Systems of Linear Equations, RREF

Announcements

Ï If anyone has not been able to access the class website please

email me at [email protected]

Ï Two corrections made to Tuesday's lecture slides (Jan 12).

See slides 27, 28 (I had written R3-R1 instead of R3-2R1) and

39 (I didn't make it completely RREF before saving).

Ï If you spot any typos in the slides any time, please bring it to

my attention asap.

Ï Lst day to drop this class with refund is tomorrow.

De�nition

Echelon form (Row Echelon form, REF): A rectangular matrix is of

Echelon form (Row Echelon Form or REF) if

Ï All nonzero rows are ABOVE any rows with all zeros

Ï Each leading entry of a row in a column is to the RIGHT to

the leading entry of the row above it (results in a STEP like

shape for leading entries)

Ï All entries in a column below the leading entry are zero

De�nition

Reduced Echelon form (Reduced Row Echelon form, RREF): A

rectangular matrix is of Echelon form (Reduced Row Echelon Form

or RREF) if

Ï The leading entry in each nonzero row is 1

Ï Each leading 1 is the only nonzero element in its column.

Important

Each matrix is row equivalent to EXACTLY ONE reduced echelon

matrix. In other words, RREF for a matrix is unique.

Note: Once you obtain the echelon form of a matrix

Ï If you do any more row operations, the leading entries will not

change positions

Ï The leading entries are always in the same positions in any

echelon form starting from the same matrix (These become 1

in reduced echelon form, RREF)

Pivot Position, Pivot Column

De�nition

Given a matrix A, the location in A which corresponds to a leading

1 in the reduced echelon form is called PIVOT POSITION

De�nition

The column in A that contains a pivot position is called a PIVOT

COLUMN

1 0 −3 0

0 1 5 0

0 0 0 1

0 0 0 0

Location in A corresponding to blue circles gives pivot position

Example, Problem 4 sec 1.2

Row reduce the matrix A below to echelon form and locate the

pivot columns.

