[PDF] Matrix Multiplication - University of Plymouth

Two classical theorems on commuting matrices

AM= MA Then either M = ° or M is nonsingular Furthermore if A = A (so that M commutes with each element of ~l) then M is scalar :l PROOF Suppose that the rank of M is r, and write (II" M = P ° where P, Q are nonsingular and II' is the r X r identity matrix Then for each AE~l, (2) (II' (P-IAP) ° ° 0) _ (QAQ-I) Put (All P-IAP= A~I

EXERCISES OF MATRICES OPERATIONS Question 1

EXERCISES OF MATRICES OPERATIONS 3 (24) If A,Bare both n×nmatrices and ABis singular, then Ais singular or Bis singular (25) Ais diagonalizable, then Ais non-singular (26) Ais symmetric, then Ais non-singular

2 ALGEBRE` - Major-Prépa

6 La matrice triangulaire A admet deux valeurs propres r´eelles 1 et −1 : elle est diagonalisable a) Si M ∈ R, alors AM = M3 = MA et M ∈ C Par la question pr´ec´edente, M est donc de la forme αI + βA b) Comme A2 = I, on a (αI +βA)2 = (α2 +β2)I +2αβA et cette matrice vaut A lorsque α2 + β2 = 0 et 2αβ = 1, on a donc β

TD 03 : Matrices

(c)Montrer qu’il existe une base B de R3 dans laquelle la matrice de u est égale à 0 1 0 0 0 0 0 0 2 (d)Déterminer les endomorphismes qui commutent avec u 6 / Montrer que 0 0 0 1 1 −1 2 2 −2 est semblable à −1 0 0 0 0 0 0 0 0 7 / Montrer qu’une matrice A ∈M n(K) est semblable à la matrice dont

Rank of Matrix by Determinant

Example 1 Find the Rank of Matrix using Determinant ????= 1 2 3 2 4 7 3 6 10 ????=140−42−220−21+312−12 =1−2−2−1+30

Some Linear Algebra Notes

ij] is row (column) equivalent to a unique ma-trix in reduced (column) row echelon form The uniqueness proof is involved, see Ho man and Kunze, Linear Algebra, 2nd ed Note: the row echelon form of a matrix is not unique Why? Theorem 2 3 Let Ax= band Cx= dbe two linear systems, each of mequations in nunknowns If

Rank of Matrix by Normal Form - WordPresscom

R3 R3 –R2, ????≅ 1 0 0 2 0 0 3 1 0 C2 C2-2C1, ????≅ 1 0 0 0 0 0 0 1 0 As, Normal Form of given matrix A is having Identity Matrix of Order 2 rank (A)= r(A) = 2 ????≅ 1

64 Hermitian Matrices - Naval Postgraduate School

Ch 6: Eigenvalues 6 4 Hermitian Matrices We consider matrices with complex entries (a i;j 2C) versus real entries (a i;j 2R) 1 in R the length of a real number xis jxj= the length from the origin to the number

TD - Matrices

Exercice 17 Soit A ∈Mn (R)une matrice nilpotente d’ordre p >1 On pose B =In −A 1 Montrer que B est inversible et exprimer son inverse à l’aide de A (penser à la factorisation de I −Ap) 2 Application : B = 1 −1 0 0 0 1 −1 0 0 0 1 −1 0 0 0 1

Matrix Multiplication - University of Plymouth

Table of Contents 1 Introduction 2 Matrix Multiplication 1 3 Matrix Multiplication 2 4 The Identity Matrix 5 Quiz on Matrix Multiplication Solutions to Exercises

[PDF] exercice commutant d'une matrice

[PDF] extrait traduction

[PDF] matrice scalaire

[PDF] matrices commutatives

[PDF] quels sont les types de lecteurs

[PDF] matrice cours et exercices pdf

[PDF] matrice cours pdf

[PDF] cours determinant d'une matrice

[PDF] résumé sur les matrices pdf

[PDF] matrice d'eisenhower excel

[PDF] matrice d'eisenhower vierge

[PDF] télécharger matrice eisenhower excel

[PDF] matrice eisenhower vierge

[PDF] fichier excel matrice eisenhower

[PDF] matrice eisenhower exemple

BasicMathematics

MatrixMultiplication

R Horan & M LavelleTheaim of this document is to provide a short, self assessment programme for students who wish to learn how to multiply matrices.Copyright c?2005Email: rhoran,mlavelle@plymouth.ac.uk

LastRevision Date: November 2, 2005Version 1.0

Table of Contents

1.Introduction

2.Matrix Multiplication 1

3.Matrix Multiplication 2

4.The Identity Matrix

5.Quiz on Matrix Multiplication

Solutions to Exercises

Solutions to Quizzes

The full range of these packages and some instructions, should they be required, can be obtained from our web pageMathematics Support Materials.

