}] .$

Finally, for entry $a_{13},$ we find

$A_{13} = [\begin{array}{cc} 1 & 4 \\ 2 & 2 \end{array}] .$
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 $(- 1)^{i + j} .$ We then obtain the following pattern of the signs within each row and column:

$[\begin{array}{cccc} + & - & + & \dots \\ - & + & - & \dots \\ + & - & + & \dots \\ ⋮ \end{array}]$

In particular, the sign factor for $a_{11}$ is $(- 1)^{1 + 1} = 1,$ for $a_{12}$ is $(- 1)^{1 + 2} = - 1,$ and for $a_{13}$ is $(- 1)^{1 + 3} = 1 .$
For each entry $a_{i j}$ in the row (or column) of $A$ we chose, we multiply the entry $a_{i j}$ by the determinant of $A_{i j}$ and the sign $(- 1)^{i + j} .$ In this case, we obtain the following numbers

$a_{11} (- 1)^{1 + 1} det (A_{11}) = 1 det [\begin{array}{cc} 4 & 3 \\ 2 & 1 \end{array}] = 1 (4 - 6) = - 2$

$a_{12} (- 1)^{1 + 2} det (A_{12}) = - 2 det [\begin{array}{cc} 1 & 3 \\ 2 & 1 \end{array}] = - 2 (1 - 6) = 10$

$a_{13} (- 1)^{1 + 3} det (A_{13}) = 0$

Note that in the last calculation, since $a_{13} = 0,$ we did not have to evaluate the rest of the terms.
Finally, we find the determinant by adding all these values:

$\begin{aligned} det (A) & = a_{11} (- 1)^{1 + 1} det (A_{11}) + a_{12} (- 1)^{1 + 2} det (A_{12}) \\ + a_{13} (- 1)^{1 + 3} det (A_{13}) \\ = 8 . \end{aligned}$

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Subsection Cofactors

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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 $A_{i j}$ be the submatrix of $A = [a_{i j}]$ found by deleting the $i$ th row and $j$ th column of $A .$ The determinant of $A_{i j}$ is called the $i j$ th minor of $A$ or the minor corresponding to the entry $a_{i j} .$
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 $i j$ th cofactor $C_{i j}$ of $A$ as

$C_{i j} = (- 1)^{i + j} det (A_{i j}) .$
Finally, we define the determinant of $A .$

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Definition 17.2.

If $A = [a_{i j}]$ is an $n \times n$ matrix, the determinant of $A$ is the scalar

det (A) = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13} + \dots + a_{1 n} C_{1 n}

where $C_{i j} = (- 1)^{i + j} det (A_{i j})$ is the $i j$ -cofactor of $A$ and $A_{i j}$ is the matrix obtained by removing row $i$ and column $j$ of matrix $A .$

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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

det (A) = a_{21} C_{21} + a_{22} C_{22} + \dots + a_{2 n} C_{2 n}

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and along the 3rd column the formula is

det (A) = a_{13} C_{13} + a_{23} C_{23} + \dots + a_{n 3} C_{n 3} .

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

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Activity 17.2.

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(a)

Let $A = [\begin{array}{rcr} 1 & 2 & - 1 \\ - 2 & 0 & 4 \\ 6 & 3 & 0 \end{array}] .$ Use the cofactor expansion along the first row to calculate the determinant of $A$ by hand.

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(b)

Calculate $det (A)$ by using a cofactor expansion along the second row where $A = [\begin{array}{ccc} 1 & 4 & 2 \\ 0 & 2 & 0 \\ 2 & 5 & 3 \end{array}] .$

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(c)

Calculate the determinant of $[\begin{array}{crr} 1 & - 2 & 3 \\ 0 & 4 & - 3 \\ 0 & 0 & 8 \end{array}] .$

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(d)

Which determinant property can be used to calculate the determinant in part (c)? Explain how. (Determinant properties are included below for easy reference.)

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(e)

Consider the matrix $A = [\begin{array}{ccc} 1 & 1 & 2 \\ 0 & 2 & 1 \\ 1 & 2 & 2 \end{array}] .$ Let $B$ be the matrix which results when $c$ times row 1 is added to row 2 of $A .$ Evaluate the determinant of $B$ by hand to check that it is equal to the determinant of $A,$ which verifies one other determinant property (in a specific case).

