Higher Order Determinants

8/3/2019 Higher Order Determinants

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2.2. Higher-Order Determina nts

The 1

1 matrix [a] is invertible exactly when a

= 0.

The 2 2 matrix

a bc d

is invertible exactly when

ad bc = 0. What about a 3 3 matrix? Is there someshort of expression which will determine when such a

matrix is invertible?The answer is yes (and the answer for larger squarematrices is also yes), and it is called the d e t e r m i n a n t of the matrix, but the formula gets very complicated

very fast as the matrix gets bigger; it is easier towork on an individual basis.


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Lets proceed one step at a time, starting with them i n o r s of a matrix A.

The determinant of the matrix obtained by removingall the entries in the ith row and the jth column of Ais called the (i, j) t h m i n o r o f A and is denoted Mi,j.

This assumes you know how to find the determinant

of a smaller matrix!


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If A =

2 4 34 8 33

6 1

, then M1,2, (1, 2)th minor of A, is

the determinant of

2 4 34 8 33

6 1

=

4 33 1

, which is 4 1 3 (3) = 13.

M2,1 is the determinant of

2 4 34 8 33

6 1

=

4 3

6 1

, which is (4) 1 (6) 3 = 14.


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The (i, j) c o f a c t o r o f A, denoted Ci,j, is (1)i+jMi,j.The (1)i+j creates a checkerboard pattern, which

changes the signs of some of the minors:

+ +

+

+ +


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changes the signs of some of the minors:Minors

+ +

+

+ + 26 13 0

14

7 0

36 18 0


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changes the signs of some of the minors:Minors Cofactors

+ +

+

+ + 26 13 0

14

7 0

36 18 0 26 13 0

14

7 0

36 18 0


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+

+

+

+ +

26 13 014

7 0+ 36 18 0

26 13 0

14

7 0+ 36 18 0


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+

+

+

+ +

26+

13 0+

14

7 0

36 18 0

26

13 0

14

7 0

36 + 18 0


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Now (at last) we can find the determinant of A,which is defined to be

A1,1C1,1 + A1,2C1,2 + A1,3C1,3 + + A1,nC1,n.

The determinant of a matrix A is denoted by |A| ordet A.

Here, the determinant is (2)(26)+(4)(13) + (3)(0) =0 .

This method is called E x p a n s i o n b y M i n o r s .


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A more interesting example, with some of the detailsomitted, is:

2 7 31 3 23 7 9

= 2 +

3 27 9

+ 7

1 23 9

+ 3 +

1 33 7

= 2 (3 9 2 7) 7 (1 9 2 3)+ 3 (1 7 3 3)

= 1


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As you can see, calculating the determinant of a 3 3matrix requires calculating the determinant of 3 2 2

matrices.

Calculating the determinant of a 4 4 matrix requirescalculating the determinant of 4 3 3 matrices, eachof which requires the determinants of 3 2

2 matrices.

This makes a total of 12 2 2 matrices.Calculating the determinant of a 5 5 matrix requirescalculating the determinant of 5 4 4 matrices, which

will require the determinant of 60 2 2 matrices.


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In general, the determinant of an n n matrixrequires the determinants of

1

2n! 2 2 matrices. This

is not an efficient procedure! (n! = 1 2 . . . n. 10! isaround 3 million, 70! is bigger than a googol.)

So how can we cut down the computation time?


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First of all, we can expand along any row and getthe same answer, not just the first one. Also, we canexpand along any column. So if some row or columnhas a lot of 0s in it, we can cut down the number ofcomputations.

We need to obey the checkerboard pattern, sothe first determinant might be subtracted insteadof being added.


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3 2 8 04 17 1 20 0 4 0

6 1 7 0


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3 2 8 04 17 1

2

0 0 40

6 1 7 0


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3 2 8 04 17 1 20 0 4 0

6 1 7 0

= 0

4 17 10 0 4

6 1 7

+ 2 +

3 2 80 0 4

6 1 7

+ 0

3 2 84 17 16 1

7

+ 0 +

3 2 84 17 10 0 4

Remember:

+ +

+ ++ +

+

+


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3 2 8 04 17 1 20 0 4 0

6 1 7 0

= 2

3 2 80 0 4

6 1 7


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3 2 8 04 17 1 20 0 4 0

6 1 7 0

= 2

3 2 80 0 4

6 1 7


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3 2 8 04 17 1 20 0 4 0

6 1 7 0

= 2

3 2 80 0 4

6 1 7

= 2

0

2 81 7

+ 0 +

3 86 7

+ 4

3 26 1

Remember: + + +

+ +


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3 2 8 04 17 1 20 0 4 0

6 1 7 0

= 2

3 2 80 0 4

6 1 7

= 2 4

3 26 1


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3 2 8 04 17 1 20 0 4 0

6 1 7 0

= 2

3 2 80 0 4

6 1 7

= 2 4

3 26 1

= 2 4 (3 1 2 6) = 72

Only one 2 2 determinant had to be calculated here!


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Now suppose we have an upper triangular matrix*Maybe we want to find the determinant of:

1 3 2 4 0 2 1 2 30 0 3

2 2

0 0 0 5 10 0 0 0 8

* An u p p e r t r i a n g u l a r m a t r i x is a matrix A whereAi,j = 0 whenever j < i.


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1 3 2 4 0 2 1 2 30 0 3

2 2

0 0 0 5 10 0 0 0 8

= 1

2 1 2 30 3

2 2

0 0 5 10 0 0 8


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1 3 2 4 0 2 1 2 30 0 3

2 2

0 0 0 5 10 0 0 0 8

= 1

2

3

2 2

0 5 1

0 0 8


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1 3 2 4 0 2 1 2 30 0 3

2 2

0 0 0 5 10 0 0 0 8

= 1

2

3

5 1

0 8


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1 3 2 4 0 2 1 2 30 0 3

2 2

0 0 0 5 10 0 0 0 8

= 1

2

3

5

|8

|


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1 3 2 4 0

21 2 3

0 03

2 2

0 0 0 5 10 0 0 0 8

= 1

2

3

5

8

T h e d e t e r m i n a n t o f a n u p p e r ( o r l o w e r ) t r i a n g u l a r m a t r i x

i s t h e p r o d u c t o f t h e e n t r i e s o n t h e d i a g o n a l .

We could use this fact, if we knew how to get amatrix into an upper triangular form . . .

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Higher Order Determinants