MM2 Linear Algebra 2011 I 1x2

8/4/2019 MM2 Linear Algebra 2011 I 1x2

1/33

Mathematics

Part 1: Review of Matrices & Elements of Linear Algebra

Part 2: Multivariable Calculus

Review of Matrix Algebra


2/33

What is a Matrix?

A matrix is a rectangular array of numbers.

A =

1 2 09 1 2

1 3 4

; M = 1 2 5 7

4 1 0 8

; v =

13

26

Order of a matrixA matrix has order n m if it has n rows and m columns.

Vectors

DefinitionA (1 m) matrix is called as a row vector

v = 1 3 2 6 A (m 1) matrix is called as a column vector

v =

13

26

By default, when we talk about a vector we mean a column vector.


3/33

Matrices

The elements of a matrix A are often written aij where

i is the row number of the element

j is the column number of the element

Example

A = 1 2 09 1 21 3 4

= a11 a12 a13a21 a22 a23a31 a32 a33

Matrices

A general m n matrix will often be written

A =

a11 a12

a1n

a21 a22 a2n...

.... . .

...am1 am2 amn


4/33

Notations

Square or round brackets. Different notations... same meaning.

A =

1 2 09 1 21 3 4

A = 1 2 09 1 21 3 4

Recallthe following notation has a different meaning (determinant):

det(A) =

1 2 09 1 21 3 4

Matrices

An obvious fact: Two matrices are said to be equal if they havethe same order and all their elements are equal.

A =

1 2 09 1 2

1 3 4

; B =

1 2 09 1 2

1 5 4

; C =

1 2 0 19 1 2 3

1 3 4 6

A = B and A = C.


5/33

Types of Matrices

An upper triangular matrix has the form

a11 a12 a1n0 a22 a2n

... ... . . . ...0 0 ann

An lower triangular matrix has the form

b11 0 0b21 b22 0

......

. . ....

bn1 bn2 bnn

An diagonal matrix has the form

d11 0

0

0 d22 0...

.... . .

...0 0 dnn

Matrix addition and subtraction

Two matrices may be added or subtracted if they have the sameorder.

2 11 0

1 3

+1 0

2 13 1

=

3 13 12 4

1 3 6 11 1 2 1

1 1 1 11 1 1 1

=

0 2 5 0

2 0 1 0

2 11 0

1 3

+

1 3 6 1

1 1 2 1

not defined


6/33

Matrix addition and subtraction

Matrices are commutative and associative under addition:

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

A zero matrix is any matrix with all elements equal to zero,and isusually written 0:

0 00 00 0

;

0 0 00 0 00 0 0

A + 0 = 0 + A = A

Matrix Multiplication

It is straightforward to multiply matrices by a number k:

k1 2 0

9 1 21 3 4

= k 2k 0

9k k 2kk 3k 4k

Multiplying two matrices together is more complicated.

1 2 00 1 21 3 4

1 0 11 1 32 0 2

=?


7/33


We define the product of a row vector and a column vector asbeing the sum of the products of their components. e.g:

1 2 0

112

= 1 1 + 2 1 + 0 2 = 1

To multiply two matrices we multiply the rows of the first one by

the columns of the second one.


1 2 00 1 2

1 3 4

1 0 11 1 3

2 0 2

=

1


8/33


1 2 00 1 2

1 3 4

1 0 11 1 3

2 0 2

=

1 2


1 2 00 1 2

1 3 4

1 0 11 1 3

2 0 2

=

1 2 5


9/33


1 2 00 1 2

1 3 4

1 0 11 1 3

2 0 2

=

1 2 55

and so forth. . .


1 2 00 1 2

1 3 4

1 0 11 1 3

2 0 2

=

1 2 55 1 7

12 3 18


10/33

Rule for matrix multiplication:

The number of columns of the first matrix

must equal the number of rows of the

second.