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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.

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Theorem 17.3.

Given $n \times n$ matrices $A, B,$ the following hold:

$det (A B) = det (A) \cdot det (B),$ and in particular $det (A^{k}) = (det A)^{k}$ for any positive integer $k .$
$det (A^{T}) = det (A) .$
$A$ is invertible if and only if $det (A) \neq 0 .$
If $A$ is invertible, then $det (A^{- 1}) = (det A)^{- 1} .$
For a $2 \times 2$ matrix $A = [\begin{matrix} a & b \\ c & d \end{matrix}],$ $det (A) = a d - b c .$
If $A$ is upper/lower triangular, then $det (A)$ is the product of the entries on the diagonal.
The determinant of a matrix is the product of the eigenvalues, with each eigenvalue repeated as many times as its multiplicity.
Effect of row operations:
- Adding a multiple of a row to another does NOT change the determinant of the matrix.
- Multiplying a row by a constant multiplies the determinant by the same constant.
- Row swapping multiplies the determinant by $(- 1) .$
If the row echelon form $U$ of $A$ is obtained by adding multiples of one row to another, and row swapping, then $det (A)$ is equal to $det (U)$ multiplied by $(- 1)^{r}$ where $r$ is the number of row swappings done during the row reduction.

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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,

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.

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Subsection The Determinant of a $3 \times 3$ Matrix

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Earlier we defined the determinant of a

3 \times 3

matrix. In this section we endeavor to understand the motivation behind that definition.

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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

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

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To find the inverse of

A

we augment

A

by the

3 \times 3

identity matrix

[A | I_{3}] = [\begin{array}{cccccc} a_{11} & a_{12} & a_{13} & 1 & 0 & 0 \\ a_{21} & a_{22} & a_{23} & 0 & 1 & 0 \\ a_{31} & a_{32} & a_{33} & 0 & 0 & 1 \end{array}]

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

[\begin{array}{rrrrrr} 1 & 0 & 0 & \frac{a_{33} a_{22} - a_{32} a_{23}}{d} & - \frac{a_{33} a_{12} - a_{32} a_{13}}{d} & \frac{- a_{13} a_{22} + a_{12} a_{23}}{d} \\ 0 & 1 & 0 & - \frac{a_{33} a_{21} - a_{31} a_{23}}{d} & \frac{a_{33} a_{11} - a_{31} a_{13}}{d} & - \frac{a_{23} a_{11} - a_{21} a_{13}}{d} \\ 0 & 0 & 1 & \frac{- a_{31} a_{22} + a_{32} a_{21}}{d} & - \frac{a_{32} a_{11} - a_{31} a_{12}}{d} & \frac{a_{22} a_{11} - a_{21} a_{12}}{d} \end{array}],

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where

\begin{matrix} (17.2) & \begin{aligned} d = a_{33} a_{11} a_{22} & - a_{33} a_{21} a_{12} - a_{31} a_{13} a_{22} \\ - a_{32} a_{11} a_{23} + a_{32} a_{21} a_{13} + a_{31} a_{12} a_{23} . \end{aligned} \end{matrix}

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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 .

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 .

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To start, we collect all terms involving

a_{11}

d .

A little algebra shows that

det (A) = a_{11} (a_{33} a_{22} - a_{32} a_{23}) - a_{33} a_{21} a_{12} - a_{31} a_{13} a_{22} + a_{32} a_{21} a_{13} + a_{31} a_{12} a_{23} .

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Now let's collect the remaining terms involving

a_{12} :

det (A) = a_{11} (a_{33} a_{22} - a_{32} a_{23}) - a_{12} (a_{33} a_{21} - a_{31} a_{23}) - a_{31} a_{13} a_{22} + a_{32} a_{21} a_{13} .

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Finally, we collect the terms involving

a_{13} :

det (A) = a_{11} (a_{33} a_{22} - a_{32} a_{23}) - a_{12} (a_{33} a_{21} - a_{31} a_{23}) + a_{13} (a_{32} a_{21} - a_{31} a_{22}) .