Section 1: Introduction 3

1. Introduction

In the packageIntroduction to Matricesthe basic rules ofaddi- tionandsubtractionof matrices, as well asscalar multiplication, were introduced. The rule for themultiplication of two matricesis the subject of this package. The first example is the simplest. Recall that ifMis a matrix then the transpose ofM, written M T, is the matrix obtained fromMby writing the rows ofMas the columns ofM T.

IfA= (a1a2... an)is a1×n(row) matrix andB= (b1b2... bn)Tis an×1(column) matrix then the productABis defined as

AB=(a1a2... an)(

((b 1 b

2···b

n) ))=a 1b 1+a 2b

2+···+a

nb nThis general rule is sometimes called theinner product. N.B.Therow matrixis on the left and thecolumn matrixis on the right.

Section 1: Introduction 4

Example 1In each of the following cases, find the productAB. (a)A= (1 2),B= (4 3)T.(b)A= (1 1 1),B= (2 3 4)T. (c)A= (1-1 2 3),B= (1 1-3 2)T.

Solution

(a)AB=(1 2)? 4

3?=1×4 +2×3 = 4 + 6 = 10.

(b)AB=(1 1 1)( (2 3 4) )=1×2 +1×3 +1×4 = 2 + 3 + 4 = 9. (c)AB=(1-1 2 3)( ((1 1 -3 2) ))=1×1 +1×(-1) +2×(-3) +3×2= 1 + (-1) + (-6) + 6 = 0.

Section 1: Introduction 5

Exercise 1.For each of the cases below, calculateAB. (Click on the greenletters for solutions.) (a)A= (-2 4),B= (3 2)T, (b)A= (5 3-2),B= (3-4 2)T, (c)A= (4 4-2-3),B= (5-4 32)T.

The following observations are worth noting.•The row matrix is on the left, the column matrix is on the right.

•Therowandcolumnhave the same number of elements. •The inner productABis a1×1matrix, i.e. anumber. •Nothing has yet been said about a matrix productBA. QuizIfA= (x x1)andB= (x6 9)T, which of the following values ofxwill result inAB= 0? (a)x= 1,(b)x= 3,(c)x=-3,(d)x=-2.

Section 2: Matrix Multiplication 1 6

2. Matrix Multiplication 1

The previous section gave the rule for the multiplication of a row vectorAwith a column vectorB, theinner productAB. This section will extend this idea to more general matrices.

Suppose thatA=?a1a2... an

c

1c2... cn?andB= (b1b2... bn)T.

ThenAB=?

a1a2... an c

1c2... cn?(

((b 1 b

2···b

n) ?a 1b 1+a 2b

2+...+a

nb nc 1b 1+c 2b

2+...+c

nb n?Example 2FindABfor each of the following cases. (a)A=?1 2

3-1?,B= (4 3)T.

(b)A=?1 1 1 -2 1-3?,B= (2 3 4)T.

Section 2: Matrix Multiplication 1 7

Solution

(a)AB=? 1 2 3-1?? 4

3?=?1×4 +2×33×4 +(-1)×3?=?

10

9?(b)AB=?

1 1 1 -2 1-3?( (2 3 4) )=?1×2 +1×3 +1×4(-2)×2 +1×3 +(-3)×4? =?9 -13?The following observations onABare worth noting. •The element in thefirst rowofABis theinner productof the first rowofAwith the column matrixB. •The element in thesecond rowofABis theinner productofthe second rowofAwith the column matrixB. •The number ofcolumnsofAmust be equal to the number of rowsofB.

Section 2: Matrix Multiplication 1 8

This rule for multiplication may be extended to matrices,A, which have more than two rows. For example, ifAhad3rows then the resulting matrix,AB, would have a third row; the value of this element would be theinner productof thethird rowofAwith the column matrixB. Exercise 2.For each of the cases below, calculateAB. (Click on the greenletters for solutions.) (a)A=?-2 4

5 3?,B= (4 3)T.

(b)A=?5 3 2

4-1-1?,B= (2 3 4)T.

(c)A=( (-2 4 5 3 4-1) ),B= (4 3)T. (d)A=?4 4-2-3

3-1-1 2?,B= (5-4 3 2)T.

