Post on 30-Dec-2015
description
Recap
Row and Reduced Row Echelon Elementary Matrices
If m and n are positive integers, then an m n matrix is a rectangular array in which each entry aij of the matrix is a number. The matrix has m rows and n columns.
a1,1 a1,2 a1,3 a1,na2,1 a2, 2 a2,3 a2,na3,1 a3,2 a3,3 a3, n am ,1 am,2 am,3 am ,n
A real matrix is a matrix all of whose entries are real numbers.
i (j) is called the row (column) subscript.
An mn matrix is said to be of size (or dimension) mn.
If m=n the matrix is square of order n.
The ai,i’s are the diagonal entries.
Given a system of equations we can talk about its coefficient matrix and its augmented matrix.
These are really just shorthand ways of expressing the information in the system.
To solve the system we can now use row operations instead of equation operations to put the augmented matrix in row echelon form.
1. Interchange two rows.
2. Multiply a row by a nonzero constant.
3. Add a multiple of a row to another row.
Two matrices are said to be row equivalent if one can be obtained from the other using elementary row operations.
A matrix is in row-echelon form if:› All rows consisting entirely of zeros are
at the bottom.› In each row that is not all zeros the first
entry is a 1.› In two successive nonzero rows, the
leading 1 in the higher row is further left than the leading 1 in the lower row.
1. Write the augmented matrix of the system.
2. Use elementary row operations to find a row equivalent matrix in row-echelon form.
3. Write the system of equations corresponding to the matrix in row-echelon form.
4. Use back-substitution to find the solutions to this system.
In Gauss-Jordan elimination, we continue the reduction of the augmented matrix until we get a row equivalent matrix in reduced row-echelon form. (r-e form where every column with a leading 1 has rest zeros)1 0 0 a
0 1 0 b
0 0 1 c
A system of linear equations in which all of the constant terms is zero is called homogeneous.
All homogeneous systems have the solutions where all variables are set to zero. This is called the trivial solution.
New Stuff
Using Elementary Matrices
An n by n matrix is called an elementary matrix if it can be obtained from In by a single elementary row operation.
These matrices allow us to do row operations with matrix multiplication.
Theorem: Let E be the elementary matrix obtained by performing an elementary row operation on In. If that same row operation is performed on an m by n matrix A, then the resulting matrix is given by the product EA.
Three types of Elementary Matrices
These correspond to the three types of EROs that we can do:› Interchanging rows of I -> Type I EM› Multiplying a row of I by a constant -> Type
II EM› Adding a multiple of one row to another ->
Type III EM
Type I EM
E1 =
How is this created? Eg. 1 Suppose
A =
E1A =
=
What is AE1?
100
001
010
333231
232221
131211
aaa
aaa
aaa
100
001
010
333231
232221
131211
aaa
aaa
aaa
333231
131211
232221
aaa
aaa
aaa
Type II EM
E2 =
E2A = =
AE2 = =
300
010
001
333231
232221
221211
333 aaa
aaa
aaa
300
010
001
333231
232221
221211
aaa
aaa
aaa
333231
232221
221211
aaa
aaa
aaa
300
010
001
333231
232221
131211
3
3
3
aaa
aaa
aaa
Type III EM
E3 =
E3A = =
AE3 = =
100
010
301
333231
232221
221211
aaa
aaa
aaa
100
010
301
333231
232221
331332123111 333
aaa
aaa
aaaaaa
333231
232221
221211
aaa
aaa
aaa
100
010
301
33313231
23212221
13111211
3
3
3
aaaa
aaaa
aaaa
Let A and B be m by n matrices. Matrix B is row equivalent to A if there exists a finite number of elementary matrices E1, E2, ... Ek such that
B = EkEk-1 . . . E2E1A.
Break it down
This means that B is row equivalent to A if B can be obtained from A through a series of finite row operations.