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Now we can connect the determinant of

A

to determinants of

2 \times 2

sub-matrices of

A .

Notice that

$a_{33} a_{22} - a_{32} a_{23}$

is the determinant of the $2 \times 2$ matrix $[\begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}]$ obtained from $A$ by deleting the first row and first column.
Similarly, the expression

$a_{33} a_{21} - a_{31} a_{23}$

is the determinant of the $2 \times 2$ matrix $[\begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array}]$ obtained from $A$ by deleting the first row and second column.
Finally, the expression

$a_{32} a_{21} - a_{31} a_{22}$

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

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Putting this all together gives us formula (17.1) for the determinant of a

3 \times 3

matrix as we defined earlier.

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Subsection Two Devices for Remembering Determinants

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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 = [\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}],

we saw that

| A | = a_{11} a_{22} - a_{21} a_{22} .

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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} .

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.

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Figure 17.4. A diagram to remember the $2 \times 2$ determinant.

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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.

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Figure 17.5. A diagram to remember the $3 \times 3$ determinant.

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Subsection Examples

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

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Example 17.6.

For each of the following

Identify the sub-matrices $A_{1, j}$
Determine the cofactors $C_{1, j} .$
Use the cofactor expansion to calculate the determinant.

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(a)

$A = [\begin{array}{ccr} 3 & 6 & 2 \\ 0 & 4 & - 1 \\ 5 & 0 & 1 \end{array}]$

Solution.

With a $3 \times 3$ matrix, we will find the sub-matrices $A_{11},$ $A_{12},$ and $A_{13} .$ Recall that $A_{i j}$ is the sub-matrix of $A$ obtained by deleting the $i$ th row and $j$ th column of $A .$ Thus,

A_{11} = [\begin{array}{cr} 4 & - 1 \\ 0 & 1 \end{array}] A_{12} = [\begin{array}{cr} 0 & - 1 \\ 5 & 1 \end{array}] and A_{13} = [\begin{array}{cc} 0 & 4 \\ 5 & 0 \end{array}] .

The $i j$ th cofactor is $C_{i j} = (- 1)^{i + j} det (A_{i j}),$ so

\begin{aligned} C_{11} & = (- 1)^{2} [\begin{array}{cr} 4 & - 1 \\ 0 & 1 \end{array}] = 4 \\ C_{12} & = (- 1)^{3} [\begin{array}{cr} 0 & - 1 \\ 5 & 1 \end{array}] = - 5 \\ C_{13} & = (- 1)^{4} [\begin{array}{cc} 0 & 4 \\ 5 & 0 \end{array}] = - 20 . \end{aligned}

Then

det (A) = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13} = (3) (4) + (6) (- 5) + (2) (- 20) = - 58 .

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(b)

$A = [\begin{array}{rrcr} 3 & 0 & 1 & 1 \\ 2 & 1 & 2 & 1 \\ 1 & - 2 & 2 & - 1 \\ - 3 & 2 & 3 & 1 \end{array}]$

Solution.

With a $4 \times 4$ matrix, we will find the sub-matrices $A_{11},$ $A_{12},$ $A_{13},$ and $A_{14} .$ We see that

\begin{aligned} A_{11} & = [\begin{array}{rcr} 1 & 2 & 1 \\ - 2 & 2 & - 1 \\ 2 & 3 & 1 \end{array}] \\ A_{12} & = [\begin{array}{rcr} 2 & 2 & 1 \\ 1 & 2 & - 1 \\ - 3 & 3 & 1 \end{array}] \\ A_{13} & = [\begin{array}{rrr} 2 & 1 & 1 \\ 1 & - 2 & - 1 \\ - 3 & 2 & 1 \end{array}] \\ A_{14} & = [\begin{array}{rrc} 2 & 1 & 2 \\ 1 & - 2 & 2 \\ - 3 & 2 & 3 \end{array}] . \end{aligned}

