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[PDF] 32 Determinants and Matrix Inverses - Math at Emory

In this section, several theorems about determinants are derived p(a1) = 0, so we have p(x)=(x?a1)p1(x) by the factor theorem (see Appendix D)

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(ii) Co-factor of an element aij is given by Aij = (–1)i+j Mij (iii) Value of determinant of a matrix A is obtained by the sum of products of elements of a row

Advanced determinant calculus: A complement - ScienceDirectcom

Why are combinatorialists so fascinated by determinant evaluations? A simplistic answer to this question goes as follows Clearly, binomial coefficients ( n k )

[PDF] Math 20F Linear Algebra Lecture 19 1 Slide 1

Theorem 1 (Main properties of n × n determinants) Let because the factor in the numerator in the right hand side is precisely det(A) = ? Slide 10

[PDF] 1 Determinants: a Row Operation By-Product

reasoning that shows that the product of determinant factors comes out the same no matter This theorem is imporant for all sorts of reasons

A GENERAL THEOREM ON QUANTIC DETERMINANTS*

which contains such factors It is now proposed to examine a direct method for finding the algebraic composition of the determinant M of the coeffi-

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2 1 Determinants by Cofactor Expansion Since a common factor of any row of a matrix can be From Theorem 2 2 4, the determinants of the elementary

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Theorem If A is a square matrix containing a row (or column) of zeros, then det(A) = 0 Proof Use summation of n terms, each term being a product of n factors

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Theorem 4 7 A square matrix A is invertible if and only if det(A) is nonzero This last theorem is one that we use repeatedly in the remainder of this text

[PDF] 36 Proof of the Cofactor Expansion Theorem - mathemoryedu

determinants stated in Theorem 3 1 2 but for rows only (see Lemma 3 6 2) This is clear by property 1 because the row of zeros has a common factor u = 0

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101396_6Determinants.pdf 68

UNIT FIVE

DETERMINANTS

5.1 INTRODUCTION

In unit one the determinant of a

22 matrix was introduced and used in the evaluation of

a cross product. In this chapter we extend the definition of a determinant to any size square matrix. The determinant has a variety of applications. The value of the determinant of a square matrix A can be used to determine whether A is invertible or noninvertible. An explicit formula for A-1 exists that involves the determinant of A. Some systems of linear equations have solutions that can be expressed in terms of determinants.

5.2 DEFINITION OF THE DETERMINANT

Recall that in chapter one the determinant of the 22 matrix A =

22211211aaaa

was defined to be the number

21122211

aaaa and that the notation det (A) or A was used to represent the determinant of A. For any given nn matrix nnij a A, the notation ij A will be used to denote the )1()1(nn submatrix obtained from A by deleting the i th row and the jth column of A.

The determinant of any size square matrix

nnij a A is defined recursively as follows.

Definition of the Determinant Let

nnija A be an nn matrix. (1) If n = 1, that is A = [a 11 ], then we define det (A) = 11 a. (2) If 1k 11 1

1, we define det(A) (-1) det(A )n

kk k na

Example

5A, then by part (1) of the definition of the determinant, det (A) = 5.

If 54
32

A , then by parts (2) and (1), det (A) = (-1)1+1

(2)det[5] + (-1) 1+2 (3)det[4]

= (1)(2)(5) + (-1)(3)(4) = 10 - 12 = -2

198765432

A , then using parts (2) and (1), we calculate the det (A) as follows. 69

9865det)4()1(1875det)3()1(1976det)2()1(det(A)

312111

= (1)(2)(-57) + (-1)(3)(-51) + (1)(4)(-3) = -114 + 153 -12 = 27.

Cofactor

If A is a square matrix, the ij th cofactor of A is defined to be (-1) i+j det(A ij ).

The notation C

ij will sometimes be used to denote the ij th cofactor of A.

Example Let

987654321

A . Then C 11 = (-1) 1+1 det

9865= (1)(45 - 48) = -3 ,

C 12 = (-1) 1+2 det

9764= (-1)(36-42) = 6 and C

23
= (-1) 2+3 det

8721= (-1)(8-14) = 6.

In the definition of the determinant, part (2) consists of multiplying each first row entry of A by its cofactor and then summing these products. For this reason it is called a first row cofactor expansion.

Example Let

5324143542315432

A . Use a first row cofactor expansion to evaluate det(A).

Solution det(A) =

324435231

det)5(41)1(

524135431

det)4(31)1(

534145421

(3)det21)1(

532143423

det)2(11)1 ( =(1)(2)

3243det)4()1(5213det)2()1(5314det)3()1(

312111

+(-1)(3)

3445det)4()1(5415(2)det)1(5314det)1()1(

312111

+(1)(4)

2435det)4()1(5415(3)det)1(5213det)1()1(

312111

+(-1)(5)

2435det)2()1(3445(3)det)1(3243det)1()1(

312111

=(1)(2){(1)(3)(20-3) + (-1)(2)(15-2) + (1)(4)(9-8)} +(-1)(3){(1)(1)(20-3) + (-1)(2)(25-4) + (1)(4)(15-16)} +(1)(4){(1)(1)(15-2) + (-1)(3)(25-4) + (1)(4)(10-12)} +(-1)(5){(1)(1)(9-8) + (-1)(3)(15-16) + (1)(2)(10-12)} =(1)(2){51-26+4} + (-1)(3){17-42-4} + (1)(4){13-63-8} + (-1)(5){1+3-4} =58 + 87 - 232 + 0 = -87 70
Although the definition of the determinant uses a first row cofactor expansion, the determinant of A may be calculated by taking any row (or column) and multiplying the entries of that row (or column) by their cofactors and summing the products. This result is given in the next theorem whose proof is omitted.

