Applications in Ch3

128 Chapter 3 Determinants

3.4 Applications of Determinants

Find the adjoint of a matrix and use it to find the inverse of

the matrix.

Use Cramers Rule to solve a system of linear equations in

variables.

Use determinants to find area, volume, and the equations of

lines and planes.

THE ADJOINT OF A MATRIX

So far in this chapter, you have studied procedures for evaluating, and properties of,

determinants. In this section, you will study an explicit formula for the inverse of a

nonsingular matrix and use this formula to derive a theorem known as Cramers Rule.

You will then solve several applications of determinants.

Recall from Section 3.1 that the cofactor of a square matrix is defined as

times the determinant of the matrix obtained by deleting the th row and the

th column of The matrix of cofactors of has the form

The transpose of this matrix is called the adjoint of and is denoted by adj

That is,

Finding the Adjoint of a Square Matrix

Find the adjoint of

SOLUTION

The cofactor is given by

Continuing this process produces the following matrix of cofactors of

The transpose of this matrix is the adjoint of That is, adjsAd 5 34

1

2

6

0

3

7

1

24.A.

34

6

7

1

0

1

2

3

24

A.

C11 5 s21d2|220 122| 5 4.321

0

1

3

22

0

2

1

224

C11

A 5 321

0

1

3

22

0

2

1

224.

adjsAd 5 3C11C12.

.

.C1n

C21C22.

.

.C2n

. . .

. . .

. . .

Cn1Cn2.

.

.Cnn

4.sAd.A

3C11C21.

.

.Cn1

C12C22.

.

.Cn2

. . .

. . .

. . .

C1nC2n.

.

.Cnn

4.AA.j

is21di1 jACi j

n

n

PROOF

Begin by proving that the product of and its adjoint is equal to the product of the

determinant of and Consider the product

The entry in the th row and th column of this product is

If then this sum is simply the cofactor expansion of in its th row, which means

that the sum is the determinant of On the other hand, if then the sum is zero.

(Try verifying this.)

Because is invertible, det and you can write

or

By Theorem 2.7 and the definition of the inverse of a matrix, it follows that

Using the Adjoint of a Matrix

to Find Its Inverse

Use the adjoint of to find where

SOLUTION

The determinant of this matrix is 3. Using the adjoint of (found in Example 1), the

inverse of is

Check that this matrix is the inverse of by showing that AA21 5 I 5 A21A.A

A21 51

|A|adjsAd 5133

4

1

2

6

0

3

7

1

24 5 3

43

13

23

2

0

1

73

13

23

4.A

A

A 5 321

0

1

3

22

0

2

1

224.A21,A

1

detsAd adjsAd 5 A21.

A3 1detsAd adjsAd4 5 I.1

detsAd AfadjsAdg 5 I

sAd 0A

AfadjsAdg 5 3detsAd

0...0

0

detsAd...0

. . .

. . .

. . .

0

0...

detsAd4 5 detsAdI

i j,A.

iAi 5 j,

ai1Cj1 1 ai2Cj2 1. . . 1 ainCjn.

ji

3C11C12.

.

.C1n

C21C22.

.

.C2n

. . .

. . .

. . .

Cj1Cj2...

Cjn

. . .

. . .

. . .

Cn1Cn2.

.

.Cnn

4.A fadjsAdg 5 3a11a21...

ai1...

an1

a12a22...

ai2...

an2

. . .

. . .

. . .

. . .

a1na2n...

ain...

ann

4In.A

A

3.4 Applications of Determinants 129

REMARKTheorem 3.10 is not particularly

efficient for calculating inverses.

The Gauss-Jordan elimination

method discussed in Section 2.3

is much better. Theorem 3.10

is theoretically useful, however,

because it provides a concise

formula for the inverse of

a matrix.

REMARKIf is a matrix

then the adjoint

of is simply

Moreover, if is invertible,

then from Theorem 3.10 you

have

which agrees with the result in

Section 2.3.

