kuttler linearalgebra slides matrices matrixarithmetic handout

8/13/2019 Kuttler LinearAlgebra Slides Matrices MatrixArithmetic Handout

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A FIRST COURSEIN LINEAR ALGEBRAAn Open Text by Ken Kuttler

Matrix ArithmeticLecture Notes by Karen Seyffarth

Adapted byLYRYX SERVICE COURSE SOLUTION

Attribution-NonCommercial-ShareAlike (CC BY-NC-SA)This license lets others remix, tweak, and build upon your work

non-commercially, as long as they credit you and license their new creationsunder the identical terms.

Matrix Arithmetic Page 1/78


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Matrices

Matrix Arithmetic Matrices Page 2/78


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Matrices - Basic Denitions and NotationDenitionsLet m and n be positive integers.

An m n matrix is a rectangular array of numbers having m rows andn columns. Such a matrix is said to have size m n.A square matrix is an m m matrix.The ( i , j )-entry of a matrix is the entry in row i and column j . For amatrix A, the (i , j )-entry of A can also be written Aij .

General notation for an m n matrix, A:

A =

A11 A12 A13 . . . A1nA21 A22 A23 . . . A2nA31 A32 A33 . . . A3n

... ...

... ...

Am 1 Am 2 Am 3 . . . Amn

= [Aij ]



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Matrices - Properties and Operations

1 Equality: two matrices are equal if and only if they have the samesize and the corresponding entries are equal.

2 Zero Matrix: an m n matrix with all entries equal to zero.3 Addition: matrices must have the same size; add corresponding

entries.4 Scalar Multiplication: multiply each entry of the matrix by the

scalar.5 Negative of a Matrix: for an m n matrix A, its negative is

denoted A and A = ( 1)A.6 Subtraction: for m n matrices A and B , A B = A + ( 1)B .



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Properties of Matrix Addition

TheoremLet A , B and C be matrices of the same size. Then, the following properties hold.

A + B = B + A

(A + B ) + C = A + ( B + C )

There exists a zero matrix 0 such that A + 0 = A

There exists a matrix A such that A + ( A) = 0



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Properties of Scalar Multiplication

TheoremLet A , B be matrices of the same size, and k , p be scalars. Then, the following properties hold.

k (A + B ) = kA + kB

(k + p ) A = kA + pA

k (pA) = ( kp ) A

1A = A



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Vectors

A row matrix (or row) is a 1 n matrix, and a column matrix (or column)is an m 1 matrix.

Matrices of size m 1 or 1 n are often referred to as vectors.

Specically, an m 1 matrix is a column vector. For example

1 1

03

A 1 n matrix is a row vector. For example

6 5 1

Matrix Arithmetic Matrix Multiplication Page 8/78


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Vector Form of a Linear System

DenitionSuppose we have a system of equations given by

a11x 1 + + a1n x n = b 1...

am1x

1 + + amn x n = b m

We can express this system in vector form as follows:

x 1

a11a

21 am 1

+ x 2

a12a22

...am 2

+ + x n

a1na2n

...amn

=

b 1b 2...

b m



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Vector Form of a Linear System

ExampleWrite the following linear system in vector form.

2x 1 + 4x 2 3x 3 = 6 x 2 + 5x 3 = 0

x 1 + x 2 + 4x 3 = 1

Solution:

x 1

2

01 + x 2

4

11 + x 3

3

54 =

6

01



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Vector Times a Matrix

DenitionLet A = [Aij ] be an m n matrix and let X be an n 1 matrix with

A = [A1 An ] , X =X 1

..

.X n

Then AX is the m 1 column vector given by

X 1A1 + X 2A2 + + X n An =

n

j =1X j A j



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Vector Times a Matrix

ExampleCompute the product AX for

A = 1 4

5 0, X =

2

3Solution:

AX = 2 15 + 3 40 =

210 +

120 =

1410



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Matrix Form of a Linear SystemDenitionSuppose we have a system of equations given by

a11x 1 + + a1n x n = b 1...

am 1x 1 + + amn x n = b m

Then we can express this system in matrix form as follows.

a11 a12 a1na21 a22 a2n

..

.