If we then take two augmented matrices (A|b) and (B|c) and they are row equivalent, then Ax = b and Bx=c must be equivalent series
Break it down
If A is row equivalent to B, B is row equivalent to A
If A is row equivalent to B and B is row equivalent to C then A is row equivalent to C
Example
Compute the inverse of A for A =
322
021
341
100
010
001
|
322
021
341
102
011
001
|
360
320
341
13121
21
21
21
23
21
|
600
020
041
131
011
001
|
600
320
341
Example
Now, solve the system:
61
21
61
41
41
41
21
21
21
|
100
010
001
13121
21
21
21
21
21
|
600
020
001
8322
122
1234
321
21
321
xxx
xx
xxx
Example
We can employ the format Ax = b so x=A-
1b
We just calculated A-1 and b is the column vector
So we can easily find the values of x by multiplying the two matrices
8
12
12
Determinants
Keys to calculating Inverses
Require square matrices Each square matrix has a determinant
written as det(A) or |A| Determinants will be used to:
› characterize on-singular matrices› express solutions to non-singular systems › calculate dimension of subspaces
If A and B are square then
It is not difficult to appreciate that
If A has a row (or column) of zeros then
If A has two identical rows (or columns) then
BAAB
AAT
0A0
353
202
171
21 CC
If B is obtained from A by ERO,
interchanging two rows (or columns) then
If B is obtained from A by ERO where row (or column) of A were multiplied by a scalar k, then
)( jiji CCRR AB
AkB
If B is obtained from A by ERO where a multiple of a row (or column) of A were added to another row (or column) of A then BA
211222112221
1211 aaaaaa
aaA
That Is the determinant is equal to the product of the elements along the diagonal minus the product of the elements along the off-diagonal.
51
23A
0)6)(1()2)(3( A
Note: The matrix A is said to be invertible or non-singular if det(A)≠ 0. If det(A) = 0, then A is singular.
For a 3 by 3 matrix
Using EROs on rows 2 and 3
333231
232221
131211
aaa
aaa
aaa
11
13313311
11
12313211
11
13212311
11
12212211
131211
0
0
a
aaaa
a
aaaaa
aaaa
a
aaaaaaa
For a 3 by 3 matrix
The matrix will be row equivalent to I iff:
0
11
13313311
11
12313211
11
13212311
11
12212211
11
a
aaaa
a
aaaaa
aaaa
a
aaaa
a
For a 3 by 3 matrix
This implies that the
Det(A) =
223113322113233112332112233211332211 aaaaaaaaaaaaaaaaaa
Use EROs to find:
605
245
123
STEP 1: Apply from property 5 this
gives us
STEP 2: Convert matrix to Echelon form
31 CC
AB
506
542
321
133
122
6
2
RRR
RRR
23120
100
321
Therefore is the same
as:
matrix is now in echelon form so we can multiply elements of main diagonal to get determinant
32 RR 100
23120
321
605
245
123
12)1)(12(1
100
23120
321
Factorize the determinants of
What is ?
2
2
2
1
1
1
zz
yy
xx
931
111
421
We see that y – x is a factor of row 2 and z – x is a factor of row 3 so we factor them out from:
And we get:
133
122
RRR
RRR
22
22
2
0
0
1
xzxz
xyxy
xx
))(( xzxy xz
xy
xx
10
10
1 2
233 RRR ))(( xzxy yz
xy
xx
00
10
1 2
The matrix is now in echelon form so we can multiply elements of main diagonal to get determinant and then multiply by factors to get:
= 2
2
2
1
1
1
zz
yy
xx
))()(( yzxzxy
Now, the matrix corresponds to
Since =
931
111
421
3
1
2
z
y
x
2
2
2
1
1
1
zz
yy
xx))()(( yzxzxy
Then
= 931
111
421
12)4)(1(3)13))(2(3))(2(1(
Cofactor expansion is one method used to find the determinant of matrices of order higher than 2.
If A is a square matrix, then the minor Mi,j of the element ai,j of A is the determinant of the matrix obtained by deleting the ith row and the jth column from A.
Consider the matrix .
The minor of the entry “0” is found by deleting the row and the column associated with the entry “0”.
605
245
123
A
The minor of the entry “0” is
Note: Since the 3 x 3 matrix A has 9 elements there would be 9 minors associated with the matrix.
1)5(1)2(325
13
The cofactor Ci,j = (-1)i+jMi,j.
Since we can think of the cofactor of as nothing more than its signed minor.
oddji
evenjiji
,1
,1)1(
Find the minor and cofactor of the entry “2” for
We first need to delete the row and column corresponding to the entry “2”
605
245
123
A
The Minor of 2 is
The minor corresponds to row 1 and column 2 so applying the formula, we have
So the cofactor of the entry “2” is 40.