1 3 5 7

3 5 7 9

5 7 9 1

Caution!! This is NOT an augmented matrix of any linear system

Basic row operations as usual1 3 5 7

3 5 7 9

5 7 9 1

. R3-5R1R2-3R1

=⇒1 3 5 7

0 −4 −8 −120 −8 −16 −34

Divide R3 by 2 1 3 5 7

0 −4 −8 −120 −4 −8 −17

Do R3-R2 1 3 5 7

0 −4 −8 −120 0 0 −5

3 5 7 9

5 7 9 1

. R3-5R1R2-3R1

=⇒1 3 5 7

0 −4 −8 −120 −8 −16 −34

Divide R3 by 2

1 3 5 7

0 −4 −8 −120 −4 −8 −17

Do R3-R2 1 3 5 7

0 −4 −8 −120 0 0 −5

3 5 7 9

5 7 9 1

. R3-5R1R2-3R1

=⇒1 3 5 7

0 −4 −8 −120 −8 −16 −34

0 −4 −8 −120 −4 −8 −17

Do R3-R2

1 3 5 7

0 −4 −8 −120 0 0 −5

3 5 7 9

5 7 9 1

. R3-5R1R2-3R1

=⇒1 3 5 7

0 −4 −8 −120 −8 −16 −34

0 −4 −8 −120 −4 −8 −17

Do R3-R2 1 3 5 7

0 −4 −8 −120 0 0 −5

Make sure we have only 0 above and below the leading 1

Divide R2 by -4 and divide R3 by -5 to get 1 as leading term. This

1 3 5 7

0 1 2 3

0 0 0 1

1 3 5 7

0 1 2 3

0 0 0 1

.R1-3R2

=⇒1 0 −1 −20 1 2 3

0 0 0 1

gives 1 3 5 7

0 1 2 3

0 0 0 1

1 3 5 7

0 1 2 3

0 0 0 1

.R1-3R2

=⇒1 0 −1 −20 1 2 3

0 0 0 1

gives 1 3 5 7

0 1 2 3

0 0 0 1

1 3 5 7

0 1 2 3

0 0 0 1

.R1-3R2

=⇒1 0 −1 −20 1 2 3

0 0 0 1

1 0 −1 −2

0 1 2 3

0 0 0 1

. R1+2R3

=⇒1 0 −1 0

0 1 2 3

0 0 0 1

1 0 −1 −2

0 1 2 3

0 0 0 1

. R1+2R3

=⇒1 0 −1 0

0 1 2 3

0 0 0 1

1 0 −1 0

0 1 2 3

0 0 0 1

.R2-3R3

1 0 −1 0

0 1 2 0

0 0 0 1

Pivot positions in blue circle

1 0 −1 0

0 1 2 3

0 0 0 1

.R2-3R3

1 0 −1 0

0 1 2 0

0 0 0 1

Pivot positions in blue circle

Answer

Ï We have pivot positions at the blue circles

Ï We need to �nd the pivot columns

Ï These are the corresponding columns in the original matrix,

that is Column 1, Column 2 and Column 4 in A (not the

columns in RREF)

Answer

Ï We have pivot positions at the blue circles

Ï We need to �nd the pivot columns

Ï These are the corresponding columns in the original matrix,

that is Column 1, Column 2 and Column 4 in A (not the

columns in RREF)

Solutions of Linear systemsThe row reduction steps gives solution of linear system when

applied to augmented matrix of the linear system

Suppose the following matrix is the RREF of the augmented matrix

of a linear system. 1 0 8 5

0 1 4 7

0 0 0 0

What is the corresponding system of equations? (Note that this

system has 3 variables since the augmented matrix has 4 columns)

x + 8z = 5

y + 4z = 7

Solutions of Linear systemsThe row reduction steps gives solution of linear system when

applied to augmented matrix of the linear system

Suppose the following matrix is the RREF of the augmented matrix

of a linear system. 1 0 8 5

0 1 4 7

0 0 0 0

What is the corresponding system of equations? (Note that this

system has 3 variables since the augmented matrix has 4 columns)

x + 8z = 5

y + 4z = 7

Basic Variables, Free variables

Ï Here x and y correspond to the pivot columns of the matrix.

They are called BASIC variables

Ï The remaining variable z which corresponds to the non-pivot

column is called a FREE variable

Ï For a consistent system, we can express the solution of the

system by solving for the basic variables in terms of free

variables

Ï This is possible because RREF makes sure that each basic

variable lies in one and only one equation.

Solution to the system is thus...

x = 5 − 8z

y = 7 − 4z

z free

Ï When we say that z is free, we mean that we can assign any

value for z . Depending on your choice of z , x and y will get

�xed.

Ï For each choice of z , there is a di�erent solution set.

Ï Every solution of the system is determined by choice of z

We thus have a GENERAL solution of the system.

Existence and Uniqueness theorem

For a linear system to be consistent,

Ï The rightmost column of an augmented matrix MUST NOT

be a pivot column to avoid the "0=non-zero" situation.

Ï If the linear system is consistent, (i)the solution is either

unique (no free variables) or (ii) in�nitely many solutions (at

least one free variable)

Problem 2, section 1.2

Which of the following matrices are REF and which others are in

A=1 1 0 1

0 0 1 1

0 0 0 0

B =1 1 0 0

0 1 1 0

0 0 1 1

A=1 1 0 1

0 0 1 1

0 0 0 0

B =1 1 0 0

0 1 1 0

0 0 1 1

A=1 1 0 1

0 0 1 1

0 0 0 0

B =1 1 0 0

0 1 1 0

0 0 1 1

A=1 1 0 1

0 0 1 1

0 0 0 0

B =1 1 0 0

0 1 1 0

0 0 1 1

1 0 0 0

1 1 0 0

0 1 1 0

0 0 1 1

Neither (The leading element in a row must be to the right of the

leading element in the row above, no step pattern)