To calculate the $i j$ th cofactor $C_{i j} = (- 1)^{i + j} det (A_{i j}),$ we need to calculate the determinants of the $A_{1 j} .$ Using the device for calculating the determinant of a $3 \times 3$ matrix we have that

\begin{aligned} det (A_{11}) & = det ([\begin{array}{rrr} 1 & 2 & 1 \\ - 2 & 2 & - 1 \\ 2 & 3 & 1 \end{array}]) \\ = (1) (2) (1) + (2) (- 1) (2) + (1) (- 2) (3) \\ - (1) (2) (2) - (1) (- 1) (3) - (2) (- 2) (1) \\ = - 5, \end{aligned}

\begin{aligned} det (A_{12}) & = det ([\begin{array}{rcr} 2 & 2 & 1 \\ 1 & 2 & - 1 \\ - 3 & 3 & 1 \end{array}]) \\ = (2) (2) (1) + (2) (- 1) (- 3) + (1) (1) (3) \\ - (1) (2) (- 3) - (2) (- 1) (3) - (2) (1) (1) \\ = 23, \end{aligned}

\begin{aligned} det (A_{13}) & = det ([\begin{array}{rrr} 2 & 1 & 1 \\ 1 & - 2 & - 1 \\ - 3 & 2 & 1 \end{array}]) \\ = (2) (- 2) (1) + (1) (- 1) (- 3) + (1) (1) (2) \\ - (1) (- 2) (- 3) - (2) (- 1) (2) - (1) (1) (1) \\ = - 2, \end{aligned}

and

\begin{aligned} det (A_{14}) & = det ([\begin{array}{rrc} 2 & 1 & 2 \\ 1 & - 2 & 2 \\ - 3 & 2 & 3 \end{array}]) \\ = (2) (- 2) (3) + (1) (2) (- 3) + (2) (1) (2) \\ - (2) (- 2) (- 3) - (2) (2) (2) - (1) (1) (3) \\ = - 37 . \end{aligned}

Then

\begin{aligned} C_{11} & = (- 1)^{2} det (A_{11}) = - 5 \\ C_{12} & = (- 1)^{3} det (A_{12}) = - 23 \\ C_{13} & = (- 1)^{4} det (A_{13}) = - 2 \\ C_{14} & = (- 1)^{5} det (A_{13}) = 37 \end{aligned}

and so

\begin{aligned} det (B) & = b_{11} C_{11} + b_{12} C_{12} + b_{13} C_{13} + b_{14} C_{14} \\ = (3) (- 5) + (0) (- 23) + (1) (- 2) + (1) (37) \\ = 20 . \end{aligned}

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Example 17.7.

Show that for any $2 \times 2$ matrices $A$ and $B,$

det (A B) = det (A) det (B) .

Solution.

Let $A = [\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}]$ and $B = [\begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array}] .$ Then

A B = [\begin{array}{cc} a_{11} b_{11} + a_{12} b_{21} & a_{11} b_{12} + a_{12} b_{22} \\ a_{21} b_{11} + a_{22} b_{21} & a_{21} b_{12} + a_{22} b_{22} \end{array}] .

\begin{aligned} det (A B) & = (a_{11} b_{11} + a_{12} b_{21}) (a_{21} b_{12} + a_{22} b_{22}) \\ - (a_{11} b_{12} + a_{12} b_{22}) (a_{21} b_{11} + a_{22} b_{21}) \\ = (a_{11} b_{11} a_{21} b_{12} + a_{11} b_{11} a_{22} b_{22} + a_{12} b_{21} a_{21} b_{12} + a_{12} b_{21} a_{22} b_{22}) \\ - (a_{11} b_{12} a_{21} b_{11} + a_{11} b_{12} a_{22} b_{21} + a_{12} b_{22} a_{21} b_{11} + a_{12} b_{22} a_{22} b_{21}) \\ = a_{11} b_{11} a_{22} b_{22} + a_{12} b_{21} a_{21} b_{12} - a_{11} b_{12} a_{22} b_{21} - a_{12} b_{22} a_{21} b_{11} . \end{aligned}

Also,

\begin{aligned} det (A) det (B) & = (a_{11} a_{22} - a_{12} a_{21}) (b_{11} b_{22} - b_{12} b_{21}) \\ = a_{11} a_{22} b_{11} b_{22} - a_{11} a_{22} b_{12} b_{21} - a_{12} a_{21} b_{11} b_{22} + a_{12} a_{21} b_{12} b_{21} . \end{aligned}

We conclude that $det (A B) = det (A) det (B)$ if $A$ and $B$ are $2 \times 2$ matrices.

