MTH 215: Introduction to Linear Algebra

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MTH 215: Introduction to Linear Algebra

Lecture 5

Jonathan A. Ch´ avez Casillas

¹

1University of Rhode Island Department of Mathematics

September 24, 2017

Jonathan Ch´avez

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1

Cofactors and n × n Determinants

2

Triangular Matrices

3

Properties of the Determinant

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Determinant of a 2 × 2 Matrix

Definition

Let A =

a b

c d

. Then thedeterminantof 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 b

c d

, we often write

a b c d

, i.e., usevertical barsinstead ofsquare brackets.

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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.

Definition

Let A = [aij] be an n × n matrix. Thesignof 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.

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The Minor of a Matrix

Definition

Let A = [aij] be an n × n matrix. Theij^th minorof A, denoted as minor (A)_ij, is the determinant of the n − 1 × n − 1 matrix which results from deleting the i^th row and the j^th column of A.

A =







a11 a12 · · · a1j · · · a1n

a21 a22 · · · a2j · · · a2n

.. .

.. . ai 1 ai 2 · · · aij · · · ain

..

. ... ... ... an1 an2 · · · anj · · · ann







For any matrix A, minor (A)_ij is found by first removing thei^th rowandj^th column, and taking the determinant of the remaining matrix.

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The Minor of a Matrix

Example

Find minor (A)12of

A =

"

₁ ₁ ₃

2 4 1

5 2 6

#

Solution

First, remove the1^st rowand2^nd columnfrom A.

A =

"

₁ ₁ ₃

2 4 1

5 2 6

#

The resulting matrix is A =

2 1

5 6

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Solution (continued)

A =

2 1

5 6

Using our previous definition, we can calculate the determinant of this matrix to be (2)(6) − (5)(1) = 12 − 5 = 7

Therefore,minor (A)12= 7.

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The Cofactors of a Matrix

Definition

Theij^thcofactorof A is

cof(A)_ij= (−1)^{i +j}minor (A)_ij

Example (continued)

Let

A =

"

₁ ₁ ₃

2 4 1

5 2 6

#

Find cof(A)12.

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Solution

By 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

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Cofactor Expansion

Using these definitions, we can now define thedeterminant of the n × n matrix A:

Definition

det A = a11cof(A)11+ a12cof(A)12+ a13cof(A)13+ · · · + a1ncof(A)1n

This is called thecofactor expansion of det A along row 1.

In other words,

det (A) =

n

X

j=1

aijcof (A)_ij=

n

X

i =1

aijcof (A)_ij

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

Cofactor expansion is also calledLaplace Expansion.

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Cofactor Expansion

Example

Let A =

"

1 1 3

2 4 1

5 2 6

#

. Find det A.

Solution

Using 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

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Solution (continued)

A =

"

₁ ₁ ₃

2 4 1

5 2 6

#

Now try cofactor expansion along column 2.

det A = 1cof₁₂(A) + 4cof₂₂(A) + 2cof₃₂(A)

= 1(−1)³

2 1 5 6

+ 4(−1)⁴

1 3 5 6

+ 2(−1)⁵

1 3 2 1

= −1(12 − 5) + 4(6 − 15) − 2(1 − 6)

= −(7) + 4(−9) − 2(−5)

= −7 − 36 + 10

= −33

We get the same answer!

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The Determinant is Well Defined

Theorem

The determinant of an n × n matrix A can be computed using cofactor expansion alongany row or column of 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 the opportunity to save ourselves some work!

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Problem

Let A =







0 1 −2 1

5 0 0 7

0 1 −1 0

3 0 0 2







. Find det A.

Solution

Cofactor expansion along row 1 yields

det A = 0 × cof(A)11+ 1 × cof(A)12+ (−2) × cof(A)13+ 1 × cof(A)14

= cof(A)12− 2 × cof(A)13+ cof(A)14, whereas cofactor expansion along, row 3 yields

det A = 0 × cof(A)31+ 1 × cof(A)32+ (−1) × cof(A)33+ 0 × cof(A)34

= 1cof(A)32+ (−1)cof(A)33,

so, in the first case we must computethreecofactors, but justtwocofactors in the second case.

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Solution (continued)

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

A =







0 1 −2 1

5 0 0 7

0 1 −1 0

3 0 0 2







det A = 1 × cof(A)32+ (−1) × cof(A)33

= 1 × (−1)⁵

0 −2 1

5 0 7

3 0 2

+ (−1) × (−1)⁶

0 1 1

5 0 7

3 0 2

= −

0 −2 1

5 0 7

3 0 2

−

0 1 1

5 0 7

3 0 2

Each of the two determinants above can easily be evaluated usingcofactor expansion along column 2.