When you multiply 2 matrices the order matters

Matrix multiplication is associative

A(BC) = (AB)C

but is non-commutative, i.e., in general

AB = BA


11/33


Because in general AB = BA, we must always specify on whichside we are doing the matrix multiplication:

X =

2 01 1

A =

1 03 1

Multiplying X on the left by A:

AX =

1 03 1

2 01 1

=

2 05 1

Multiplying X on the right by A:

XA =

2 01 1

1 03 1

=

2 04 1


Matrix multiplication has some other strange properties:

AB = 0 does not necessarily mean that A = 0 or B = 01 10 0

2 0

2 0

=

0 00 0

AD = AC does not necessarily mean that D = C


12/33


The product of two matrices is a matrix whose elements can bewritten in a compact form as:

[AB]ij =Nk=1

aikbkj

where N is the number of columns of A and the number of rows

of B.

Identity Matrix

The identity matrix I is the matrix with the property

AI = IA = A

I =

1 00 1

for 2 2 matrices

I =

1 0 00 1 0

0 0 1

for 3 3 matrices


13/33

Inverse of a Matrix

The inverse of a square matrix A is written A1 and has theproperty

A1A = AA1 = I

A matrix only has one inverse

This means that if it exists, the inverse of a matrix is unique.

Inverse of a Matrix

Inverse of a 2 2 matrix.

if A =

a bc d

, A1 =

1

ad bc

d b

c a

Remarks:

ad bc is the determinant of the matrix.

The inverse is defined only if det(A) = ad bc = 0.

If det(A) = 0, A is said to be a singular matrix.

For larger square matrices there is a more general way tocompute inverses. Well see that shortly.

The sum of the diagonal elements of A is called the trace ofA: tr(A) = a + d.


14/33

Transpose of a Matrix

The transpose of a matrix is obtained by interchanging the rowsand the columns:

If A =

1 3 6 1

1 1 2 1

, AT =

1 13 16 21 1

If A and B have the same order

(A + B)T

= AT

+ BT

(AT)T = A

(AB)T = BTAT

Matrix Algebra

Using the rules for addition, subtraction, multiplication andinverses, as well as the special matrices I and 0, we can re-arrangematrix equations.

Example

Given

A =

1 02 1

; B =

1 22 1

; C =

0 00 1

Rearange the equation

A + 2BX = C

to find X.


15/33

Matrix Algebra

Formally we have

A + 2BX = C

2BX = C A

BX =1

2(C A)

X =1

2B1(C A) (multiply on the left)

B =1 2

2 1

B1

=1

1 4 1 2

2 1

= 1

5 1 2

2 1

C A =

1 02 0

X =

1

10

1 2

2 1

1 02 0

=

1

10

3 04 0

Matrix Algebra

Example

Given

B =

1 11 1

; C =

4 20 1

Rearange the equation

X+ XB = C

to find X.


16/33

Matrix Algebra

X+ XB = C

X(I+ B) = C

X = C(I + B)1 (not (I + B)1C)

I + B =

2 11 2

(I + B)1 =1

3 2 1

1 2 X =

1

3

4 20 1

2 1

1 2

=

1

3

6 01 2

Linear Systems


17/33

Linear system of equations

A system of equations is a set of mathematical statementscontaining several unknown quantities.

If unknowns can be found so that all the statements aresatisfied then we say that the system has a solution.

There are three possibilities when trying to find the solution:

1 There is too little information to find the unknowns.

2

There is too much information to find the unknowns.3 There is exactly the right amount of information.


Example

From the two statements

1

There are twice as many sheep as people in Australia.2 There are 20 million more sheep than people in Australia.

we can construct the system of equationss = 2p

s = p + 20 p = 20, s = 40 (in millions)

Unique solution


18/33


Example

From the two statements

1 There are 20 million more sheep than people in Australia

2 There is the same number of sheep as people in Australia

we can construct the system of equations

s = p + 20s = p

No solution The two statements are inconsistent

Linear system of equations Graphical interpretation


19/33


An equation for several unknowns x1, x2 . . . xn is linear if it can beput in the form

a1x1 + a2x2 + + anxn = b

where the coefficients a1 . . . an and b are constants.A set of linear equations is known as a linear system.


A linear system of m equations in n unknowns is a set of equationsof the form

a11x1 + a12x2 + + a1nxn = b1

a21x1 + a22x2 +

+ a2nxn = b2...

am1x1 + am2x2 + + amnxn = bm

A solution is a set of numbers (x1, x2, . . . , xn) that satisfiesall m equations.

This set of numbers can be gathered into a vector.