🔗

Subsection Summary

The determinant of an $n \times n$ matrix $A = [a_{i j}]$ is found by taking the cofactor expansion of $A$ along the first row. That is

$det (A) = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13} + \dots + a_{1 n} C_{1 n},$

where
- $A_{i j}$ is the sub-matrix of $A$ found by deleting the $i$ th row and $j$ th column of $A .$
- $C_{i j} = (- 1)^{i + j} det (A_{i j})$ is the $i j$ th cofactor of $A .$
The matrix $A$ is invertible if and only if $det (A) \neq 0 .$

🔗

Exercises Exercises

🔗

1.

Use the cofactor expansion to explain why multiplying each of the entries of a $3 \times 3$ matrix $A$ by 2 multiplies the determinant of $A$ by 8.

🔗

2.

Use the determinant criterion to determine for which $c$ the matrix $A = [\begin{array}{crc} 1 & 1 & 2 \\ 1 & 0 & c \\ 2 & - 1 & 2 \end{array}]$ is invertible.

🔗

3.

Let $A$ be a square matrix.

🔗

(a)

Explain why $det (A^{2}) = [det (A)]^{2}$

🔗

(b)

Expand on the argument from (a) to explain why $det (A^{k}) = [det (A)]^{k}$ for any positive integer $k .$

🔗

(c)

Suppose that $A$ is an invertible matrix and $k$ is a positive integer. Must $A^{k}$ be an invertible matrix? Why or why not?

🔗

4.

Let $A$ be an invertible matrix. Explain why $det (A^{- 1}) = \frac{1}{det (A)}$ using determinant properties.

🔗

5.

Simplify the following determinant expression using determinant properties:

det (P A^{4} P^{- 1} A^{T} (A^{- 1})^{3})

🔗

6.

Find the eigenvalues of the following matrices. Find a basis for and the dimension of each eigenspace.

🔗

(a)

$A = [\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}]$

🔗

(b)

$A = [\begin{array}{ccc} 2 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 1 & 2 \end{array}]$

🔗

7.

Label each of the following statements as True or False. Provide justification for your response.

🔗

(a) True/False.

For any two $n \times n$ matrices $A$ and $B,$ $det (A + B) = det A + det B .$

🔗

(b) True/False.

For any square matrix $A,$ $det (- A) = - det (A) .$

🔗

(c) True/False.

For any square matrix $A,$ $det (- A) = det (A) .$

🔗

(d) True/False.

The determinant of a square matrix with all non-zero entries is non-zero.

🔗

(e) True/False.

If the determinant of $A$ is non-zero, then so is the determinant of $A^{2} .$

🔗

(f) True/False.

If the determinant of a matrix $A$ is 0, then one of the rows of $A$ is a linear combination of the other rows.

🔗

(g) True/False.

For any square matrix $A,$ $det (A^{2}) > det (A) .$

🔗

(h) True/False.

If $A$ and $B$ are $n \times n$ matrices and $A B$ is invertible, then $A$ and $B$ are invertible.

🔗

(i) True/False.

If $A^{2}$ is the zero matrix, then the only eigenvalue of $A$ is 0.

🔗

(j) True/False.

If 0 is an eigenvalue of $A,$ then 0 is an eigenvalue of $A B$ for any $B$ of the same size as $A .$

🔗

(k) True/False.

Suppose $A$ is a $3 \times 3$ matrix. Then any three eigenvectors of $A$ will form a basis of $R^{3} .$

🔗

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

Area (R) .

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 $R_{1} \cap R_{2} = \emptyset,$ then $Area (R_{1} \cup R_{2}) = Area (R_{1}) + Area (R_{2}) .$
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 $v$ is a function $T_{v}$ such that the image $T_{v} (R)$ of a region $R$ is defined as

$T_{v} (R) = {r + v : r \in R} .$

Since area is translation invariant, $Area (T_{v} (R)) = Area (R) .$
The area of a one-dimensional object like a line segment is $0 .$

🔗

Now we turn our attention to areas of parallelograms. Let

u

and

v

be vectors in

R^{2} .