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Solution (continued)

det A = −

0 −2 1

5 0 7

3 0 2

−

0 1 1

5 0 7

3 0 2

= −(−2)(−1)³

5 7 3 2

− 1(−1)³

5 7 3 2

= −2(10 − 21) + 1(10 − 21)

= −2(−11) + (−11)

= 22 − 11

= 11.

Therefore, det A = 11.

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A Row or Column of Zeros

Example

Let

A =







−8 1 0 −4

5 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 Fact

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

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Determinants of a Triangular Matrices

Definitions

1 An n × n matrix A is calledupper triangularif all entries

below

the main diagonal are zero.

2 An n × n matrix A is calledlower triangularif all entries

above

the main diagonal are zero.

3 An n × n matrix A is calledtriangularif it is upper triangular or lower triangular.

Theorem

If 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.

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Example

det

"

₁ ₂ ₃

0 5 6

0 0 9

#

= 1 × det

5 6

0 9

= 1 × 5 × det

9

= 1 × 5 × 9

= 45.

Notice that 45 is the product of the entries on the main diagonal.

"

₁ ₂ ₃

0 5 6

0 0 9

#

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Elementary Row Operations and Determinants

Example

Let

A =

"

₂ ₀ ₋₃

0 4 0

1 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.

LetB1,B2, andB3be obtained from A byinterchanging rows 1 and 2,multiplying row 3 by −3, andadding twice row 1 to row 3, respectively, i.e.,

B1=

"

2 0 −3 1 0 −2

0 4 0

#

, B2=

"

2 0 −3

0 4 0

−3 0 6

#

, B3=

"

2 0 −3

0 4 0

5 0 −8

#

.

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

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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.

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Theorem

Let 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. Then det (B) = k det (A) .

3 Let B be a matrix which results from adding a multiple of a row to another row. Then det (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 forelementary column operation. If A is a matrix, then anelementary column operationon A is simply the corresponding elementary row operation performed on the transpose of A, A^T.

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Computing the Determinant

Example

det

"

₋₃ ₅ ₋₆

1 −1 3

2 −4 1

#

=

0 2 3

1 −1 3

2 −4 1

row 1 + 3×(row 2)

=

0 2 3

1 −1 3

0 −2 −5

row 3 − 2×(row 2)

= (1)(−1)²⁺¹

2 3

−2 −5

cofactor expansion: column 1

= −(−10 + 6)

= 4.

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Example

det





3 1 2 4

−1 −3 8 0

1 −1 5 5

1 1 2 −1





⁼

7 5 10 0

−1 −3 8 0

6 4 15 0

1 1 2 −1





= (−1)(−1)⁸

7 5 10

−1 −3 8

6 4 15

= −

0 −16 66

−1 −3 8

0 −14 63

= −(−1)(−1)³

−16 66

−14 63

= −

−2 3

−14 63

= −(−126 + 42)

= 84.

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Problem

If det

"

a1 a2 a3

b1 b2 b3

c1 c2 c3

#

= 4, find det

"

−b1 −b2 −b3

a1+ 2b1 a2+ 2b2 a3+ 2b3

3c1 3c2 3c3

#

.

Solution

−b₁ −b₂ −b₃

a1+ 2b1 a2+ 2b2 a3+ 2b3

3c1 3c2 3c3

= (−1)(3)

b1 b2 b3

a1+ 2b1 a2+ 2b2 a3+ 2b3

c1 c2 c3

= (−3)

b1 b2 b3

a1 a2 a3

c1 c2 c3

= (−3)(−1)

a1 a2 a3

b1 b2 b3

c1 c2 c3

= (−3)(−1) × 4

= 12.

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Problem

Let A =

"

₂ ₃ ₅

3 5 9

1 2 4

#

. Find det A.

Solution

det A =

2 3 5

3 5 9

1 2 4

=

3 5 9

1 2 4

=

0 0 0

3 5 9

1 2 4

= 0.

Notice:

row1 + row3 − row2 =

0 0 0 Hence the determinant equals 0.

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Problem

Let

A =

"

_a _b _c

p q r

x y z

#

and B =

"

_{2a + p} _{2b + q} _{2c + r} 2p + x 2q + y 2r + z 2x + a 2y + b 2z + c

#

.

Show that det B = 9 det A.