Section 3: Matrix Multiplication 2 9

3. Matrix Multiplication 2

The extension of the concept of matrix multiplication to matrices, A,B, in whichAhas more than one row andBhas more than one column is now possible. The product matrixABwill have the same number of columns asBand each column is obtained by taking the product ofAwith each column ofB, in turn, as shown below.

LetA=(

(4 1 2 3 -1 2) )andB=?2-1

3 2?and letb

1,b

2be the first and

second columns ofBrespectively. Then Ab 1=( (4 1 2 3 -1 2) 2 3?=( (11 13 4) )andAb 2=( (4 1 2 3 -1 2) -1 2?=( (-2 4 5)

ThusAB=(

(4 1 2 3 -1 2) 2-1 3 2? (11-2 13 4 4 5)

Section 3: Matrix Multiplication 2 10

Exercise 3.For each of the cases below, calculateAB. (Click on the greenletters for solutions.) (a)A=?-2 4

5 3?,B=?-2 4

5 3?. (b)A=?3 2

7 5?,B=?5-2

-7 3?. (c)A=( (-2 4 5 3 4-1) ),B=?-2 4 5 3?. (d)A=?5 3 2

4-1-1?,B=(

(-2 4 5 3 4-1) NBThe rules for finding the product of two matrices are summarised on the next page.

Section 3: Matrix Multiplication 2 11

•IfAism×nandBisn×rthen the productABexists. •The resulting matrix ism×r.((m×n)(n????×r)=m×r) •The element in theithrow,jthcolumn of the matrixABis the inner productof theithrow ofAwith thejthcolumn ofB. Example 3Find the element in the2ndrow3rdcolumn ofABif

A=?1 2

-1 3?andB=?1 4-2

3-1 2?.

SolutionSinceAis2×2andBis2×3, the productABexists and is a2×3matrix. The required element is the inner product of the second rowofAwith thethird columnofB, i.e. (-1)×(-2) +3×2 = 2 + 6 = 8.

Section 3: Matrix Multiplication 2 12

Exercise 4.If

A=( (1-2 4 5

7-8-6 2

2 3-1-2)

)andB=( ((-2 1-3 0-2-5 4-3-7

0 0 6)

))find theijelement,i.e.the element in theith rowjth column, of ABfor the following cases. (Click on thegreenletters for solutions.) (a)i= 3,j= 2,(b)i= 2,j= 3,(c)i= 1,j= 2, (d)i= 2,j= 1,(e)i= 3,j= 1,(f)i= 1,j= 3, QuizWhich of the following is the element in the3rd row,3rd column, of the matrixABin the above exercise? (a)26,(b)-26,(c)-12,(d)12.

Section 4: The Identity Matrix 13

4. The Identity Matrix

IfAandBare two matrices, the productABcan be found if the number ofcolumnsofAequals the number ofrowsofB. IfAis2×3 andBis3×5thenABcan be calculated butBAdoes not exist. The orderin which matrices are multiplied together matters. Even when ABandBAboth exist it is usually the case thatAB?=BA. There is one particular matrix, theidentity matrix, which has very special multiplication properties. Then×nidentity matrixis the n×nmatrix with1s and0sas shown below. Example 4The2×2,3×3and4×4identity matrices are 1 0

0 1?,(

(1 0 0 0 1 0

0 0 1)

((1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1)

Section 4: The Identity Matrix 14

The most important property of the identity matrix is revealed in the following exercise.Exercise 5.If theidentity matrixis denoted byIand the matrixM is

M=?1 2 4

7 8 6?,

use the appropriate identity matrix to calculate the following matrix products. (Click on thegreenletters for solutions.) (a)IM, whereIis the2×2identity matrix,(b)MI, whereIis the3×3identity matrix. In matrix multiplication the identity matrix,I, behaves exactly like the number1in ordinary multiplication. This was seen in the previous exercise. For part(a),the matrixIis the2×2identity matrix; in part(b),Iwas3×3; they satisfy the equationIM=M=MI.

Section 4: The Identity Matrix 15

Example 5The matricesA,Bare

A=( (4 3 2 5 6 3

3 5 2)

)andB=( (3-4 3 1-2 2 -7 11-9)

CalculateABandBA.

SolutionUsing the rules of matrix multiplication,

AB=( (4 3 2 5 6 3

3 5 2)

(3-4 3 1-2 2 -7 11-9) (1 0 0 0 1 0

0 0 1)

)=I. BA=( (3-4 3 1-2 2 -7 11-9)quotesdbs_dbs16.pdfusesText_22