mth 215: introduction to linear algebra - chapter...

DeterminantsApplications of determinants

MTH 215: Introduction to Linear AlgebraChapter 3

Jonathan A. Chavez Casillas1

1University of Rhode IslandDepartment of Mathematics

February 13, 2019

Jonathan Chavez


1 DeterminantsCofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices

2 Applications of determinantsA Formula for the InverseCramer’s RulePolynomial Interpolation

Jonathan Chavez


Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices

Determinant of a 2 × 2 Matrix

Definition

Let A =[

a bc d

]. Then the determinant of A is defined as

det A = ad − bc

What about the determinant of an n × n matrix for other values of n?

Notation. For det[

a bc d

], we often write

∣∣∣∣ a bc d

∣∣∣∣, i.e., use vertical bars instead of square

brackets.

Jonathan Chavez



How do we find the determinant of an n × n matrix?The determinant of an n × n matrix is defined recursively, using determinants of (n − 1)× (n − 1)submatrices, and requires some new definitions and notation.

DefinitionLet A = [aij ] be an n × n matrix. The sign of the (i , j) position is (−1)i+j .Thus the sign is 1 if (i + j) is even, and −1 if (i + j) is odd.

Jonathan Chavez



The Minor of a MatrixDefinitionLet A = [aij ] be an n × n matrix. The ij th minor of A, denoted as minor (A)ij , is the determinant of then − 1× n − 1 matrix which results from deleting the i th row and the j th column of A.

A =

a11 a12 · · · a1j · · · a1na21 a22 · · · a2j · · · a2n

......

......

ai1 ai2 · · · aij · · · ain...

......

...

an1 an2 · · · anj · · · ann

For any matrix A, minor (A)ij is found by first removing the i th row and j th column, and taking thedeterminant of the remaining matrix.

Jonathan Chavez



ExampleFind minor(A)12 of

A =

[1 1 32 4 15 2 6

]

SolutionFirst, remove the 1st row and 2nd column from A.

A =

[1 1 32 4 15 2 6

]

The resulting matrix is A =[

2 15 6

]. Using our previous definition, we calculate the determinant of this matrix as:

(2)(6) − (5)(1) = 12 − 5 = 7

Therefore, minor(A)12 = 7.

Jonathan Chavez



The Cofactors of a MatrixDefinitionThe ij th cofactor of A is

cof(A)ij = (−1)i+j minor (A)ij

Example (continued)Find cof(A)12 if

A =

[1 1 32 4 15 2 6

]

SolutionBy the definition, we know that cof(A)12 = (−1)1+2minor (A)12. From earlier, we know that minor (A)12 = 7.

Therefore, cof(A)12 = (−1)1+2minor (A)12 = (−1)37 = −7

Jonathan Chavez



Cofactor ExpansionUsing these definitions, we can now define the determinant of the n × n matrix A:

Definitiondet A = a11cof(A)11 + a12cof(A)12 + a13cof(A)13 + · · ·+ a1ncof(A)1nThis is called the cofactor expansion of det A along row 1.

In other words,

det (A) =n∑

j=1

aij cof (A)ij =n∑

i=1

aij cof (A)ij

The first formula consists of expanding the determinant along the i th row and the second expands thedeterminant along the j th column.

Cofactor expansion is also called Laplace Expansion.

Jonathan Chavez



Cofactor ExpansionExample

Let A =

[1 1 32 4 15 2 6

]. Find det A.

SolutionUsing cofactor expansion along row 1,

det A = 1cof11(A) + 1cof12(A) + 3cof13(A)

= 1(−1)2∣∣∣ 4 1

2 6

∣∣∣ + 1(−1)3∣∣∣ 2 1

5 6

∣∣∣ + 3(−1)4∣∣∣ 2 4

5 2

∣∣∣= 1(24 − 2) − 1(12 − 5) + 3(4 − 20)= 22 − 7 + 3(−16)= 22 − 7 − 48= −33

Jonathan Chavez



Solution (continued)

A =

[1 1 32 4 15 2 6

]

Now try cofactor expansion along column 2.

We get the same answer!

