Mathematics 13: Lecture 12 - Furman Universitymath.furman.edu/~dcs/courses/math13/lectures/lecture-12.pdfResult from previous lecture I The following statements are equivalent for

Mathematics 13: Lecture 12Elementary Matrices and Finding Inverses

Dan Sloughter

Furman University

January 30, 2008

Dan Sloughter (Furman University) Mathematics 13: Lecture 12 January 30, 2008 1 / 24

Result from previous lecture

I The following statements are equivalent for an n × n matrix A:

I A is invertible.I The only solution to A~x = ~0 is the trivial solution ~x = ~0.I The reduced row-echelon form of A is In.I A~x = ~b has a solution for ~b in Rn.I There exists an n × n matrix C for which AC = In.



I The following statements are equivalent for an n × n matrix A:I A is invertible.

I The only solution to A~x = ~0 is the trivial solution ~x = ~0.I The reduced row-echelon form of A is In.I A~x = ~b has a solution for ~b in Rn.I There exists an n × n matrix C for which AC = In.



I The following statements are equivalent for an n × n matrix A:I A is invertible.I The only solution to A~x = ~0 is the trivial solution ~x = ~0.

I The reduced row-echelon form of A is In.I A~x = ~b has a solution for ~b in Rn.I There exists an n × n matrix C for which AC = In.



I The following statements are equivalent for an n × n matrix A:I A is invertible.I The only solution to A~x = ~0 is the trivial solution ~x = ~0.I The reduced row-echelon form of A is In.

I A~x = ~b has a solution for ~b in Rn.I There exists an n × n matrix C for which AC = In.



I The following statements are equivalent for an n × n matrix A:I A is invertible.I The only solution to A~x = ~0 is the trivial solution ~x = ~0.I The reduced row-echelon form of A is In.I A~x = ~b has a solution for ~b in Rn.

I There exists an n × n matrix C for which AC = In.



I The following statements are equivalent for an n × n matrix A:I A is invertible.I The only solution to A~x = ~0 is the trivial solution ~x = ~0.I The reduced row-echelon form of A is In.I A~x = ~b has a solution for ~b in Rn.I There exists an n × n matrix C for which AC = In.


Computing an inverse

I The proof of the previous theorem provides a method for findinginverses.

I Namely: given an n × n matrix A, let ~c1, ~c2, . . . , ~cn be the solutionsof A~x = ~ej , for j = 1, 2, . . . , n.

I Then A−1 is the matrix with columns ~c1, ~c2, . . . , ~cn.

I Note: this requires solving n systems of n equations in n unknowns.

I Note: we can do this at once by writing an augmented matrixconsisting of A and In and reducing A to In.

I That is, the same row operations which take A to In take In to A−1.










































Example

I We will find the inverse of

A =

1 2 −12 2 41 3 −3

.

I Using elementary row operations, we find1 2 −1 1 0 02 2 4 0 1 01 3 −3 0 0 1

−→1 2 −1 1 0 0

0 −2 6 −2 1 00 1 −2 −1 0 1


Example

I We will find the inverse of

A =

1 2 −12 2 41 3 −3

.

I Using elementary row operations, we find1 2 −1 1 0 02 2 4 0 1 01 3 −3 0 0 1

−→1 2 −1 1 0 0

0 −2 6 −2 1 00 1 −2 −1 0 1


Example (cont’d)

I Continuing:

−→

1 2 −1 1 0 0

0 1 −3 1 −12 0

0 1 −2 −1 0 1

−→1 0 5 −1 1 0

0 1 −3 1 −12 0

0 0 1 −2 12 1

−→

1 0 0 9 −32 −5

0 1 0 −5 1 3

0 0 1 −2 12 1

.

I Hence

A−1 =

9 −32 −5

−5 1 3

−2 12 1

.


Example (cont’d)

I Continuing:

−→

1 2 −1 1 0 0

0 1 −3 1 −12 0

0 1 −2 −1 0 1

−→1 0 5 −1 1 0

0 1 −3 1 −12 0

0 0 1 −2 12 1

−→

1 0 0 9 −32 −5

0 1 0 −5 1 3

0 0 1 −2 12 1

.

