Elementary Linear AlgebraAnton & Rorres, 9th Edition

Lecture Set – 02Chapter 2: Determinants

2

Chapter Content

Determinants by Cofactor Expansion Evaluating Determinants by Row Reduction Properties of the Determinant Function A Combinatorial Approach to Determinants

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2-1 Minor and Cofactor

Definition Let A be mn

The (i,j)-minor of A, denoted Mij is the determinant of the (n-1) (n-1) matrix formed by deleting the ith row and jth column from A

The (i,j)-cofactor of A, denoted Cij, is (-1)i+j Mij

Remark Note that Cij = Mij and the signs (-1)i+j in the definition of

cofactor form a checkerboard pattern:

...

...

...

...

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2-1 Example 1

Let

The minor of entry a11 is

The cofactor of a11 is C11 = (-1)1+1M11 = M11 = 16

Similarly, the minor of entry a32 is

The cofactor of a32 is C32 = (-1)3+2M32 = -M32 = -26

8 4 1

6 5 2

4 1 3

A

168 4

6 5

8 4 1

6 5 2

4 1 3

11

M

266 2

4 3

8 4 1

6 5 2

4 1 3

32

M

5

2-1 Cofactor Expansion The definition of a 3×3 determinant in terms of minors and

cofactors det(A) = a11M11 +a12(-M12)+a13M13

= a11C11 +a12C12+a13C13

this method is called cofactor expansion along the first row of A

Example 2

10)11)(1()4(3 4 5

4- 20

2 5

3 2 1

2 4

3 43

2 4 5

3 4 2

0 1 3

)det(

A

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2-1 Cofactor Expansion det(A) =a11C11 +a12C12+a13C13 = a11C11 +a21C21+a31C31

=a21C21 +a22C22+a23C23 = a12C12 +a22C22+a32C32

=a31C31 +a32C32+a33C33 = a13C13 +a23C23+a33C33

Theorem 2.1.1 (Expansions by Cofactors) The determinant of an nn matrix A can be computed by multiplying the

entries in any row (or column) by their cofactors and adding the resulting products; that is, for each 1 i, j n

det(A) = a1jC1j + a2jC2j +… + anjCnj

(cofactor expansion along the jth column)

and

det(A) = ai1Ci1 + ai2Ci2 +… + ainCin

(cofactor expansion along the ith row)

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2-1 Example 3 & 4 Example 3

cofactor expansion along the first column of A

Example 4 smart choice of row or column

det(A) = ?

1)3(5)2)(2()4(3 3 4

0 1 5

2 4

0 1 )2(

2 4

3 43

2 4 5

3 4 2

0 1 3

)det(

A

1002

12-01

2213

1-001

A

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2-1 Adjoint of a Matrix If A is any nn matrix and Cij is the cofactor of aij, then the

matrix

is called the matrix of cofactors from A. The transpose of this matrix is called the adjoint of A and is denoted by adj(A)

Remarks If one multiplies the entries in any row by the corresponding

cofactors from a different row, the sum of these products is always zero.

nnnn

n

n

C CC

C CC

C CC

21

22221

11211

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2-1 Example 5 Let

a11C31 + a12C32 + a13C33 = ?

Let

333231

232221

131211

aaa

aaa

aaa

A

131211

232221

131211

'

aaa

aaa

aaa

A

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2-1 Example 6 & 7

Let

The cofactors of A are: C11 = 12, C12 = 6, C13 = -16, C21 = 4, C22 = 2, C23 = 16, C31 = 12, C32 = -10, C33 = 16

The matrix of cofactor and adjoint of A are

The inverse (see below) is

042

361

123

A

16 10 12

16 2 4

16 6 12

16 16 16

10 2 6

12 4 12

)(adj A

16 16 16

10 2 6

12 4 12

64

1)(adj

)det(

11 AA

A

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Theorem 2.1.2 (Inverse of a Matrix using its Adjoint) If A is an invertible matrix, then

Show first that )(adj

)det(

11 AA

A

)det(2211 ACaCaCa ininiiii

IAAA )det()(adj

)(02211 jiCaCaCa jninjiji

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Theorem 2.1.3 If A is an n × n triangular matrix (upper triangular, lower

triangular, or diagonal), then det(A) is the product of the entries on the main diagonal of the matrix; det(A) = a11a22…ann

E.g.

44434241

333231

2221

11

0

00

000

aaaa

aaa

aa

a

A

?

40000

89000

67600

15730

38372

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2-1 Prove Theorem 1.7.1c A triangular matrix is invertible if and only if its diagonal

entries are all nonzero

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2-1 Prove Theorem 1.7.1d The inverse of an invertible lower triangular matrix is

lower triangular, and the inverse of an invertible upper triangular matrix is upper triangular

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Theorem 2.1.4 (Cramer’s Rule) If Ax = b is a system of n linear equations in n unknowns such

that det(I – A) 0 , then the system has a unique solution. This solution is

where Aj is the matrix obtained by replacing the entries in the column of A by the entries in the matrix b = [b1 b2 ··· bn]T

)det(

)det(,,

)det(

)det(,

)det(

)det( 22

11 A

Ax

A

Ax

A

Ax n

n

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2-1 Example 9 Use Cramer’s rule to solve

Since

Thus,

832

30643

62

321

321

31

xxx

xxx

x x

8 2 1

30 4 3

6 0 1

,

3 8 1

6 30 3

2 6 1

,

3 2 8

6 4 30

2 0 6

,

3 2 1

6 4 3

2 0 1

321 AAAA

11

38

44

152

)det(

)det(,

11

18

44

72

)det(

)det(,

11

10

44

40

)det(

)det( 33

22

11

A

Ax

A

Ax

A

Ax

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Exercise Set 2.1 Question 13

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Exercise Set 2.1 Question 23

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Exercise Set 2.1 Question 2 7