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Solution

det B =

2a + p 2b + q 2c + r 2p + x 2q + y 2r + z 2x + a 2y + b 2z + c

R1+(−2)R3

−−−−−−−→

p − 4x q − 4y r − 4z 2p + x 2q + y 2r + z 2x + a 2y + b 2z + c

R₂+(−2)R₁

−−−−−−−→

p − 4x q − 4y r − 4z

9x 9y 9z

2x + a 2y + b 2z + c

1 9R2

−−→ 9

p − 4x q − 4y r − 4z

x y z

2x + a 2y + b 2z + c

R1+(4)R2

−−−−−→ 9

p q r

x y z

2x + a 2y + b 2z + c

R3+(−2)R2

−−−−−−−→ 9

p q r

x y z

a b c

R1<−>R3

−−−−−−→ −9

a b c

x y z

p q r

R2<−>R3

−−−−−−→ 9

a b c

p q r

x y z

= 9 det A.

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Determinants and Scalar Multiplication

Problem

Suppose A is a 3 × 3 matrix with det A = 7. What is det(−2A)?

Solution

Write A =

"

a11 a12 a13

a21 a22 a23

a31 a32 a33

#

. Then −2A =

"

−2a11 −2a12 −2a13

−2a21 −2a22 −2a23

−2a₃₁ −2a₃₂ −2a₃₃

#

.

det(−2A) =

−2a11 −2a12 −2a13

−2a₂₁ −2a₂₂ −2a₂₃

−2a31 −2a32 −2a33

= (−2)

a11 a12 a13

−2a₂₁ −2a₂₂ −2a₂₃

−2a31 −2a32 −2a33

= (−2)(−2)

a11 a12 a13

a21 a22 a23

−2a31 −2a32 −2a33

= (−2)(−2)(−2)

a11 a12 a13

a21 a22 a23

a31 a32 a33

= (−2)³det A = (−8) × 7 = −56.

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Solution (continued)

Think about the matrix −2A as the matrix obtained from A be multiplyingeach of its rows by −2. This involvesthree elementary row operations, each of which contributes a factor of −2 to the determinant, and thusdet A = (−2) × (−2) × (−2) × 7 = (−2)³× 7.

Theorem

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

det(kA) = kⁿdet A.

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Determinants of Inverses

Theorem

An n × n matrix A is invertible if and only ifdet A , 0. In this case, det(A⁻¹) = 1

det A.

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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 −₁₀¹. We can verify this directly.

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

A⁻¹= 1

−10

4 3

6 2

.

Therefore, det A⁻¹=

−1 10

²

4 3 6 2

=

−1 10

²

(8 − 18) = 1

100× (−10) = −1 10.

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Problem

Find all values of c for which A =

"

_c ₁ ₀

0 2 c

−1 c 5

#

is invertible.

Solution

det A =

c 1 0

0 2 c

−1 c 5

= c

2 c c 5

+ (−1)

1 0 2 c

= c(10 − c²) − c

= c(9 − c²)

= c(3 − c)(3 + c).

Since A is invertiblewhen det(A) , 0, A is invertible for all c , 0, 3, −3.

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Determinants of Products and Transposes

Theorem

Let A and B be n × n matrices. Then

det (AB) = det A × det B.

Theorem

If A is an n × n matrix, then the determinant of its transpose is given by det(A^T) = det A.

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Problem

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

det A = −1, det B = 2, and det C = 1.

Find det(2A²(B⁻¹)(C^T)³B(A⁻¹)).

Solution

det(2A²(B⁻¹)(C^T)³B(A⁻¹)) = 2⁴(det A)² 1

det B(det C )³(det B) 1 det A

= 16(det A)(det C )³

= 16 × (−1) × 1³

= −16.

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Problem

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

det(2A⁻¹) = −4 = det(A³(B⁻¹)^T).

Solution

First,

det(2A⁻¹) = −4 2³det(A⁻¹) = −4

1

det A = −4 8 = −1

2. Therefore, det A = −2. Using this fact,

det(A³(B⁻¹)^T) = −4 (det A)³det(B⁻¹) = −4 (−2)³det(B⁻¹) = −4 (−8) det(B⁻¹) = −4

1

det B = −4

−8 =1 2

Therefore, det B = 2.

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Problem

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

Solution

Since det(−A) = (−1)ⁿdet A, det(−A) = det A if and only if (−1)ⁿdet A = det A.

When is this possible?

1 (−1)ⁿdet A = det A whenever det A = 0.

2 If det A , 0, then (−1)ⁿdet A = det A only if (−1)ⁿ= 1, i.e., only if n is even.

Therefore, det(−A) = det A only ifdet A = 0 or n is even.

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MTH 215: Introduction to Linear Algebra