I Hence

A−1 =

9 −32 −5

−5 1 3

−2 12 1

.


Example

I Suppose we wish to find an inverse for

A =

2 1 −4−4 −1 6−2 2 −2

.

I Using elementary row operations, we find that 2 1 −4 1 0 0−4 −1 6 0 1 0−2 2 −2 0 0 1

−→2 1 −4 1 0 0

0 1 −2 2 1 00 3 −6 1 0 1


Example

I Suppose we wish to find an inverse for

A =

2 1 −4−4 −1 6−2 2 −2

.

I Using elementary row operations, we find that 2 1 −4 1 0 0−4 −1 6 0 1 0−2 2 −2 0 0 1

−→2 1 −4 1 0 0

0 1 −2 2 1 00 3 −6 1 0 1


Example (cont’d)

I Continuing, we have

−→

2 1 −4 1 0 00 1 −2 2 1 00 0 0 −5 −3 1

.

I Note: this shows that it is not possible to reduce A to I3 usingelementary row operations.

I It follows that A does not have an inverse.


Example (cont’d)


−→

2 1 −4 1 0 00 1 −2 2 1 00 0 0 −5 −3 1

.




Example (cont’d)


−→

2 1 −4 1 0 00 1 −2 2 1 00 0 0 −5 −3 1

.




Inverses in Octave

I To find the inverse of

A =

1 2 3−1 0 1

4 2 1

,

we can either

I apply rref to the augmented matrix 1 2 3 1 0 0−1 0 1 0 1 0

4 2 1 0 0 1

,

I or use inv(A).


Inverses in Octave


A =

1 2 3−1 0 1

4 2 1

,

we can eitherI apply rref to the augmented matrix 1 2 3 1 0 0

−1 0 1 0 1 04 2 1 0 0 1

,

I or use inv(A).


Inverses in Octave


A =

1 2 3−1 0 1

4 2 1

,

we can eitherI apply rref to the augmented matrix 1 2 3 1 0 0

−1 0 1 0 1 04 2 1 0 0 1

,

I or use inv(A).


Inverses in Octave (cont’d)

I In either case, we find that

A−1 =

−1.0000 2.0000 1.00002.5000 −5.5000 −2.0000−1.0000 3.0000 1.0000

I Use the menus to find inverses in wxMaxima.

I In wxMaxima, A.B multiplies the matrices A and B.

I In wxMaxima,

A^^3

raises A to the third power.




A−1 =

−1.0000 2.0000 1.00002.5000 −5.5000 −2.0000−1.0000 3.0000 1.0000



I In wxMaxima,

A^^3





A−1 =

−1.0000 2.0000 1.00002.5000 −5.5000 −2.0000−1.0000 3.0000 1.0000



I In wxMaxima,

A^^3





A−1 =

−1.0000 2.0000 1.00002.5000 −5.5000 −2.0000−1.0000 3.0000 1.0000



I In wxMaxima,

A^^3



Definition

I An elementary matrix is any matrix which may be obtained byapplying an elementary row operation to an identity matrix.

I Examples: Each of the following is an elementary matrix:

2 0 00 1 00 0 1

,

1 0 00 0 10 1 0

, and

1 0 0 00 1 0 0−5 0 1 0

0 0 0 1

.


Definition

I An elementary matrix is any matrix which may be obtained byapplying an elementary row operation to an identity matrix.

I Examples: Each of the following is an elementary matrix:

2 0 00 1 00 0 1

,

1 0 00 0 10 1 0

, and

1 0 0 00 1 0 0−5 0 1 0

0 0 0 1

.


Example

I Let

A =

1 2 3 14 5 6 27 8 9 1

and E =

1 0 00 3 00 0 1

.

I Note: E is an elementary matrix obtained by multiplying the secondrow of I3 by 3.

I Note:

EA =

1 0 00 3 00 0 1

1 2 3 14 5 6 27 8 9 1

=

1 2 3 112 15 18 67 8 9 1

,

which is what we would obtain applying the same elementary rowoperation to A.


Example

I Let

A =

1 2 3 14 5 6 27 8 9 1

and E =

1 0 00 3 00 0 1

.