0 1 1 1 1

0 0 2 2 2

0 0 0 0 3

0 0 0 0 0

1 0 0 0

1 1 0 0

0 1 1 0

0 0 1 1

0 1 1 1 1

0 0 2 2 2

0 0 0 0 3

0 0 0 0 0

1 0 0 0

1 1 0 0

0 1 1 0

0 0 1 1

0 1 1 1 1

0 0 2 2 2

0 0 0 0 3

0 0 0 0 0

1 0 0 0

1 1 0 0

0 1 1 0

0 0 1 1

0 1 1 1 1

0 0 2 2 2

0 0 0 0 3

0 0 0 0 0

Find the general solutions of the systems whose augmented matrix

is given below

A= 1 −7 0 6 5

0 0 1 −2 −3−1 7 −4 2 7

Do R3+R1 to get new R3

B =1 −7 0 6 5

0 0 1 −2 −30 0 −4 8 12

Divide last row by -4

B =1 −7 0 6 5

0 0 1 −2 −30 0 1 −2 −3

is given below

A= 1 −7 0 6 5

0 0 1 −2 −3−1 7 −4 2 7

B =1 −7 0 6 5

0 0 1 −2 −30 0 −4 8 12

B =1 −7 0 6 5

0 0 1 −2 −30 0 1 −2 −3

is given below

A= 1 −7 0 6 5

0 0 1 −2 −3−1 7 −4 2 7

B =1 −7 0 6 5

0 0 1 −2 −30 0 −4 8 12

B =1 −7 0 6 5

0 0 1 −2 −30 0 1 −2 −3

Do R3-R2 to get new R3

B =1 −7 0 6 5

0 0 1 −2 −30 0 0 0 0

Our system of equations thus is

x1 − 7x2 + 6x4 = 5

x3 − 2x4 = −3

Column 1 and Column 3 have the pivot positions. The

corresponding variables x1 and x3 are basic variables. So, x2 and x4are free variables.

To get the general solution, express the basic variables in terms of

the free variables. This gives

x1 = 5 − 6x4 + 7x2

x3 = −3 + 2x4

x2 and x4 free

Choose h and k such that the system has (a) no solution (b)

exactly one solution and (c) in�nitely many solutions

x1 + 3x2 = 2

3x1 + hx2 = k

Solution: As usual, start with the augmented matrix

. R2-3R1

[1 3 2

0 h−9 k −6

Ï If h= 9 AND k = 6, we have 0=0 which means there is a free

variable. So in�nite number of solutions for this choice.

Ï If h= 9 and k 6= 6 we have "0=non-zero" situation. So

inconsistent for this choice of h and k .

Ï If h 6= 9 and for any value of k we have exactly one solution.

. R2-3R1

[1 3 2

0 h−9 k −6

Ï If h= 9 AND k = 6, we have 0=0 which means there is a free

variable. So in�nite number of solutions for this choice.

Ï If h= 9 and k 6= 6 we have "0=non-zero" situation. So

inconsistent for this choice of h and k .

Ï If h 6= 9 and for any value of k we have exactly one solution.

Section 1.3, Vector Equations

Ï A vector is nothing but a list of numbers (for the time being).

Ï In the 2 dimensional x −y plane, this will look like u=[1

−2]or w=

]Ï Such a matrix with only 1 column is called a column vector or

just vector

Ï Adding (or subtracting) 2 vectors is easy, just subtract the

corresponding entries. So if u=[1

]and v=

−2]then

]Ï You can multiply a vector with a number, it scales the vector

accordingly. So if u=[1

]then 5u=

Some Geometry

Ï The column vector =[a

]identi�es the point (a,b) on the x−y

Ï The vector u=[3

]is an arrow from the origin (0,0) to (3,2)

x(0, 0)

(3, 2)

(-2, 1)

Some Geometry

Ï The vector u=[3

x(0, 0)

(3, 2)

(-2, 1)

Some Geometry

Ï The vector u=[3

x(0, 0)

(3, 2)

(-2, 1)

Some Geometry

Ï The vector u=[3

x(0, 0)

(3, 2)

(-2, 1)

Some Geometry

Ï The vector u=[3

x(0, 0)

(3, 2)

(-2, 1)

Parallelogram rule for vector addition

If u and v are points in a plane, then u+v is the fourth vertex of a

parallelogram whose other vertices are u, 0 and v

parallelogram whose other vertices are u, 0 and vy

Scalar Multiplication

Multiplying a vector by a number scales it correspondingly.

Multiplying a vector by a number scales it correspondingly.y

Vectors in 3-D

Vectors in the x −y −z plane is a 3X1 column matrix. Here u=234

is illustrated.

Vectors in 3-D

is illustrated.

Vectors in 3-D

is illustrated.

Systems of Linear Equations, RREF

Education

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