Jonathan Chavez



The Determinant is Well Defined

TheoremThe determinant of an n × n matrix A can be computed using cofactor expansion along any row or columnof A.

What is the significance of this theorem?This theorem allows us to choose any row or column for computing cofactor expansion, which gives us theopportunity to save ourselves some work!

Jonathan Chavez



Problem

Let A =

0 1 −2 15 0 0 70 1 −1 03 0 0 2

. Find det A.

SolutionCofactor expansion along row 1 yields

whereas cofactor expansion along, row 3 yields

so, in the first case we must compute three cofactors, but just two cofactors in the second case.

Jonathan Chavez



Solution (continued)Therefore, we can save ourselves some work by using cofactor expansion along row 3 rather than row 1.

A =

0 1 −2 15 0 0 70 1 −1 03 0 0 2

Each of the two determinants above can easily be evaluated using cofactor expansion along column 2.Jonathan Chavez




Jonathan Chavez



A Row or Column of Zeros

ExampleLet

A =

−8 1 0 −45 7 0 −7

12 −3 0 8−3 11 0 2

.

By choosing column 3 for cofactor expansion, we get det A = 0, i.e.,

det A = 0× cof(A)13 + 0× cof(A)23 + 0× cof(A)33 + 0× cof(A)43 = 0.

Important FactIf A is an n × n matrix with a row or column of zeros, then det A = 0.

Jonathan Chavez



Determinants of Inverses

TheoremAn n × n matrix A is invertible if and only if det A , 0. In this case,

det(A−1) = 1det A .

Jonathan Chavez



Example (Illustration of the above Theorem.)

Let A =[

2 −3−6 4

].

Then det A = 2(4)− (−3)(−6) = −10.

Since det A , 0, the above Theorem tell us that A is invertible, and that det(A−1) should be equal to − 110 .

We can verify this directly.

Using the formula for the inverse of a 2× 2 matrix

A−1 = 1−10

[4 36 2

].

Therefore, det A−1 =(− 1

10

)2∣∣∣∣ 4 3

6 2

∣∣∣∣ =(− 1

10

)2(8− 18) = 1

100 × (−10) = − 110 .

Jonathan Chavez



Problem

Find all values of c for which A =

[c 1 00 2 c

−1 c 5

]is invertible.

Solution

Jonathan Chavez



Determinants of Products and Transposes

TheoremLet A and B be n × n matrices. Then

det (AB) = det A× det B.

TheoremIf A is an n × n matrix, then the determinant of its transpose is given by

det(AT ) = det A.

Jonathan Chavez



ProblemSuppose A, B and C are 4 × 4 matrices with

det A = −1, det B = 2, and det C = 1.

Find det(2A2(B−1)(CT )3B(A−1)).

Solution

Jonathan Chavez



ProblemSuppose A is a 3 × 3 matrix. Find det A and det B if

det(2A−1) = −4 = det(A3(B−1)T ).

Solution

Jonathan Chavez



ProblemLet A be an n × n matrix. Find all conditions that ensure det(−A) = det A.

Solution

Jonathan Chavez



Elementary Row Operations and DeterminantsExampleLet

A =

[2 0 −30 4 01 0 −2

].

Computing det A by cofactor expansion along row (or column) 2 yields

det A = 4(−1)4∣∣∣ 2 −3

1 −2

∣∣∣ = 4(−1) = −4.

Let B1, B2, and B3 be obtained from A by interchanging rows 1 and 2, multiplying row 3 by −3, and adding twice row 1 torow 3, respectively, i.e.,

B1 =

[2 0 −31 0 −20 4 0

], B2 =

[2 0 −30 4 0

−3 0 6

], B3 =

[2 0 −30 4 05 0 −8

].

We are interested in how elementary operations affect the determinant. Computing det B1, det B2, and det B3:

Jonathan Chavez



Example (continued)

det B1 = 4(−1)5∣∣∣∣ 2 −3

1 −2

∣∣∣∣ = (−4)(−1) = 4 = (−1) det A.

det B2 = 4(−1)4∣∣∣∣ 2 −3−3 6

∣∣∣∣ = 4(12− 9) = 4× 3 = 12 = −3 det A.

det B3 = 4(−1)4∣∣∣∣ 2 −3

5 −8

∣∣∣∣ = 4(−16 + 15) = 4(−1) = −4 = det A.