...am 1 am 2 amn

x 1x 2...x n

=

b 1b 2...b m

This means that a system of equations can be written as the matrixequation

AX = B .Matrix Arithmetic Matrix Multiplication Page 13/78


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Matrix Form of a Linear System

ExampleWrite the following linear system in matrix form.

2x 1 + 4x 2 3x 3 = 6 x 2 + 5x 3 = 0

x 1 + x 2 + 4x 3 = 1

Solution:

2 4 3

0 1 51 1 4

x 1

x 2x 3

=

6

01



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The Form AX=B

TheoremEvery system of linear equations can be written as AX = B.The system has a solution if and only if B is a linear combination of the columns of A.

The vector X = x 1 x 2 . . . x n T gives the solution to the system AX = B if and only if x 1 , x 2 , . . . , x n are a solution to the vector equation

x 1A1 + x 2A2 + x n An = B ,

where A i are the columns of A.



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The Form AX=B

ExampleLet

A =1 0 2 12 1 0 13 1 3 1

Can B =111

be expressed as a linear combination of the columns of

A? If so, nd a linear combination that does so.

Solution: Solve the system AX = B for X = x 1 x 2 x 3 x 4T .



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Example (continued)To do this, put the augmented matrix A B in reduced row-echelonform.

1 0 2 1 12 1 0 1 13 1 3 1 1

1 0 0 1 170 1 0 1 570 0 1 1 37

Since there are innitely many solutions, simply choose a value for x 4.Taking x 4 = 0 gives us

111

= 1

7

123

5

7

0 1

1+

3

7

203

This linear combination can be written as

B = 1

7A1

5

7A2 +

3

7A3 + 0A4



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Matrix Multiplication

DenitionLet A be an m n matrix and let B be an n p matrix of the form

B = [B 1 B p ]

where B 1 , ..., B p are the n 1 columns of B . Then AB is dened asfollows:

AB = A [B 1 B p ] = [AB 1 AB p ]

Note that AB has size m p .



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ExampleFind the product AB for the matrices

A = 1 0 32 1 1 and B = 1 1 2

0 2 41 0 0

Solution: AB has columns

AB 1 = 1 0 3

2 1 1 1

01

, AB 2 = 1 0 3

2 1 11

20

,

and AB 3 = 1 0 32 1 1

240

Thus, AB = 4 1 2 1 4 0 .


l l b h


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Multiplication by the Zero Matrix

ExampleCompute the product A0 for the matrix

A = 1 23 4

and the 2 2 zero matrix given by

0 = 0 00 0

Solution: In this product, we compute

1 23 4

0 00 0 =

0 00 0

Hence, A0 = 0.


Th E f P d


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The Entry of a ProductDenitionLet A be an m n matrix and B an n p matrix. Then the (i , j )-entry of AB is given by

Ai 1B 1 j + Ai 2B 2 j + + Ain B nj =n

k =1

Aik B kj = AB ij

Example (revisited)Use the above denition to compute the (1 , 2)-entry of the product

1 0 32 1 1

1 1 20 2 41 0 0

Solution: AB 12 = ( 1)(1) + (0)( 2) + (3)(0) = 1



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Properties of Matrix Multiplication

Matrix Arithmetic Properties of Matrix Multiplication Page 22/78

C tibilit f M t i M lti li ti


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Compatibility for Matrix Multiplication

DenitionLet A and B be matrices, such that A is an m n matrix.

In order for the product AB to exist, the number of rows in B mustbe equal to the number of columns in A.AB is dened if and only if B is an n p matrix for some p .When dened, AB is an m p matrix.

If the product is dened, then A and B are said to be compatible for

(matrix) multiplication.



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Example (revisited)As we saw earlier,

2 3 1 0 3

2 1 1

3 3 1 1 2

0 2 41 0 0

=

2 3 4 1 2 1 4 0

Note that the product

3 3 1 1 2

0 2 4

1 0 0

2 3 1 0 3

2 1 1

does not exist.


l


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ExampleLet

A =1 2

3 01 4

and B = 1 1 2 0

3 2 1 3

Does AB exist? If so, compute it.Does BA exist? If so, compute it.

AB =7 5 4 6

3 3 6 0 11 7 2 12

BA does not exist



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ExampleLet

G = 11 and H = 1 0

Does GH exist? If so, compute it.Does HG exist? If so, compute it.

GH = 1 01 0

HG = 1

In this example, GH and HG both exist, but they are not equal. Theyarent even the same size!