51

ad 2 bc 3 d

2c

2b

a4

A21 51

|A|adjsAd

A

adjsAd 5 3 d2c

2b

a4.A

A 5 3acb

d4,2 3 2A

THEOREM 3.10 The Inverse of a Matrix Given by Its Adjoint

If is an invertible matrix, then A21 51

detsAdadjsAd.n 3 nA

The adjoint of a matrix is useful for finding the inverse of as indicated in the

next theorem.

A,A

CRAMERS RULE

Cramers Rule, named after Gabriel Cramer (17041752), uses determinants to solve a

system of linear equations in variables. This rule applies only to systems with

unique solutions. To see how Cramers Rule works, take another look at the solution

described at the beginning of Section 3.1. There, it was pointed out that the system

has the solution

and

when Each numerator and denominator in this solution can be

represented as a determinant, as follows.

The denominator for and is simply the determinant of the coefficient matrix

of the original system. The numerators for and are formed by using the column

of constants as replacements for the coefficients of and in These two

determinants are denoted by and as follows.

and

You have and This determinant form of the solution is called

Cramers Rule.

Using Cramers Rule

Use Cramers Rule to solve the system of linear equations.

SOLUTION

First find the determinant of the coefficient matrix.

Because you know the system has a unique solution, and applying Cramers

Rule produces

and

The solution is and x2 5 21.x1 5 2

x2 5|A2||A|

5|43 1011|

2145

14

2145 21.

x1 5|A1||A|

5|1011 2225|

2145

228

2145 2

|A| 0,

5 214 |A| 5 |43 2225|

3x1 2 5x2 5 11

4x1 2 2x2 5 10

x2 5|A2||A|

.x1 5|A1||A|

|A2| 5 |a11a21 b1b2||A1| 5 |b1b2 a12a22||A2|,|A1|

|A|.x2x1x2x1

Ax2x1

a11a22 2 a21a12 0x2 5|a11a21 b1b2||a11a21 a12a22|

,x1 5|b1b2 a12a22||a11a21 a12a22|

,

a11a22 2 a21a12 0.

x2 5b2a11 2 b1a21

a11a22 2 a21a12x1 5

b1a22 2 b2a12a11a22 2 a21a12

a21x1 1 a22x2 5 b2

a11x1 1 a12x2 5 b1

nn


Cramers Rule generalizes easily to systems of linear equations in variables. The

value of each variable is given as the quotient of two determinants. The denominator is

the determinant of the coefficient matrix, and the numerator is the determinant of the

matrix formed by replacing the column corresponding to the variable being solved for

with the column representing the constants. For instance, the solution for in the system

is

PROOF

Let the system be represented by Because is nonzero, you can write

If the entries of are then but

the sum (in parentheses) is precisely the cofactor expansion of which means that

and the proof is complete.

Using Cramers Rule

Use Cramers Rule to solve the system of linear equations for

SOLUTION

The determinant of the coefficient matrix is

Because you know that the solution is unique, so apply Cramers Rule to solve

for as follows.

x 5|102 2024 2314|

105

s1ds21d5|12 224|10

5s1ds21ds28d

105

4

5

x,|A| 0,

|A| 5 |2123 2024 2314| 5 10. 3x 2 4y 1 4z 5 2

2x 1 z 5 0

2x 1 2y 2 3z 5 1

x.

x i 5 |Ai|y|A|,Ai,

x i 51

|A|sb1C1i 1 b2C2i 1 . . . 1 bnCnid,b1, b2, . . . , bn,B

X 5 A21B 51

|A| adjsAdB 5 3

x1

x2...

xn

4.|A|AX 5 B.

x3 5|A3||A|

5|a11a21a31 a12a22a32 b1b2b3||a11a21a31 a12a22a32 a13a23a33|

.

