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Chapter 7

The Determinant via Elimination

The fast way to compute the determinant of a large matrix is via elimination.

The fast formula for the determinant

Theorem 7.1. Via forward elimination,

 

det A =

± (product of the pivots) if there is a pivot in each column,

 

 

0

otherwise.

where

 

 

±

= + if we make an even number of row swaps during forward elimination,

otherwise.

 

 

 

In particular, A is invertible just when det A 6= 0.

Forward elimination takes

 

 

!

 

!

A = 0

7

to U =

2

3

2

3

 

0

7

with one row swap so det A = −(2)(7) = −14.

The fast formula isn’t actually any faster for small matrices, so for a 2 × 2 or 3 × 3 you wouldn’t use it. But we need the fast formula anyway; each of the two formulas gives di erent insight.

Proof. We can see how the determinant changes during elimination: adding multiples of rows to other rows does nothing, swapping rows changes sign.

7.1 Use the fast formula to find the determinant of

2 5 5

A = 2 5

7

 

 

 

 

2 6 11

53

54

 

 

 

 

 

 

The Determinant via Elimination

7.2 Just by looking, find

 

 

 

 

 

 

 

 

1001

 

1002

1003

 

1004

2002

 

2004

2006

 

2008

 

 

 

 

 

 

 

 

 

det 2343

 

6787

1938

 

4509 .

 

 

 

 

 

 

 

 

 

9873

 

7435

2938

 

9038

7.3 Prove that a square matrix is invertible just when its determinant is not

zero.

 

 

 

 

 

 

 

 

Review problems

 

 

 

 

 

 

 

 

7.4 Find the determinant of

 

 

 

 

 

 

 

 

1

0

0

 

 

 

0

1

0

1

 

 

−1

 

0

1

−1

 

 

−1

 

 

 

 

 

 

 

 

 

1

0

1

 

 

 

0

 

 

 

 

 

 

 

 

 

7.5 Find the determinant of

01

 

 

 

 

 

0

 

 

1

0

 

 

 

 

 

2

2

 

 

0

 

 

−1

0

 

 

1

−1

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

0

1

 

 

1

 

 

 

 

 

 

 

 

 

7.6 Find the determinant of

 

 

1

 

 

 

 

−1

 

0

 

2

 

1

−1

 

2

 

1

 

 

1

 

 

 

 

7.7 Find the determinant of

0

 

 

1

 

 

 

0

 

 

0

 

2

0

 

 

 

 

 

2

 

2

1

 

7.8 Find the determinant of

 

 

 

 

 

2

 

 

 

 

 

 

 

0

2

 

 

2

 

1

−1

 

 

 

0

 

2

1

 

7.9 Find the determinant of

 

 

 

 

 

 

0

 

−1

2

 

 

 

 

 

 

 

 

0

 

−1

−1

 

 

 

0

 

1

0

 

 

 

 

 

 

 

 

 

 

 

Determinants Multiply

55

7.10Prove that a square matrix with a zero row has determinant 0.

7.11Prove that det P A = (−1)N det A if P is the permutation matrix of a product of N transpositions.

7.12Use the fast formula to find the determinant of

0 2 1

A = 3

1 2

 

 

3 5 2

7.13 Prove that the determinant of any lower triangular square matrix

 

L21

L22

 

 

 

 

 

 

L11

 

 

 

 

 

 

L =

L31

L32

L33

 

 

 

 

 

L42

.

..

 

 

 

 

L41

L43

 

 

 

 

 

 

 

 

 

 

 

 

.

.

.

.

.

..

 

 

..

..

..

..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ln1

Ln2

Ln3 . . .

Ln(n

1)

 

 

 

 

 

 

 

Lnn

(with zeroes above the diagonal) is the product of the diagonal terms: det L =

L11L22 . . . Lnn.

Determinants Multiply

Theorem 7.2. det (AB) = det(A) det(B), for any square matrices A and B of the same size.

Proof. Suppose that det A = 0. By the fast formula, A is not invertible. Problem 5.10 on page 43 tells us that therefore AB is not invertible, and both det(AB) and det(A) det(B) are 0. So we can safely suppose that det A 6= 0. Via Gauss-Jordan elimination, any invertible matrix is a product of matrices each of which adds a multiple of one row to another, or scales a row, or swaps two rows. Write A as a product of such matrices, and peel o one factor at a time, applying lemma 6.4 on page 48 and proposition 6.7 on page 51.