The general effects of elementary row operations on the determinant are summarized in the next theorem.

Jonathan Chavez



TheoremLet A be an n × n matrix and B be an n × n matrix as defined below.

1 Let B be a matrix which results from switching two rows of A. Then det (B) = − det (A) .

2 Let B be a matrix which results from multiplying some row of A by a scalar k. Thendet (B) = k det (A) .

3 Let B be a matrix which results from adding a multiple of a row to another row. Thendet (A) = det (B).

4 If A contains a row which is a multiple of another row of A, then det (A) = 0

An analogous theorem holds for elementary column operation. If A is a matrix, then an elementary columnoperation on A is simply the corresponding elementary row operation performed on the transpose of A, AT .

Jonathan Chavez



Determinants and Scalar MultiplicationProblemSuppose A is a 3 × 3 matrix with det A = 7. What is det(−2A)?

SolutionWrite A =

[a11 a12 a13a21 a22 a23a31 a32 a33

]. Then

Jonathan Chavez



Solution (continued)Think about the matrix −2A as the matrix obtained from A be multiplying each of its rows by −2. Thisinvolves three elementary row operations, each of which contributes a factor of −2 to the determinant, andthus det A = (−2)× (−2)× (−2)× 7 = (−2)3 × 7.

TheoremIf A is an n × n matrix and k is any scalar, then

det(kA) = kn det A.

Jonathan Chavez



Computing the Determinant

Example

det

[ −3 5 −61 −1 32 −4 1

]=

∣∣∣∣∣ 0 2 31 −1 32 −4 1

∣∣∣∣∣ row 1 + 3×(row 2)

=

∣∣∣∣∣ 0 2 31 −1 30 −2 −5

∣∣∣∣∣ row 3 − 2×(row 2)

= (1)(−1)2+1∣∣∣∣ 2 3−2 −5

∣∣∣∣ cofactor expansion: column 1

= −(−10 + 6)= 4.

Jonathan Chavez



Example

det

3 1 2 4−1 −3 8 0

1 −1 5 51 1 2 −1

=

Jonathan Chavez



Problem

If det

[a1 a2 a3b1 b2 b3c1 c2 c3

]= 4, find det

[−b1 −b2 −b3

a1 + 2b1 a2 + 2b2 a3 + 2b33c1 3c2 3c3

].

Solution∣∣∣∣∣ −b1 −b2 −b3a1 + 2b1 a2 + 2b2 a3 + 2b3

3c1 3c2 3c3

∣∣∣∣∣ =

Jonathan Chavez



Problem

Let A =

[2 3 53 5 91 2 4

]. Find det A.

SolutionNotice:

row1 + row3 − row2 =[

0 0 0]

Hence

Jonathan Chavez



ProblemLet

A =

[ a b cp q rx y z

]and B =

[ 2a + p 2b + q 2c + r2p + x 2q + y 2r + z2x + a 2y + b 2z + c

].

Show that det B = 9 det A.

Jonathan Chavez



Solution

Jonathan Chavez



Using Row OperationsProblemFind det(A) if:

A =

[5 1 21 3 22 6 0

]

Solution

Jonathan Chavez




Jonathan Chavez



Determinants of a Triangular Matrices

Definitions1 An n × n matrix A is called upper triangular if all entries below the main diagonal are zero.2 An n × n matrix A is called lower triangular if all entries above the main diagonal are zero.3 An n × n matrix A is called triangular if it is upper triangular or lower triangular.

TheoremIf A = [aij ] is an n × n triangular matrix, then

det A = a11 × a22 × a33 × · · · × ann,

i.e., det A is the product of the entries of the main diagonal of A.

Jonathan Chavez



Example

det

[ 1 2 30 5 60 0 9

]= 1× det

[5 60 9

]= 1× 5× det

[9]

= 1× 5× 9= 45.