I Note:

EA =

1 0 00 3 00 0 1

1 2 3 14 5 6 27 8 9 1

=

1 2 3 112 15 18 67 8 9 1

,



Example

I Let

A =

1 2 3 14 5 6 27 8 9 1

and E =

1 0 00 3 00 0 1

.


I Note:

EA =

1 0 00 3 00 0 1

1 2 3 14 5 6 27 8 9 1

=

1 2 3 112 15 18 67 8 9 1

,



Example (cont’d)

I Let

E =

1 0 03 1 00 0 1

,

the elementary 3× 3 matrix obtained by adding 3 times the first rowto the second row of I3.

I Now

EA =

1 0 03 1 00 0 1

1 2 3 14 5 6 27 8 9 1

=

1 2 3 17 11 15 57 8 9 1

,

which is what we would obtain from applying the same row operationto A.


Example (cont’d)

I Let

E =

1 0 03 1 00 0 1

,

the elementary 3× 3 matrix obtained by adding 3 times the first rowto the second row of I3.

I Now

EA =

1 0 03 1 00 0 1

1 2 3 14 5 6 27 8 9 1

=

1 2 3 17 11 15 57 8 9 1

,



Example (cont’d)

I Let

E =

1 0 00 0 10 1 0

,

the elementary 3× 3 matrix obtained by interchanging the second andthird rows of I3.

I Then

EA =

1 0 00 0 10 1 0

1 2 3 14 5 6 27 8 9 1

=

1 2 3 17 8 9 14 5 6 2

,



Example (cont’d)

I Let

E =

1 0 00 0 10 1 0

,

the elementary 3× 3 matrix obtained by interchanging the second andthird rows of I3.

I Then

EA =

1 0 00 0 10 1 0

1 2 3 14 5 6 27 8 9 1

=

1 2 3 17 8 9 14 5 6 2

,



Row operations

I Suppose an elementary row operation is performed on an m × nmatrix A to obtain the matrix B.

I Let E be the elementary matrix obtained by performing the same rowoperation on Im.

I Then B = EA.


Row operations



I Then B = EA.


Row operations



I Then B = EA.


Inverses of elementary matrices

I Every elementary matrix E is invertible, and E−1 is itself anelementary matrix:

I If E multiplies a row by a nonzero scalar c , E−1 multiplies that samerow by 1

c .I If E multiplies row i by scalar c and adds it to row j , E−1 multiplies

row i by −c and adds it to row j .I If E interchanges rows i and j , E−1 interchanges rows i and j (and, so,

E = E−1).





c .

I If E multiplies row i by scalar c and adds it to row j , E−1 multipliesrow i by −c and adds it to row j .

I If E interchanges rows i and j , E−1 interchanges rows i and j (and, so,E = E−1).






row i by −c and adds it to row j .

I If E interchanges rows i and j , E−1 interchanges rows i and j (and, so,E = E−1).






row i by −c and adds it to row j .I If E interchanges rows i and j , E−1 interchanges rows i and j (and, so,

E = E−1).


Theorem

I Suppose an m × n matrix A is reducible to an m × n matrix B by aseries of elementary row operations. Then:

I B = EkEk−1 · · ·E1A, where E1, E2, . . . , Ek are elementary matrices.I B = UA for some invertible matrix U.I U is the matrix obtained from Im by performing the row operations on

Im which reduce A to B.


Theorem


I B = EkEk−1 · · ·E1A, where E1, E2, . . . , Ek are elementary matrices.

I B = UA for some invertible matrix U.I U is the matrix obtained from Im by performing the row operations on



Theorem


I B = EkEk−1 · · ·E1A, where E1, E2, . . . , Ek are elementary matrices.I B = UA for some invertible matrix U.

I U is the matrix obtained from Im by performing the row operations onIm which reduce A to B.


Theorem


I B = EkEk−1 · · ·E1A, where E1, E2, . . . , Ek are elementary matrices.I B = UA for some invertible matrix U.I U is the matrix obtained from Im by performing the row operations on



Example

I Let

A =

2 −1 4 61 2 1 25 0 9 14

.