a11x1a21x1a31x1

1

1

1

a12x2a22x2a32x2

1

1

1

a13x3a23x3a33x3

5

5

5

b1b2b3

x3

nn


THEOREM 3.11 Cramers Rule

If a system of linear equations in variables has a coefficient matrix with

a nonzero determinant then the solution of the system is

. . . ,

where the th column of is the column of constants in the system of equations.Aii

xn 5detsAnddetsAd

x2 5detsA2ddetsAd

,x1 5detsA1ddetsAd

,

|A|,Ann

REMARKTry applying Cramers Rule to

solve for and You will see

that the solution is and

z 5 285.

y 5 232

z.y

AREA, VOLUME, AND EQUATIONS OF LINES AND PLANES

Determinants have many applications in analytic geometry. One application is in

finding the area of a triangle in the -plane.

PROOF

Prove the case for Assume that and that lies above the

line segment connecting and as shown in Figure 3.1. Consider the three

trapezoids whose vertices are

The area of the triangle is equal to the sum of the areas of the first two trapezoids minus

the area of the third trapezoid. So,

If the vertices do not occur in the order or if the vertex is not

above the line segment connecting the other two vertices, then the formula above

may yield the negative of the area. So, use and choose the correct sign to give a

positive area.

Finding the Area of a Triangle

Find the area of the triangle whose vertices are

and

SOLUTION

It is not necessary to know the relative positions of the three vertices. Simply evaluate

the determinant

and conclude that the area of the triangle is square units.32

12|124 023 111| 5 232

s4, 3d.s2, 2d,s1, 0d,

sx3, y3dx1 # x3 # x2

512|x1x2x3 y1y2y3 111|.

512sx1y2 1 x2y3 1 x3y1 2 x1y3 2 x2y1 2 x3y2d

Area 512sy1 1 y3dsx3 2 x1d 1

12sy3 1 y2dsx2 2 x3d 2

12sy1 1 y2dsx2 2 x1d

Trapezoid 3: sx1, 0d, sx1, y1d, sx2, y2d, sx2, 0d.Trapezoid 2: sx3, 0d, sx3, y3d, sx2, y2d, sx2, 0dTrapezoid 1: sx1, 0d, sx1, y1d, sx3, y3d, sx3, 0d

sx2, y2d,sx1, y1dsx3, y3dx1 # x3 # x2yi > 0.

xy


Area of a Triangle in the -Plane

The area of a triangle with vertices

and

is

where the sign is chosen to give a positive area.sd

Area 5 12 det3

x1

x2

x3

y1

y2

y3

1

1

14

sx3, y3dsx1, y1d, sx2, y2d,

xy

Figure 3.1

x

(x3, y3)

(x1, y1)

(x1, 0) (x3, 0) (x2, 0)

(x2, y2)

y

Suppose the three points in Example 5 had been on the same line. What would have

happened had you applied the area formula to three such points? The answer is that the

determinant would have been zero. Consider, for instance, the three collinear points

and as shown in Figure 3.2. The determinant that yields the area of

the triangle that has these three points as vertices is

If three points in the -plane lie on the same line, then the determinant in the formula

for the area of a triangle is zero. The following generalizes this result.

The test for collinear points can be adapted to another use. That is, when you are

given two points in the -plane, you can find an equation of the line passing through

the two points, as follows.

Finding an Equation of the Line

Passing Through Two Points

Find an equation of the line passing through the two points

and

SOLUTION

Let and Applying the determinant formula for the

equation of a line produces

To evaluate this determinant, expand by cofactors in the first row to obtain the following.