The parallelogram

P (u, v)

defined by

u

and

v

with point

Q

as basepoint is the set

P (u, v) = {\vec{O Q} + r u + s v : 0 \leq r, s \leq 1} .

🔗

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

🔗
Figure 17.8. A parallelogram and a translated, rotated parallelogram.

🔗

u = [u_{1} u_{2}]^{T}

and

v = [v_{1} v_{2}]^{T},

then we will also represent

P (u, v)

P ([\begin{array}{cc} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array}]) .

🔗

Since area is translation and rotation invariant, we can translate our parallelogram by

- \vec{O Q}

to place its basepoint at the origin, then rotate by an angle

θ

(as shown at left in Figure 17.8. This transforms the vector

v

to a vector

v^{'}

and the vector

u

to a vector

u^{'}

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

u

along the

x

-axis, and

v

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 $u$ and $v,$ both shown in Figure 17.9. The parallelogram will be determined by the lengths of these vectors.

Figure 17.9. Parallelograms formed by $u$ and $v$

🔗

(a)

In the situation depicted at left in Figure 17.9, use geometry to explain why $Area (P (u, v)) = h | u | .$

Hint.

What can we say about the triangles $O D B$ and $E A C ?$

🔗

(b)

In the situation depicted at right in Figure 17.9, use geometry to again explain why $Area (P (u, v)) = h | u | .$ (Hint: What can we say about $Area (A E C)$ and $Area (O D B) ?$ )

🔗

The result of Project Activity 17.3 is that the area of

P (u, v)

is given by

h | u |,

where

h

is the height of the parallelogram determined by dropping a perpendicular from the terminal point of

v

to the line determined by the vector

u .

🔗

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 (u, v) .

🔗

Project Activity 17.4.

Let $u$ and $v$ be vectors that determine a parallelogram in $R^{2} .$

Figure 17.10. Parallelograms formed by $k u$ and $v$ and by $u$ and $v + k u .$

🔗

(a)

Explain why

\begin{matrix} (17.3) & Area (P (u, v)) = Area (P (v, u)) \end{matrix}

🔗

(b)

If $k$ is any scalar, then $k u$ either stretches or compresses $u .$ Use this idea, and the result of Project Activity 17.3, to explain why

\begin{matrix} (17.4) & Area (P (k u, v)) = Area (P (u, k v)) = | k | Area (P (u, v)) \end{matrix}

for any real number $k .$ A representative picture of this situation is shown at left in Figure 17.9 for a value of $k > 1 .$ You will also need to consider what happens when $k < 0 .$

🔗

(c)

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

\begin{matrix} (17.5) & Area (P (u + k v, v)) = Area (P (u, v + k u)) = Area (P (u, v)) \end{matrix}

for any real number $k .$ 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

u

and

v .

🔗

Project Activity 17.5.

Let $u = [u_{1} u_{2}]^{T}$ and $v = [v_{1} v_{2}]^{T} .$ We will now demonstrate that

Area (P (u, v)) = | det (| \begin{array}{cc} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array}]) | .

Before we begin, note that if both $u_{1}$ and $v_{1}$ are $0,$ then $u$ and $v$ are parallel. This makes $P (u, v)$ a line segment and so $Area (P (u, v)) = 0 .$ But if $u_{1} = v_{1} = 0,$ it is also the case that

det (| \begin{array}{cc} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array}]) = u_{1} v_{2} - u_{2} v_{1} = 0

as well. So we can assume that at least one of $u_{1},$ $v_{1}$ is not $0 .$ Since $P (u, v) = P (v, u),$ we can assume without loss of generality that $u_{1} \neq 0 .$

🔗

(a)

Explain using properties (17.4) and (17.5) as appropriate why

Area (P (u, v)) = Area (P (u, [0 v_{2} - \frac{v_{1}}{u_{1}} u_{2}])) .