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ExampleLet

P = 1 0

2 1 and Q = 1 1

0 3

Does PQ exist? If so, compute it.Does QP exist? If so, compute it.

PQ = 1 1 2 1

QP = 1 1

6 3In this example, PQ and QP both exist and are the same size, butPQ = QP .



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FactThe four previous examples illustrate an important property of matrix multiplication.

In general, matrix multiplication is not commutative, i.e., the order of the matrices in the product is important.

In other words, in general AB = BA.



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ExampleLet

U = 2 0

0 2 and V = 1 2

3 4

Does UV exist? If so, compute it.Does VU exist? If so, compute it.

UV = 2 46 8

VU = 2 46 8

In this particular example, the matrices commute, i.e., UV = VU .




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TheoremThe following hold for matrices A , B , and C and for scalars r and s,

A (rB + sC ) = r (AB ) + s (AC )

(B + C ) A = BA + CA

A (BC ) = ( AB ) C

Notice that this applies for Matrix-Vector multiplication as well, since wecan consider a vector as an n 1 or 1 m matrix.


Elementary Proofs


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Elementary Proofs

ExampleLet A and B be m n matrices, and let C be an n p matrix. Prove thatif A and B commute with C , then A + B commutes with C .

Proof.

We are given that AC = CA and BC = CB . Consider (A + B )C .

(A + B )C = AC + BC = CA + CB

= C (A + B )Since (A + B )C = C (A + B ), A + B commutes with C .



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Matrix Transposition


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Matrix Transposition

DenitionIf A is an m n matrix, then its transpose, denoted AT , is the n mmatrix given by

AT = [Aij ]T = [ A ji ]

Theorem (Properties of the Transpose of a Matrix)Let A be an m n matrix and let B be a n p matrix. Then

(AB )T = B T AT

If r and s are scalars,

(rA + sB )T = rAT + sB T

Matrix Arithmetic Transpose of a Matrix Page 33/78

Symmetric Matrices


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Symmetric Matrices

DenitionLet A = [Aij ] be an m n matrix. The entries A11 , A22 , A33 , . . . are calledthe main diagonal of A.

DenitionThe matrix A is called symmetric if and only if AT = A. Note that this

immediately implies that A is a square matrix.

Examples

2 3 3 17 , 1 0 5

0 2 115 11 3

,

0 2 5 1

2 1 3 05 3 2 7 1 0 7 4

are symmetric matrices, and each is symmetric about its main diagonal.



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Example

Show that if A and B are symmetric matrices, then AT + 2 B is symmetric.

Proof.

(AT + 2 B )T = ( AT )T + (2 B )T = A + 2 B T

= AT + 2 B , since AT = A and B T = B .

Since (AT + 2 B )T = AT + 2 B , AT + 2 B is symmetric.


Skew Symmetric Matrices


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yDenitionAn n n matrix A is said to be skew symmetric if AT = A.

ExampleLet

A =0 9 4

9 0 3 4 3 0

Show that A is skew symmetric.Solution: AT is given by

0 9 49 0 34 3 0

Notice that AT = A so A is skew symmetric.



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The Identity Matrix and the Inverse of a Matrix

Matrix Arithmetic The Identity and Inverses Page 37/78

The Identity Matrix


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y

DenitionThe n n identity matrix, denoted I n is the matrix having ones on its maindiagonal and zeros elsewhere, and is dened for all n 2.

Denition

Let n 2. For each j , 1 j n, we denote by E j the j th column of I n .

TheoremSuppose A is an m n matrix and I n is the n n identity matrix. ThenAI n = A. If I m is the m m identity matrix, it also follows that I m A = A.


The Inverse of a Matrix


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DenitionLet A be an n n matrix. Then B is an inverse of A if and only if

AB = I n and BA = I n .Note that since A and I n are both n n, B must also be an n n matrix.

Example

Let A = 1 23 4 and B = 2 132 12

. Then

AB = 1 00 1

andBA = 1 00 1

so B is an inverse of A.Matrix Arithmetic The Identity and Inverses Page 39/78

Not every square matrix has an inverse


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Not every square matrix has an inverse.

Example

A = 0 10 1

has no inverse.

Why?