So, an equation of the line is x 2 3y 5 210.

x 2 3y 1 10 5 0

xs1d 2 ys3d 1 1s10d 5 0

x|43 11| 2 y| 221 11| 1 1| 221 43| 5 0| x221 y43 111| 5 0.

sx2, y2d 5 s21, 3d.sx1, y1d 5 s2, 4d

s21, 3d.s2, 4d

xy

xy

12|024 123 111| 5 0.

s4, 3d,s2, 2d,s0, 1d,


Test for Collinear Points in the -Plane

Three points and are collinear if and only if

det3x1

x2

x3

y1

y2

y3

1

1

14 5 0.

sx3, y3dsx2, y2d,sx1, y1d,

xy

Two-Point Form of the Equation of a Line

An equation of the line passing through the distinct points and

is given by

det3x

x1

x2

y

y1

y2

1

14 5 0.1

sx2, y2dsx1, y1d

Figure 3.2

x

1 2 3 4

2

3

(2, 2)

(4, 3)

(0, 1)

y

The formula for the area of a triangle in the plane has a straightforward

generalization to three-dimensional space, which is presented without proof as follows.

Finding the Volume of a Tetrahedron

Find the volume of the tetrahedron whose vertices are

and as shown in Figure 3.3.

Figure 3.3

SOLUTION

Using the determinant formula for the volume of a tetrahedron produces

So, the volume of the tetrahedron is 12 cubic units.

If four points in three-dimensional space lie in the same plane, then the determinant

in the formula for the volume of a tetrahedron is zero. So, you have the following test.

5 212. 16|0432 4052 1025 1111| 5 16s272d

yx

(4, 0, 0)

(2, 2, 5)

(3, 5, 2)

(0, 4, 1)

5

55

z

s2, 2, 5d,s3, 5, 2d,s4, 0, 0d,s0, 4, 1d,


Test for Coplanar Points in Space

Four points and are coplanar

if and only if

det3x1

x2

x3

x4

y1

y2

y3

y4

z1

z2

z3

z4

1

1

1

14 5 0.

sx4, y4, z4dsx3, y3, z3d,sx2, y2, z2d,sx1, y1, z1d,

Volume of a Tetrahedron

The volume of a tetrahedron with vertices

and is

where the sign is chosen to give a positive volume.sd

Volume 5 16 det3

x1

x2

x3

x4

y1

y2

y3

y4

z1

z2

z3

z4

1

1

1

14

sx4, y4, z4dsx3, y3, z3d,sx2, y2, z2d,sx1, y1, z1d,

An adaptation of this test is the determinant form of an equation of a plane passing

through three points in space, as follows.

Finding an Equation of the Plane

Passing Through Three Points

Find an equation of the plane passing through the three points

and

SOLUTION

Using the determinant form of the equation of a plane produces

To evaluate this determinant, subtract the fourth column from the second column to obtain

Now, expanding by cofactors in the second row yields

This produces the equation 4x 2 3y 1 5z 5 23.

xs4d 2 sy 2 1ds3d 1 zs5d 5 0.

x| 221 21| 2 sy 2 1d|2122 21| 1 z|2122 221| 5 0| x02122 y 2 10221 z021 1111| 5 0.| x02122 y 130 z021 1111| 5 0.s22, 0, 1d.

s21, 3, 2d,s0, 1, 0d,


Three-Point Form of the Equation of a Plane

An equation of the plane passing through the distinct points

and is given by

det3x

x1x2x3

y

y1y2y3

z

z1z2z3

1

1

1

14 5 0.

sx3, y3, z3dsx2, y2, z2d,sx1, y1, z1d,

LINEAR ALGEBRA APPLIED

According to Keplers First Law of Planetary Motion, the

orbits of the planets are ellipses, with the sun at one focus

of the ellipse. The general equation of a conic section (such

as an ellipse) is

To determine the equation of the orbit of a planet, an

astronomer can find the coordinates of the planet along its

orbit at five different points where 2, 3, 4, and

5, and then use the determinant

|x2

x21x22x23x24x25

xy

x1y1x2y2x3y3x4y4x5x5

y2

y21y22y23y24y25

x

x1x2x3x4x5

y

y1y2y3y4y5

1

1

1

1

1

1|.i 5 1,sxi, yi d,

ax2 1 bxy 1 cy2 1 dx 1 ey 1 f 5 0.

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