20/33


The augmented matrix associated with the system is

a11 a12 a1n b1a21 a22 a2n b2

......

. . ....

am1 am2 amn bm

A system with zeros below the main diagonal is said to be in

row-echelon form.A system in row-echelon form can be solved easily byback-substitution.


Example

1 2 1 110 3 1 250 0 2 70

x1 + 2x2 + x3 = 11

3x2 + x3 = 252x3 = 70

From the third row we get: 2x3 = 70 x3 = 35

From row 2 we get: 3x2 + x3 = 25 x2 =10

3Finally we get: x1 + 2x2 + x3 = 11

x1 + 203

+ 35 = 11 x1 = 923


21/33

Row-echelon form

A matrix is in row echelon form ifAll rows that contain only zeros are grouped at the bottom ofthe matrix.

In any two consecutive non-zero rows, the leftmost non-zeroentry in the lower row occurs farther to the right than theleftmost nonzero entry in the upper row.

We call a row (column) nonzero if it contains at least one nonzero

entry.

Row-echelon form

A matrix in row-echelon form has a staircase pattern like thefollowing matrix:

0 0 0 0 0 0 0 0 0 0 0 0

where the greek letters correspond to non-zero values and the starsmay or may not be zero.

For the augmented matrix in row echelon form, each leftmostnonzero entry corresponds to a basic variable.

For instance, for the matrix above the basic variables are x1,x2 and x5.


22/33

Row-echelon form

The row-echelon form can give us some important informationabout a system. If a system is consistent (i.e., if it has a solution)

if all variables are basic variables the solution is unique.

if there is at least one non-basic variable there are infinitelymany solutions.

Row-echelon formThe number of solutions (unique, none, infinitely many) can befound by looking at the augmented matrix in row-echelon form:

a11 a12 a1n b1a21 a22 a2n b2

......

. . ....

0 0

amn bm

unique solution if a11, a22, . . . , amn = 0

a11 a12 a1n b1a21 a22 a2n b2

......

. . ....

0 0 0 bm

no solution if bm = 0

a11 a12 a1n b1a21 a22 a2n b2

.

.....

.. .

.

..0 0 0 0

infinitely many solutions

The number of non-zero rows in row-echelon form is the rank of the


23/33

Gauss elimination method for solving a linear system of

equations

Any system of equations is left unchanged by:1 Multiplying a row by a non-zero scalar.

2 Adding or subtracting two rows.

3 Swapping any two rows.

These are known as row operations.

In Gauss Elimination, we use row-operations to change the

system to row-echelon form.We can then solve the system by back-substitution.

Gauss elimination

Use Gauss elimination to solve the system

x1 + x2 + 2x3 = 23x1 x2 + x3 = 6

x1 + 3x2 + 4x3 = 4

The augmented matrix is 1 1 2 23 1 1 6

1 3 4 4

We need to use row operations to put the augmented matrix inrow-echelon form.


24/33

Gauss elimination

1 1 2 23 1 1 6

1 3 4 4

Example of a linear system: Balancing a chemical reaction

Consider the following chemical reaction:

C2H6 + O2 CO2 + H2O

Balancing the reaction means finding positive integers n1, n2, n3,and n4 such that

n1C2H6 + n2O2 n3CO2 + n4H2O

has the same number of atoms of each element on each side of theequation.

We can cast the problem as a linear system.


25/33

Example of a linear system: Balancing a chemical reaction

n1C2H6 + n2O2 n3CO2 + n4H2O

Inverse of a Matrix

We saw how to find the inverse of a 2 2 matrix.

If we can find the inverse of a n n matrix we could solve ageneral system

AX = b X = A1b


26/33

Inverse of a 3 3 Matrix. Method 1: row operations

1 Write the LHS of the system in matrix form.

1 2 21 2 13 2 1

2 Augment the matrix with the identity matrix.

1 2 2 1 0 01 2 1 0 1 0

3 2 1 0 0 1

3 Use row operations to transform the left half into the identity.The right half will then be the inverse.


Example

Solve

x1 2x2 + 2x3 = 1

x1 2x2 + x3 = 1

3x1 2x2 + x3 = 1

by computing the inverse of the matrix associated with the linearsystem.