Uniqueness of Inverse


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TheoremSuppose A is an n n matrix such that an inverse A 1 exists. Then there

is only one such inverse matrix. That is, given any matrix B such that AB = BA = I , B = A 1.

Example (revisited)

For A = 1 23 4 and B = 2 132 12

, we saw that

AB = 1 00 1

andBA = 1 00 1

We said that B is an inverse of A. Now we can call B the inverse of A.Matrix Arithmetic The Identity and Inverses Page 41/78

Invertible Matrix


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DenitionsLet A be a square matrix, i.e., an n n matrix.

The inverse of A, if it exists, is denoted A 1, and

AA 1 = I = A 1A.

If A has an inverse, then we say that A is invertible.



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Finding the Inverse

Matrix Arithmetic Finding the Inverse Page 43/78

Finding the Inverse


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ProblemSuppose that A is any n n matrix.

How do we know whether or not A 1

exists? If A 1 exists, how do we nd it?

Answer: The Matrix Inverse Algorithm!


The Matrix Inverse Algorithm


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Example

LetA = 2 75 18

Find the inverse of A, if it exists.Solution: Begin with a matrix obtained from A by augmenting A with the2 2 identity matrix. We write this as

A I = 2 7 1 05 18 0 1

Then perform elementary row operations on this matrix until the left half of the matrix is transformed into I 2.




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Example (continued)

2 7 1 0

5 18 0 1

1 7212 0

5 18 0 1

1 7212 0

0 12

52 1

1 72

12 0

0 1 5 2 1 0 18 70 1 5 2




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Example (continued)We have successfully transformed the left-hand side of A I into I using elementary row operations, and thus

A I I A 1

Therefore A 1 exists, andA 1 = 18 75 2

You can verify this by calculating AA 1 and A 1A.




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TheoremSuppose A is an n n matrix. To nd A 1 if it exists, form the augmented n 2n matrix

[A|I ]

If possible do row operations until you obtain an n 2n matrix of the form[I |B ]

When this has been done, B = A 1 . In this case, we say that A is

invertible . If it is impossible to row reduce to a matrix of the form [I |B ] ,then A has no inverse.


Example


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Example

Find, if possible, the inverse of 1 0 1

2 1 3 1 1 2

.

Using the matrix inversion algorithm

1 0 1 1 0 0 2 1 3 0 1 0 1 1 2 0 0 1

1 0 1 1 0 00 1 1 2 1 00 1 1 1 0 1

1 0 1 1 0 00 1 1 2 1 00 0 0 1 1 1

From this, we see that A has no inverse.


Systems of Linear Equations and Inverses


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Suppose that a system of n linear equations in n variables is written inmatrix form as AX = B , and suppose that A is invertible.

ExampleThe system of linear equations

2x 7y = 3

5x 18y = 8can be written in matrix form as AX = B :

2 75 18

x y =

38

Note that in this example, A is invertible.

How do we know this?


Systems of Linear Equations and Inverses


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Since A is invertible, A 1 exists and has the property thatAA 1 = I = A 1A, and thus

AX = B A 1(AX ) = A 1B

(A

1A)X = A

1B IX = A 1B X = A 1B ,

i.e., AX

= B

has the unique solution given by X

= A 1B

. This gives usanother way to solve linear systems!



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Properties of the Inverse

Matrix Arithmetic Properties of the Inverse Page 52/78

The Inverse of the Transpose and a Product


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TheoremLet A and B be invertible matrices. Then,

1 (AT ) 1 = ( A 1)T

2 If A and B are n n, then (AB ) 1 = B 1A 1Note that if A 1 , A2 , . . . , Ak are invertible n n matrices, the product A1A2 Ak is also invertible, and (A1A2 Ak ) 1 = A

1k A

1k 1 A

12 A

11 .


The Inverse of the Transpose and a ProductProof:


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Proof:1 Suppose A is an invertible matrix. Then

AT (A 1)T = ( A 1A)T = I T = I ,

and(A 1)T AT = ( AA 1)T = I T = I .

This means that (AT

) 1

= ( A 1

)T

.2 Suppose A and B are invertible n n matrices. Then

(AB )(B 1A 1) = A(BB 1)A 1 = AIA 1 = AA 1 = I ,

and

(B 1A 1)(AB ) = B 1(A 1A)B = B 1IB = B 1B = I .