If

0

2

5

, B =

2

2

0

,

A =

 

1

4

6

 

1

0

0

 

 

0

0

3

 

7

5

4

 

 

 

 

 

 

 

 

 

 

then it is hard to compute out AB, and then compute out det AB. But det AB = det A det B = (1)(2)(3)(1)(2)(4) = 48.

56 The Determinant via Elimination

Transpose

The transpose of a matrix A is the matrix At whose entries are Atij = Aji (switching rows with columns). Flip over the diagonal:

 

 

10

2

 

, At = 10

 

5 .

A =

3

40

3

 

 

5

6

 

2

40

6!

 

 

 

 

 

 

 

 

7.14

Find the transpose of

4

 

6 .

 

A =

5

 

 

1

2

3

 

 

0

0

0

7.15

Prove that

 

 

 

 

(AB)t = BtAt.

(The transpose of the product is the product of the transposes, in the reverse order.)

7.16 Prove that the transpose of any permutation matrix is a permutation matrix. How is the permutation of the transpose related to the original permutation?

Corollary 7.3.

det A = det At

Proof. Forward elimination gives U = V A, U upper triangular and V a product of permutation and strictly lower triangular matrices. Tranpose: Ut = AtV t. But V t is a product of permutation and strictly upper triangular matrices, with

the same number of row

swaps as V , so det V t = det V = 1. The matrix Ut

 

t

±

t

(by

is lower triangular, so det U

 

is the product of the diagonal entries of U

 

problem 7.13 on the preceding page), which are the diagonal entries of U, so det Ut = det U.

Expanding Down Any Column or Across Any Row

Consider the “checkerboard pattern”

+

+

. . .

+

+ . . ..

.

.

 

 

.

.

 

 

.

.

 

 

Theorem 7.4. We can compute the determinant of any square matrix A by picking any column (or any row) of A, writing down plus and minus signs from the same column (or row) of the checkboard pattern matrix, writing down the

Summary

57

entries of A from that column (or row), multiplying each of these entries by the determinant obtained from deleting the row and column of that entry, and adding all of these up.

For

A =

 

3

2

1

 

 

 

1

4

5

 

,

 

6

7

2

 

 

 

 

 

 

 

 

if we expand along the second row, we get

 

 

3

2

1

 

 

 

 

 

det A = − (1) det

1

4

5

 

 

 

 

 

 

 

 

 

 

6

7

2

 

 

 

 

 

 

 

3

2

1

 

+ (4) det

1

4

5

 

 

 

 

 

 

6 7 2

 

 

3

2

1

 

− (5) det

1

4

5

 

 

 

 

 

 

6 7 2

Proof. By swapping columns (or rows), we change signs of the determinant. Swap columns (or rows) to get the required column (or row) to slide over to become the first column (or row). Take the sign changes into account with the checkboard pattern: changing all plus and minus signs for each swap.

7.17 Use this to calculate the determinant of

 

 

 

 

 

3

 

4

0

0

 

 

1

 

2

0

1

.

A =

0

 

0

0

2

 

 

 

 

 

 

 

 

 

 

 

1702

1

 

 

 

839

 

493

Summary

Determinants

(a)scale when you scale across a row (or down a column),

(b)add when you add across a row (or down a column),

(c)switch sign when you swap two rows, (or when you swap two columns),

(d)don’t change when you add a multiple of one row to another row (or a multiple of one column to another column),

58

The Determinant via Elimination

(e)don’t change when you transpose,

(f)multiply when you multiply matrices. The determinant of

(a)an upper (or lower) triangular matrix is the product of the diagonal entries.

(b)a permutation matrix is (−1)# of transpositions.

(c)a matrix is not zero just when the matrix is invertible.

(d)any matrix is det A = (−1)N det U, if A is taken by forward elimination with N row swaps to a matrix U.

7.18If A is a square matrix, prove that

det Ak = (det A)k

for k = 1, 2, 3, . . . .

7.19 Use this last exercise to find

det A2222444466668888

where

A =

1

1234567890!.

 

0

1

7.20 If A is invertible, prove that

det A−1 = det1 A.

Review problems

7.21What are all of the di erent ways you know to calculate determinants?

7.22How many solutions are there to the following equations?

x1 + 1010x2 + 130923x3 = 2839040283 2x2 + 23932x3 = 2390843248 3x3 = 98234092384

7.23 Prove that no matter which entry of an n × n matrix you pick (n > 1), you can find some invertible n × n matrix for which that entry is zero.

Bases and Subspaces

59

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