🔗

(b)

Let $v_{1} = {[0 v_{2} - \frac{v_{1}}{u_{1}} u_{2}]}^{T} .$ Recall that our alternate representation of $P (u, v))$ allows us to write

Area (P (u, v_{1})) = Area (P ([\begin{array}{cc} u_{1} & u_{2} \\ 0 & v_{2} - \frac{v_{1}}{u_{1}} u_{2} \end{array}])) .

This should seem very suggestive. We are essentially applying the process of Gaussian elimination to our parallelogram matrix to reduce it to a diagonal matrix. From there, we can calculate the area. The matrix form should indicate the next step — applying an operation to eliminate the entry in the first row and second column. To do this, we need to consider what happens if $v_{2} - \frac{v_{1}}{u_{1}} u_{2} = 0$ and if $v_{2} - \frac{v_{1}}{u_{1}} u_{2} \neq 0 .$

🔗

(i)

Assume that $v_{2} - \frac{v_{1}}{u_{1}} u_{2} = 0 .$ Explain why $Area (P (u, v)) = 0 .$ Then explain why $Area (P (u, v)) = 0 = det ([\begin{array}{cc} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array}]) .$

🔗

(ii)

Now we consider the case when $v_{2} - \frac{v_{1}}{u_{1}} u_{2} \neq 0 .$ Complete the process as in part (a), using properties (17.4) and (17.5) (compare to Gaussian elimination) to continue to reduce the problem of calculating $Area (P (u, v))$ to one of calculating $Area (P (e_{1}, e_{2})) .$ Use this process to conclude that

Area (P (u, v)) = | det ([\begin{array}{cc} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array}]) | .

🔗

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

u_{1},

u_{2},

\dots,

u_{n}

R^{n},

we let

P (u_{1}, u_{2}, \dots, u_{n}) = {\vec{O Q} + x_{1} u_{1} + x_{2} u_{2} + \dots + x_{n} u_{n} : 0 \leq x_{i} \leq 1 for each i} .

🔗

n = 2,

then

P (u_{1}, u_{2})

is the parallelogram determined by

u_{1}

and

u_{2}

with basepoint

Q .

n = 3,

then

P (u_{1}, u_{2}, u_{3})

is the parallelepiped with basepoint

Q

determined by

u_{1},

u_{2},

and

u_{3} .

In higher dimensions the sets

P (u_{1}, u_{2}, \dots, u_{n})

are called parallelotopes, and we use the notation

Vol (P (u_{1}, u_{2}, \dots, u_{n}))

for their volume. The

n

-dimensional volumes of these paralleotopes satisfy the following properties:

\begin{aligned} Vol & (P (u_{1}, u_{2}, \dots, u_{i - 1}, u_{i}, u_{i + 1}, \dots, u_{j - 1}, u_{j}, u_{j + 1}, \dots, u_{n})) \\ (17.6) & = Vol (P (u_{1}, u_{2}, \dots, u_{i - 1}, u_{j}, u_{i + 1}, \dots, u_{j - 1}, u_{i}, u_{j + 1}, \dots, u_{n})) \end{aligned}

🔗

for any

i

and

j .

\begin{matrix} (17.7) & Vol (P (u_{1}, u_{2}, \dots, u_{i - 1}, k u_{i}, u_{i + 1}, \dots, u_{n})) = | k | Vol (P (u_{1}, u_{2}, \dots, u_{n})) \end{matrix}

🔗

for any real number

k

and any

i .

\begin{matrix} (17.8) & Vol (P (u_{1}, u_{2}, \dots, u_{i - 1}, u_{i} + k u_{j}, u_{i + 1}, \dots, u_{n})) = Vol (P (u_{1}, u_{2}, \dots, u_{n})) \end{matrix}

🔗

for any real number

k

and any distinct

i

and

j .

🔗

Project Activity 17.6.

We now show that $Vol (P (u_{1}, u_{2}, u_{3}))$ is the absolute value of the determinant of $[\begin{matrix} u_{1} \\ u_{2} \\ u_{3} \end{matrix}] .$ For easier notation, let $u = [u_{1} u_{2} u_{3}]^{T},$ $v = [v_{1} v_{2} v_{3}]^{T},$ and $w = [w_{1} w_{2} w_{3}]^{T} .$ As we argued in the 2-dimensional case, we can assume that all terms that we need to be nonzero are nonzero, and we can do so without verification.