This means that (AB ) 1 = B 1A 1.Matrix Arithmetic Properties of the Inverse Page 54/78

Properties of Inverses


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TheoremSuppose A is an n n matrix.

1 I is invertible, and I 1 = I .2

If A is invertible, so is A 1

, and (A 1

) 1

= A.3 If A is invertible, so is Ak , and (Ak ) 1 = ( A 1)k .4 If A is invertible and p R is nonzero, then pA is invertible, and

(pA) 1 = 1p A 1.



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TheoremLet A be an n n matrix; X and B are n 1 vectors. The following conditions are equivalent.

1 A is invertible.2 AX = 0 has only the trivial solution, X = 0 .3 A can be transformed to I n by elementary row operations.4 The system AX = B has at least one solution X for any choice of B.5 There exists an n n matrix C with the property that AC = I n .



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Elementary Matrices

Matrix Arithmetic Elementary Matrices Page 59/78

Elementary Matrices


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DenitionAn elementary matrix is a matrix obtained from an identity matrix byperforming a single elementary row operation.

The type of an elementary matrix is given by the type of row operation

used to obtain the elementary matrix. Recall the types of row operations.Elementary Row Operations:

Type I: Interchange two rows.Type II: Multiply a row by a nonzero number.Type III: Add a (nonzero) multiple of one row to a different row.


Elementary Matrices


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Example

E =

1 0 0 00 0 0 10 0 1 0

0 1 0 0

, F =

1 0 0 00 1 0 00 0 2 0

0 0 0 1

, G =

1 0 0 00 1 0 0

3 0 1 0

0 0 0 1

,

are examples of elementary matrices of types I, II and III, respectively.

Let A =

1 12 2

3 34 4

; compute EA, FA, and GA.


Elementary Matrices

E l ( i d)


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Example (continued)

EA =

1 0 0 00 0 0 10 0 1 00 1 0 0

1 12 23 34 4

=

1 14 43 32 2

FA =

1 0 0 00 1 0 00 0 2 00 0 0 1

1 12 23 34 4

=

1 12 2

6 64 4

GA =

1 0 0 00 1 0 0

3 0 1 00 0 0 1

1 12 23 34 4

=

1 12 20 04 4


Multiplication by an Elementary Matrix


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TheoremTo perform any of the three row operations on a matrix A it suffices to

take the product EA, where E is the elementary matrix obtained by using the desired row operation on the identity matrix.


Example

Let A = 4 1 and C = 1 3 . Find elementary matrices E and


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Let A 1 3 and C 2 5 . Find elementary matrices E and

F so that C = FEA.

Note. The statement of the problem tells you that C can be obtainedfrom A by a sequence of two elementary row operations, represented byelementary matrices E and F .

A = 4 11 3

E 1 34 1

F 1 32 5 = C

E = 0 11 0 and F = 1 0 2 1

Therefore 1 32 5 = 1 0 2 1

0 11 0

4 11 3 .

You can check your work by doing the matrix multiplication.Matrix Arithmetic Elementary Matrices Page 64/78

Inverses of Elementary Matrices

Example


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ExampleWithout using the matrix inversion algorithm, what is the inverse of the

elementary matrix

G =

1 0 0 00 1 0 0

3 0 1 00 0 0 1

?

Hint. What row operation can be applied to G to transform it to I 4?The row operation G I 4 is to add three times row one to row three,and thus

G

1 =

1 0 0 00 1 0 03 0 1 00 0 0 1

Check by computing G 1G .


Inverses of Elementary Matrices


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Example (continued)Similarly,

E 1 =

1 0 0 00 0 0 10 0 1 0

0 1 0 0

1

=

1 0 0 00 0 0 10 0 1 0

0 1 0 0and

F 1 =

1 0 0 00 1 0 00 0 2 00 0 0 1

1

=

1 0 0 00 1 0 00 0 12 00 0 0 1


The Form B = UA


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Suppose A is an m n matrix and that B can be obtained from A by a

sequence of k elementary row operations. Then there exist elementarymatrices E 1 , E 2 , . . . E k such that

B = E k (E k 1( (E 2(E 1A)) ))

Since matrix multiplication is associative, we haveB = ( E k E k 1 E 2E 1)A

or, more concisely, B = UA where U = E k E k 1 E 2E 1.