Notice that 45 is the product of the entries on the main diagonal.[ 1 2 30 5 60 0 9

]

Jonathan Chavez


A Formula for the InverseCramer’s RulePolynomial Interpolation

Table of Contents

1 DeterminantsCofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices

2 Applications of determinantsA Formula for the InverseCramer’s RulePolynomial Interpolation

Jonathan Chavez



The Cofactor Matrix

DefinitionLet A = [aij ] be an n × n matrix. The cofactor matrix of A, is the matrix

[cof(A)ij ] ,

i.e., the matrix whose (i , j)-entry is the (i , j)-cofactor of A.

Reminder: the (i , j)-cofactor

cof(A)ij = (−1)i+jminor (A)ij ,

where minor (A)ij is the determinant of the matrix obtained from A by deleting row i and column j.

Jonathan Chavez



ProblemFind the cofactor matrix [cof(A)ij ] of the matrix

A =

[4 0 31 9 70 6 4

].

SolutionFor each i and j, 1 ≤ i, j ≤ 3, we need to compute cof(A)ij , so there are 9 cofactors to compute.

cof(A)11 = (−1)1+1 det A11 =∣∣∣ 9 7

6 4

∣∣∣ = 9 × 4 − 6 × 7 = 36 − 42 = −6.

cof(A)12 = (−1)1+2 det A12 =∣∣∣ 1 7

0 4

∣∣∣ = −(4 − 0) = −4.

cof(A)13 = (−1)1+3 det A12 =∣∣∣ 1 9

0 6

∣∣∣ = (6 − 0) = 6.

Computing the six remaining cofactors results in the cofactor matrix[−6 −4 618 16 −24

−27 −25 36

].

Jonathan Chavez



Definition (The Adjugate)If A is an n × n matrix, then the adjugate of A is defined by

adj A =[

cof(A)ij]T

,

where cof(A)ij is the (i, j)-cofactor of A, i.e., adj A is the transpose of the cofactor matrix.

Example

A =

[4 0 31 9 70 6 4

], has cofactor matrix

[−6 −4 618 16 −24

−27 −25 36

].

Therefore, the adjugate of A is

adj A =

[−6 −4 618 16 −24

−27 −25 36

]T

=

[−6 18 −27−4 16 −25

6 −24 36

].

Jonathan Chavez



Problem

Find adj A when A =

[ 2 1 35 −7 13 0 −6

].

Solution

adj A =

[ 42 6 2233 −21 1321 3 −19

].

Jonathan Chavez



The Adjugate of a 2 × 2 MatrixExampleLet A be a 2 × 2 matrix, say A =

[a bc d

].

Then

adj A =[

cof(A)11 cof(A)12cof(A)21 cof(A)22

]T=

[(−1)2d (−1)3c(−1)3b (−1)4a

]T

=[

d −b−c a

].

We’ve seen this matrix before: if det A , 0, then

A−1 =1

det A

[d −b

−c a

]=

1det A

adj A.

This actuallyholds true for all matrices.

Jonathan Chavez



The Adjugate Formula

TheoremIf A is an n × n matrix, then

A(adj A) = (det A)I = (adjA)A.

Furthermore, if det A , 0, then we get a formula for A−1, i.e.,

A−1 = 1det A adj A.

Inverting a matrix using the adjugateExcept in the case of a 2× 2 matrix, the adjugate formula is a very inefficient method for computing theinverse of a matrix; the matrix inversion algorithm is much more practical. However, the adjugate formulais of theoretical significance.

Jonathan Chavez



Using the Adjugate to Find the Inverse of a MatrixExample

Let A =

[ 4 0 31 9 70 6 4

]. As we saw earlier, det A = −6 , 0, so A is invertible, and

adj A =

[ −6 18 −27−4 16 −25

6 −24 36

].

A−1 = 1det A adj A = 1

−6

[ −6 18 −27−4 16 −25

6 −24 36

]=

1 −3 9

223 − 8

3256

−1 4 −6

.

You can check this by computing AA−1.

Jonathan Chavez



ProblemLet A be an n × n invertible matrix. Show that det(adj A) = (det A)n−1.

Solution

Even if A is not invertible, det(adj A) = (det A)n−1, but the proof is more complicated.

