DETERMINANTS BY ROW AND COLUMN EXPANSION - University of New Mexico

[Pages:4]DETERMINANTS BY ROW AND COLUMN EXPANSION

TERRY A. LORING

1. EXPANDING ALONG A ROW OR COLUMN

The book does a good job explaining how you can expand on a row. Please see the book.

You can expand on any column or row. You use a checkerboard pattern to figure the signs:

+-+- -+-+ +-+- -+-+

and the "cross out" a row and a columns, several times, to write the big determinant as a sum/difference of many (many) smaller determinants.

Again, see the book.

2. AN EXAMPLE Let's compute det(A) in several ways, where

0 1 4 32

A

=

3

6

1 0

4 2

40

32

3 1 6 60

1

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TERRY A. LORING

2.1. Expand on the second row (twice).

0 1 4 32 3 1 4 40 6 0 2 32 3 1 6 60

X0 1 4 32

0 X1 4 32

0 1 X4 32

0 1 4 X3X2

= -3

X3 X6

X1 0

X4 2

X4X0 32

+

X3 6

X1 X0

X4 2

X4X0 32

-4

X3 6

X1 0

X4 X2

X4X0 32

+ 40

X3 6

X1 0

X4 2

4XX0 X1X32

X3 1 6 60

3 X1 6 60

3 1 X6 60

3 1 6 X6X0

1 4 32

0 4 32

0 1 32

014

= -3 0 2 32 + 6 2 32 - 4 6 0 32 + 40 6 0 2

1 6 60

3 6 60

3 1 60

316

1 X4 32

1 4 X3X2

= -3 2 X0 X1 X1X6 - 32 X0 X1 X1X6

1 X6 60

1 6 X6XX0

X0 4 32

0 X4 32

0 4 X3X2

+ -6 X3 X1 X1X6 + 2 X3 X1 X1X6 - 32 X3 X1 X1X6

X3 6 60

3 X6 60

3 6 X6X0

X0 1 32

0 1 X3X2

-4 -6 X3 X0 X1X6 - 32 X3 X0 X1X6

X3 1 60

3 1 X6X0

X0 1 4

0 1 X4

+40 -6 X3 X0 X1 - 2 X3 X0 X1

X3 1 6

3 1 X6

= -3

2

1 1

32 60

- 32

1 1

4 6

+

-6

4 6

32 60

+2

0 3

32 60

- 32

0 3

4 6

1 32

01

-4 -6 1 60 - 32 3 1

+ 40

-6

1 1

4 6

-2

0 3

1 1

= -3 (2(28) - 32(2)) + (-6(48) + 2(-96) - 32(-12)) -4 (-6(28) - 32(-3)) + 40 (-6(2) - 2(-3))

= -3 (-8) + (-96) - 4 (-72) + 40 (-6) = -24

It is not customary to write down the matrices with the crossed out row and column. I just thought one complete example would help you.

2.2. Mixing Row and Column Operations with Expansion. Column operations work just like row operations for determinants. So if all you want is the determinant, and you see patterns in the columns, take advantage. The idea is to create lots of zeros so expanding is not so painful.

DETERMINANTS BY ROW AND COLUMN EXPANSION

3

In this computation, I do:

?

a

type

II

column

operation

(

1 3

C1

C1)

? a type III row operation

? type III column operation

? expand along the second row

? expand along the second row:

0 1 4 32 3 1 4 40 6 0 2 32 3 1 6 60

0 1 4 32

=

3

1 2

1 0

4 2

40 32

1 1 6 60

0 1 4 32

=

3

1 2

0 0

0 2

8 32

1 1 6 60

0 1 4 32

=

3

1 2

0 0

0 2

0 16

1 1 6 52

1 4 32

= -3 0 2 16 1 6 52

= -3

2 6

16 52

+

4 2

32 16

= -3 (8 - 0)) = -24

2.3. Row ops to upper triangular. There is no rule that says you must write down each row operation. By not doing so, you shorten your write-up. The trade-off is that it is harder to follow and harder to find errors.

Here I do a bunch of row operations, keeping a running total (product) of the factors introduced by each row operation. Notice I start by swapping two rows to get a nonzero number at the top left for a pivot, then divide that by two so it is

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TERRY A. LORING

easy to use type I operations to create all zeros below that first pivot. Etc.

0 1 4 32 3 1 4 40 6 0 2 32 3 1 6 60

6 0 2 32

=

-

3 0

1 1

4 4

40 32

3 1 6 60

3 0 1 16

=

-2

3 0

1 1

4 4

40 32

3 1 6 60

3 0 1 16 0 1 3 24 = -2 0 1 4 32 0 1 5 44

E-mail address: loring@math.unm.edu UNIVERSITY OF NEW MEXICO

3 0 1 16

=

-2

0 0

1 0

3 1

24 8

0 0 2 20

3 0 1 16 0 1 3 24 = -2 0 0 1 8 000 4

= -2 ? 3 ? 1 ? 1 ? 4 = -24.

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