I Applying row reduction, we have2 −1 4 6 1 0 01 2 1 2 0 1 05 0 9 14 0 0 1

−→1 2 1 2 0 1 0

2 −1 4 6 1 0 05 0 9 14 0 0 1


Example

I Let

A =

2 −1 4 61 2 1 25 0 9 14

.

I Applying row reduction, we have2 −1 4 6 1 0 01 2 1 2 0 1 05 0 9 14 0 0 1

−→1 2 1 2 0 1 0

2 −1 4 6 1 0 05 0 9 14 0 0 1


Example (cont’d)

I Continuing,

−→

1 2 1 2 0 1 00 −5 2 2 1 −2 00 −10 4 4 0 −5 1

−→

1 2 1 2 0 1 0

0 1 −25 −2

5 −15

25 0

0 −10 4 4 0 −5 1

−→

1 0 95

145

25

15 0

0 1 −25 −2

5 −15

25 0

0 0 0 0 −2 −1 1

.


Example (cont’d)

I Hence if

B =

1 0 95

145

0 1 −25 −2

5

0 0 0 0

,

the reduced row-echelon form of A, and

U =

25

15 0

−15

25 0

−2 −1 1

,

then U is invertible and B = UA.


Example (cont’d)

I Note: U is not unique.

I For example, using Octave to perform row reductions to reducedrow-echelon form, we obtain

V =

0.00000 0.00000 0.200000.00000 0.50000 −0.100001.00000 0.50000 −0.50000

with the property that V is invertible and B = VA.

I In particular, this shows that we may have UA = VA and yet U 6= V .


Example (cont’d)



V =

0.00000 0.00000 0.200000.00000 0.50000 −0.100001.00000 0.50000 −0.50000




Example (cont’d)



V =

0.00000 0.00000 0.200000.00000 0.50000 −0.100001.00000 0.50000 −0.50000




Theorem

I If A is an invertible matrix, then there exist elementary matrices E1,E2, . . . Ek such that

A = E1E2 · · ·Ek .

I Reason: Since A is reducible to I , there exist elementary matrices D1,D2, . . . , Dk such that

I = DkDk−1 · · ·A.

I HenceA = D−1

1 D−12 · · ·D

−1k .

I So let Ei = D−1i , i = 1, 2, . . . , k .


Theorem


A = E1E2 · · ·Ek .


I = DkDk−1 · · ·A.

I HenceA = D−1

1 D−12 · · ·D

−1k .

I So let Ei = D−1i , i = 1, 2, . . . , k .


Theorem


A = E1E2 · · ·Ek .


I = DkDk−1 · · ·A.

I HenceA = D−1

1 D−12 · · ·D

−1k .

I So let Ei = D−1i , i = 1, 2, . . . , k .


Theorem


A = E1E2 · · ·Ek .


I = DkDk−1 · · ·A.

I HenceA = D−1

1 D−12 · · ·D

−1k .

I So let Ei = D−1i , i = 1, 2, . . . , k .


Example

I If

A =

[3 21 −1

],

then I2 = D4D3D2D1A, where

D1 =

[0 11 0

], D2 =

[1 0−3 1

], D3 =

[1 00 1

5

], and D4 =

[1 10 1

].


Example (cont’d)

I It follows that A = E1E2E3E4, where

E1 =

[0 11 0

], E2 =

[1 03 1

], E3 =

[1 00 5

], and E4 =

[1 −10 1

].


Theorem

I If R and S are both reduced row-echelon forms of a matrix A, thenR = S .

I We will not prove this, but the proof is based on the followingobservations:

I There exist invertible matrices P and Q such that R = PA andS = QA.

I Hence A = Q−1S and A = P−1R, so S = QP−1R.I That is, S = UR for some invertible matrix U.


Theorem






Theorem






Theorem




I Hence A = Q−1S and A = P−1R, so S = QP−1R.

I That is, S = UR for some invertible matrix U.


Theorem






Documents

Mathematics 13: Lecture 12 - Furman Universitymath.furman.edu/~dcs/courses/math13/lectures/lecture-12.pdfResult from previous lecture I The following statements are equivalent for