Jonathan Chavez



Cramer’s Rule

If A is an n × n invertible matrix, then the solution to AX = B can be given in terms of determinants ofmatrices.

Theorem (Cramer’s Rule)Let A be an n× n invertible matrix, and consider the system AX = B, where X =

[x1 x2 · · · xn

]T .We define Ai to be the matrix obtained from A by replacing column i with B.Then for each value of i , 1 ≤ i ≤ n,

xi = det Aidet A

Jonathan Chavez



Example (Cramer’s Rule)Solve the following system of linear equations using Cramer’s Rule.

3x1 + x2 − x3 = −15x1 + 2x2 = 2x1 + x2 − x3 = 1

First, x1 = det A1det A , where

A =

[ 3 1 −15 2 01 1 −1

]and A1 =

[ −1 1 −12 2 01 1 −1

].

Computing the determinants of these two matrices,

det A = −4 and det A1 = 4,

and thus x1 = 4−4 = −1.

Jonathan Chavez



Example (continued)Secondly, x2 = det A2

det A where det A = −4 and

det A2 =

∣∣∣∣∣ 3 −1 −15 2 01 1 −1

∣∣∣∣∣ = −14,

and thus x2 = −144 = 7

2 . Finally, x3 = det A3det A , where det A = −4 and

det A3 =

∣∣∣∣∣ 3 1 −15 2 21 1 1

∣∣∣∣∣ = −6,

and thus x3 = −6−4 = 3

2 . Therefore, the solution to the system is given by

X =

[ x1x2x3

]=

−17232

.

You can check this by substituting these values into the original system.Jonathan Chavez



Polynomial InterpolationProblemGiven data points (0, 1), (1, 2), (2, 5) and (3, 10), find an interpolating polynomial p(x) of degree at most three, and thenestimate the value of y corresponding to x = 3

2 .

SolutionWe want to find the coefficients r0, r1, r2 and r3 of

p(x) = r0 + r1x + r2x2 + r3x3

so that p(0) = 1, p(1) = 2, p(2) = 5, and p(3) = 10.

p(0) = r0 = 1p(1) = r0 + r1 + r2 + r3 = 2p(2) = r0 + 2r1 + 4r2 + 8r3 = 5p(3) = r0 + 3r1 + 9r2 + 27r3 = 10

Jonathan Chavez



Solution (continued)Solve this system of four equations in the four variables r0, r1, r2 and r3. 1 0 0 0 1

1 1 1 1 21 2 4 8 51 3 9 27 10

→ · · · → 1 0 0 0 1

0 1 0 0 00 0 1 0 10 0 0 1 0

Therefore r0 = 1, r1 = 0, r2 = 1, r3 = 0, and so

p(x) = 1 + x2.

The estimate for the value of y corresponding to x = 32 is

y = p(3

2

)= 1 +

(32

)2= 13

4 ,

resulting in the point ( 32 , 13

4 ).

Jonathan Chavez



TheoremGiven n data points (x1, y1), (x2, y2), . . . , (xn, yn) with the xi distinct, there is a unique polynomial

p(x) = r0 + r1x + r2x2 + · · · + rn−1xn−1

such that p(xi ) = yi for i = 1, 2, . . . , n. The polynomial p(x) is called the interpolating polynomial for the data points.

To find p(x), set up a system of n linear equations in the n variables r0, r1, r2, . . . , rn−1.

r0 + r1x1 + r2x21 + · · · + rn−1xn−1

1 = y1r0 + r1x2 + r2x2

2 + · · · + rn−1xn−12 = y2

.

.

....

.

.

.

r0 + r1xn + r2x2n + · · · + rn−1xn−1

n = yn

The fact that the xi are distinct guarantees that the coefficient matrix1 x1 x2

1 · · · xn−11

1 x2 x22 · · · xn−1

2...

.

.

....

.

.

....

1 xn x2n · · · xn−1

n

has determinant not equal to zero, and so the system has a unique solution, i.e., there is a unique interpolating polynomialfor the data points.

Jonathan Chavez

mth 215: introduction to linear algebra - chapter...

Documents