27/33


The augmented matrix is

1 2 2 1 0 01 2 1 0 1 03 2 1 0 0 1

R1 R1 2R3R3 R1 R2

5 2 0 1 0 21 2 1 0 1 00 0 1 1 1 0

R2 R2 R3

5 2 0 1 0 21 2 0 1 2 0

0 0 1 1 1 0

R1 R1 + R2R2 5R2 +R1

4 0 0 0 2 20 8 0 4 10 2

0 0 1 1 1 0

R1 R1/4R2 R2/8

1 0 0 0 1/2 1/20 1 0 1/2 5/4 1/4

0 0 1 1 1 0


Therefore the inverse is

0 1/2 1/21/2

5/4 1/41 1 0

and the solution of the system is

x1x2x3

=

0 1/2 1/21/2 5/4 1/4

1 1 0

11

1

=

01/2

0


28/33

Cramers rule for solving a linear system

Given the system

a11 a12a21 a22

x1x2

=

b1b2

the solution can be expressed in terms of determinants:

x1 =

b1 a12b2 a22

a11 a12a21 a22

; x2 =

a11 b1a21 b2

a11 a12a21 a22

Of course this requires that the determinant of the system isnon-zero.

Cramers rule for solving a linear system

Cramers rule works for higher order systems:

a11 a12 a13a21 a22 a23

a31 a32 a33

x1x2

x3

=

b1b2

b3

x1 =

b1 a12 a13b2 a22 a23b3 a32 a33

a11 a12 a13a21 a22 a23a31 a32 a33

; x2 =

a11 b1 a13a21 b2 a23a31 b3 a33

a11 a12 a13a21 a22 a23a31 a32 a33

; x3 =

a11 a12 b1a21 a22 b2a31 a32 b3

a11 a12 a13a21 a22 a23a31 a32 a33

But how do we compute a general determinant?


29/33

Determinant

There are two (main) methods for calculating the determinant of an n matrix:

1 Cofactor expansion.

2 Row operations.

Determinant Cofactor expansion

Steps to expand using cofactors with respect to a given row:

1 Multiply each coefficients of this row by (1)i+j where i is itsrow number and j its column number.

2 Multiply each coefficients of this row with the determinant of

the minor matrix i.e., the matrix obtained by deleting thecoefficients row and column.a11 a12 a13a21 a22 a23a31 a32 a33

= a11a22 a23a32 a33

a12a21 a23a31 a33

+ a13a21 a22a31 a32

The cofactors are the signed determinants of the small matrices.You can perform the cofactor expansion with respect to any row orcolumn.


30/33

Determinant Cofactor expansion

Calculate the determinant of

1 2 21 2 1

3 2 1

Inverse of a 3 3 Matrix. Method 2: Cofactors

1 Calculate the determinant of the matrix det(A).

2 If det(A) = 0 the inverse is given by

A1 =1

det(A)CT

where C is the matrix of the cofactors of A.


31/33

Determinant Row operations

Interchanging of two rows multiplies the determinant by -1.

Adding a multiple of one row to another leaves thedeterminant unchanged.

Multiplying an entire row by a number k multiplies thedeterminant by k.

These rules also hold for column operations.

Determinant Useful tips

Things to look for before trying to calculate a determinant

If any row or column contains only zeros then the determinantis zero.

If any two rows or columns are identical then the determinantis zero.

If any two rows or columns are multiples of each other thenthe determinant is zero.


32/33

Determinant Special case

To find the determinant of a matrix in row-echelon form, multiplydown the diagonal

1 2 30 5 20 0 1

= 1 5 1 = 5

General properties of determinant

det(A + B) = det(A)+det(B).

det(AB) = det(A)det(B).

det(A)det(A1)=1.A has an inverse if and only if det(A) = 0.

Steps to evaluate determinants using row reduction:

1 Reduce the determinant to row-echelon form (keeping track ofrow-swaps and multiplication by numbers).

2 Multiply the coefficients along the diagonal.


33/33

Linear Systems Summary

We have seen 3 methods to solve a linear system

1 Gauss elimination.

2 Using the inverse matrix.

3 Cramers rule.

Date post:	07-Apr-2018
Category:	Documents
Upload:	aaron-lee
View:	221 times
Download:	0 times

MM2 Linear Algebra 2011 I 1x2

Documents