🔗

(a)

Explain how property (17.7) shows that $Vol (P (u, v, w))$ is equal to

Vol (P ([\begin{array}{ccc} u_{1} & u_{2} & u_{3} \\ 0 & \frac{1}{u_{1}} (v_{2} u_{1} - v_{1} u_{2}) & \frac{1}{u_{1}} (v_{3} u_{1} - v_{1} u_{3}) \\ 0 & \frac{1}{u_{1}} (w_{2} u_{1} - w_{1} u_{2}) & \frac{1}{u_{1}} (w_{3} u_{1} - w_{1} u_{3}) \end{array}])) .

Hint.

Think about how these properties are related to row operations.

🔗

(b)

Now let

v_{1} = {[0 \frac{1}{u_{1}} (v_{2} u_{1} - v_{1} u_{2}) \frac{1}{u_{1}} (v_{3} u_{1} - v_{1} u_{3})]}^{T}

and

w_{1} = {[0 \frac{1}{u_{1}} (w_{2} u_{1} - w_{1} u_{2}) \frac{1}{u_{1}} (w_{3} u_{1} - w_{1} u_{3})]}^{T} .

Explain how property (17.7) shows that $Vol (P (u, v, w))$ is equal to

Vol (P ([\begin{array}{ccc} u_{1} & u_{2} & u_{3} \\ 0 & \frac{1}{u_{1}} (v_{2} u_{1} - v_{1} u_{2}) & \frac{1}{u_{1}} (v_{3} u_{1} - v_{1} u_{3}) \\ 0 & 0 & d \end{array}])),

where

d = \frac{1}{u_{1} v_{2} - u_{2} v_{1}} (u_{1} (v_{2} w_{3} - v_{3} w_{2}) - u_{2} (v_{1} w_{3} - v_{3} w_{1}) + u_{3} (v_{1} w_{2} - v_{2} w_{1})) .

🔗

(c)

Just as we saw in the 2-dimensional case, we can proceed to use the diagonal entries to eliminate the entries above the diagonal without changing the volume to see that

Vol (P (u, v, w)) = Vol (P ([\begin{array}{ccc} u_{1} & 0 & 0 \\ 0 & \frac{1}{u_{1}} (v_{2} u_{1} - v_{1} u_{2}) & 0 \\ 0 & 0 & d \end{array}])) .

Complete the process, applying appropriate properties to explain why

Vol (P (u, v, w)) = x Vol (P (e_{1}, e_{2}, e_{3}))

for some constant $x .$ Find the constant and, as a result, find a specific expression for $Vol (P (u, v, w))$ involving a determinant.

🔗

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.

Tucker, Alan. (1993). The Growing Importance of Linear Algebra in Undergraduate Mathematics. The College Mathematics Journal, 1, 3-9.

Section 17 The Determinant

Focus Questions

Subsection Application: Area and Volume

Subsection Introduction

Preview Activity 17.1.

(a)

(b)

(c)

(d)

(e)

Subsection The Determinant of a Square Matrix

Subsection Cofactors

Definition 17.2.

Activity 17.2.

(a)

(b)

(c)

(d)

(e)

Theorem 17.3.

Subsection The Determinant of a 3×3 Matrix

Subsection Two Devices for Remembering Determinants

Subsection Examples

Example 17.6.

(a)

Solution.

(b)

Solution.

Example 17.7.

Solution.

Subsection Summary

Exercises Exercises

1.

2.

3.

(a)

(b)

(c)

4.

5.

6.

(a)

(b)

7.

(a) True/False.

(b) True/False.

(c) True/False.

(d) True/False.

(e) True/False.

(f) True/False.

(g) True/False.

(h) True/False.

(i) True/False.

(j) True/False.

(k) True/False.

Subsection Project: Area and Volume Using Determinants

Project Activity 17.3.

(a)

(b)

Project Activity 17.4.

(a)

(b)

(c)

Project Activity 17.5.

(a)

(b)

(i)

(ii)

Project Activity 17.6.

(a)

(b)

(c)

Subsection The Determinant of a $3 \times 3$ Matrix