40)2(5)6(565
25
40)40(140)1( 21 ijc
Theorem: Let A be a square matrix of order n. Then for any i,j,
Columns: and Rows:
det(A) A ai , jCi, jj1
n
det(A) A ai , jCi, j .i1
n
Given find det(A).
Cofactor is found for the first entry in column 1 “-3”
Cofactor is found for the second entry in column 1 “-5”
605
245
123
A
2402460
2411 c
1201260
1221 c
Cofactor is found for the third entry in column 1 “5”
The cofactors are then multiplied by the corresponding entry and summed.
04424
1231 c
312111 55)(3 cccA
12
6072
)0(5)12(5)24(3
Using row 2 - expansion we fix row 2 and find the minors for each entry in row 2 then apply the sign corresponding to each entries position to find the cofactors. The cofactors are then multiplied by the corresponding entry and summed.
12
209260
)10(*2)23(*4)12(*5
05
23*2
65
13*4
60
12*5
232222 24)(5 cccA
It is easy to show that
44332211
44
3433
242322
14131211
000
00
0aaaa
a
aa
aaa
aaaa
If A is square and is in Echelon form then is the product of the entries on the (main) diagonal.
00*1*2*1
0000
0100
0020
2001
Using CRAMER’S RULE we can apply this method to finding the solution to a system of linear systems that have the same number of variables as equations
There are two cases to consider
Consider the square system AX = B where A is n x n.
If then the system has either I) No solution or ii) Many solutions
If A1 is formed from A by replacing column 1 of A with column B and
I. , then the system has NO solution
II. |A| ≠ 0, then the system has a unique solution
0A
01 A
Consider the square system AX = B where A is n x n.
If |A|≠ 0 then the system has a unique solution
The unique solution is obtained by using the Cramer’s rule.
Where Ai is found from A by replacing column i of A with B.
etcA
AX
A
AX
A
AX
33
22
11
Use Cramer’s rule to write down the solution to the system
354
735
21
21
xx
xx
354
735
21
21
xx
xx371225 A
371225 A
4334
75
2653
37
2
1
A
A
37
43
37
26
22
11
A
AX
A
AX
37
43,
27
26
The adjoint of A (adj.A)) is- : , the transpose of the matrix of cofactors (a matrix of signed minors).
If then the inverse of A exists and
If , A has no inverse.
TijCadjA )(
0A
adjAA
A11
0A
Find inverse of
Det(A) =
= (3*2*0) – (3*3*1) – (-2*1*0)+(-2*1*1) +(3*1*3) –(3*1*2)
=0 – 9 – 0 – 2 + 9 – 6= -8
031
121
323
A
223113322113233112332112233211332211 aaaaaaaaaaaaaaaaaa
8110
031
893
808
1139
013
21
23
11
33
12
3231
23
01
33
03
3231
21
01
11
03
12
T
T
TijCadjA )(
1110
031
893
8
1
1
1
1
A
adjAA
A
Given that show:
a.b.c.
d.e. Solve where and
542
111
241
A
0342 IAA
0A
IAA 12133
)(3
11 AAIA
bxA Txxxx ),,( 321 Tb )1,1,1(
17168
414
8161
542
111
241
542
111
2412A
0
000
000
000
100
010
001
3
542
111
241
4
17168
414
8161
342
IAA
IIAA
IAA
IAA
3)4(
34
0342
2
Taking the determinant
Implying that and therefore; exists
IIAA
IIAA
34
3)4(
Since, Multiplying through by A
IAA
AIAA
AIAA
IAA
AAA
AAA
1213
31216
3)34(4
)34(
34
034
3
3
3
2
23
23
0342 IAA
Multiplying through by A-1
)4(3
1
43
034
)0(
034)(
034
1
1
1
1
111
2
AIA
AIA
AIA
AA
AAAAAA
IAA
142
131
243
3
1
542
111
241
400
040
004
3
1
)4(3
1
1
1
A
AIA
Since
1
1
1
3
3
3
3
1
1
1
1
142
131
243
3
1
1
x
bAx
bxA