Theorem Let A be a square nn matrix, then

n k kjkjjk ikn k ikki aa 11 )det(A)1()det(A(-1)det(A).

Example Let

013102431

A . Evaluate det (A) by (a) a second row cofactor expansion. (b) a third column cofactor expansion.

Solution

(a) det (A) =

1331det)1()1(0341(0)det)1(0143det)2()1(

322212

= (-1)(2)(-4) + (1)(0)(-12) + (-1)(1)(-8) = 8 + 0 + 8 = 16. (b) det (A) =

0231(0)det)1(1331det)1()1(1302det)4()1(

333231

= (1)(4)(2) + (-1)(1)(-8) + (1)(0)(-6) = 8 + 8 + 0 = 16. Theorem If A is a square matrix containing a row (or column) of zeros, then det(A) = 0. Proof Use a cofactor expansion along the row (or column) of zeros. Theorem If A is an nn matrix with two identical rows (or columns), then det (A) = 0. Proof The theorem is certainly true for n = 2 since 0det

12111211

aaaaaaaa. If n = 3, use a cofactor expansion along the row different from the two identical rows.

Let this row be the k

th row. Using a cofactor expansion along this row gives det (A) = (-1) k+1 (a k1 )det(A k1 ) + (-1) k+2 (a k2 )det(A k2 ) + (-1) k+3 (a k3 )det(A k3 ). But each of the submatrices A k1 , A k2 and A k3 has two identical rows so their determinants are 0, hence det (A) = 0 for any 3

3 matrix.

3n proceed as above writing det (A) as a sum of products involving submatrices

with two identical rows whose determinants are 0. i th row cofactor expansion j th column cofactor expansion 71
Triangular and Diagonal Matrices A square matrix is said to be an upper triangular matrix if all the entries below the main diagonal are zero. A square matrix is said to be a lower triangular matrix if all the entries above the main diagonal are zero.

A square matrix is said to be a

diagonal matrix if all entries not on the main diagonal are zero. A diagonal matrix is both upper triangular and lower triangular.

Example

600540321

is an upper triangular matrix.

654032001

is a lower triangular matrix.

300020001

is a diagonal matrix. It is both upper and lower triangular. Theorem If A is upper triangular, lower triangular or diagonal, nn aaa 2211
det(A). Proof Suppose A is upper triangular. To evaluate det(A) use a cofactor expansion along the first column. Since there is only one nonzero entry in the first column the expansion gives det(A) = (-1) 1+1 a 11 det(A 11 ) = a 11 det(A 11 ). Now A 11 is upper triangular so proceed as above to use a cofactor expansion along its first column to get det (A 11 ) = a 22
det(A 22
) where A 22
is A 11 with its first row and first column deleted. Combining the results gives det(A) = a 11 a 22
det(A 22
). Continuing in this fashion, we eventually get det (A) = nn aaa 2211
as required. If A is lower triangular or diagonal, the argument is similar.

Example

70)7)(5)(2(

700650432

det . det

500030002

= (2)(3)(5) = 30

Theorem det(I

n ) = 1 for all n.

Proof Since I

n is a diagonal matrix, det(I n ) = 1)1()1)(1(

Basket-Weave Method

The following method is an alternative way to evaluate the determinant of a 3

3 matrix. This method is only applicable to 33 matrices and is

sometimes called the basket-weave method.

Construct a 3

5 array by writing down the entries of the 33 matrix and then repeating

the first two columns. Calculate the products along the six diagonal lines shown in the diagram. The determinant is equal to the sum of products along diagonals labeled 1, 2 and 3 minus the sum of the products along the diagonals labeled 4, 5 and 6. a 11 a 12 a 13 a 11 a 12 a 21
a 22
a 23
a 21
a 22
a 31
a 32
a 33
a 31
a 32
1 2 3 4 5 6 72
Example Use the basket-weave method to calculate the determinant of A =.

523412321

Solution

1 2 3 1 2 det (A) = (5 + 24 + 12) - (9 + 8 + 20) 2 1 4 2 1 = (41) - (37) = 4 3 2 5 3 2

5.2 PROBLEMS

1. Use the definition of the determinant to evaluate the determinant of the given matrix.

(a)

223132121

(b)

201534312

(c)

235412311

(d)

543012101

(e)

4132342013120101

(f)

2013204023051023

(g)

1320421321401345

(h)

3014013234201203

2. Use the basket-weave method to evaluate the determinants 1(a), 1(b), 1(c) and 1(d).

3. Evaluate the determinants of the following matrices by inspection.

(a)

100650432

(b)

200360953

(c)

231035004

(d)

400030002

4. Let

146532321

A Find the following cofactors of A. (a) C 11 (b) C 12 (c) C 13 (d) C 21
(e) C 22
(f) C 32