To nd U so that B = UA, we could nd E 1 , E 2 , . . . , E k and multiply thesetogether (in the correct order), but there is an easier method for nding U .


DenitionLet A be an m n matrix. We write


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A B

if B can be obtained from A by a sequence of elementary row operations.

TheoremSuppose A is an m n matrix and that A B. Then, construct the

matrix A I m

and row reduce to the form

B U .

It follows that B = UA where U is m m. Further, U = E k E k 1 E 2E 1,where E 1 , E 2 , . . . , E k are elementary matrices which represent, in order, the elementary row operations used to carry A to B.


Example

Let A = 3 0 1 and let R be the reduced row echelon form of A


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Let A = 2 1 0 , and let R be the reduced row-echelon form of A.

Find a matrix U so that R = UA.Solution.

3 0 1 1 02 1 0 0 1

1 1 1 1 12 1 0 0 1

1 1 1 1 10 3 2 2 3

1 1 1 1 10 1 2323 1

1 0 1

313 00 1 2323 1

Starting with A I , weve obtained R U .Therefore R = UA, where

U = 13 023 1

.


Express a Matrix as a Product of Elementary MatricesExample


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Let A =1 2 4

3 6 130 1 2

. Do row operations to put A in reduced

row-echelon form, and write down the corresponding elementary matrices.

1 2 4 3 6 13

0 1 2

E 1

1 2 40 0 10 1 2

E 2

1 2 40 1 20 0 1

E 3

1 2 4

0 1 20 0 1

E 4

1 0 0

0 1 20 0 1

E 5

1 0 0

0 1 00 0 1

Notice that the reduced row-echelon form of A equals I 3. Now nd thematrices E 1 , E 2 , E 3 , E 4 and E 5.


Example1 0 0 1 0 0 1 0 0


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E 1 =1 0 03 1 0

0 0 1

, E 2 =1 0 00 0 1

0 1 0

, E 3 =1 0 00 1 0

0 0 1

E 4 =1 2 00 1 00 0 1

, E 5 =1 0 00 1 20 0 1

It follows that

(E 5(E 4(E 3(E 2(E 1A))))) = I (E 5E 4E 3E 2E 1)A = I

and thereforeA 1 = E 5E 4E 3E 2E 1 .


Example (continued)


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Example (continued)Since A 1 = E 5E 4E 3E 2E 1,

A 1 = E 5E 4E 3E 2E 1(A 1) 1 = ( E 5E 4E 3E 2E 1)

1

A = E 11 E 12 E

13 E

14 E

15 .

This example illustrates the following result.

TheoremLet A be an n n matrix. Then, A 1 exists if and only if A can be written

as the product of elementary matrices.


Express a Matrix as a Product of Elementary MatricesExample


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Express A = 4 1 3 2 as a product of elementary matrices.

4 1 3 2

E 1 1 3 3 2

E 2 1 30 11

E 3 1 30 1

E 4 1 00 1

and

E 1 = 1 10 1 ,

E 2 = 1 03 1 ,

E 3 = 1 00 111

, E 4 = 1 30 1 .

Since E 4E 3E 2E 1A = I , A 1

= E 4E 3E 2E 1, and henceA = E 11 E

12 E

13 E

14 .



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Example (continued)

Therefore,

A = 1 10 1

1 1 03 1

1 1 00 111

1 1 30 1

1

,

i.e.,

A = 1 10 1 1 0 3 1

1 00 11

1 30 1 .

Check your work by computing the product.



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More on Matrix Inverses

Matrix Arithmetic More on Matrix Inverses Page 75/78

The Inverse of a Square Matrix


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TheoremOnly square matrices can be invertible.

TheoremSuppose A and B are n n matrices such that AB = I. Then it follows that BA = I. Further, both A and B are invertible, with B = A 1 and A = B 1.


WarningIn the second Theorem, it is essential that the matrices be square.

Example


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ExampleLet

A = 1 1 0 1 4 1 and B =

1 00 01 1

.

Compute AB and compute BA.

Solution:

AB = 1 00 1 = I 2

However,

BA =1 1 00 0 00 5 1

= I 3


The Reduced Row-Echelon Form of an Invertible Matrix


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TheoremFor any matrix A the following are equivalent.

A is invertible The reduced row-echelon form of A is an identity matrix


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