linear algebra handout

YOGYAKARTA STATE UNIVERSITY

MATHEMATICS AND NATURAL SCIENCES FACULTY MATHEMATICS EDUCATION STUDY PROGRAM Topic: Linear Systems

An equation of the type a₁x₁ a₂x₂ a_nx_n b , (1) expressing b in terms of the variables x₁,x₂,,x_n and the known constants

n a a

a₁, ₂,, , is called a linear equation. A solution to the linear equation (1) is a sequence of nnumbers s₁,s₂,,s_n, which has the property that (1) is satisfied when x₁s₁,x₂ s₂,,x_n s_n are subtituted in (1).

Example 1: x₁ 2,x₂ 3,andx₃ 4 is a solution to the linear equation 13

4 3

6x₁  x₂  x₃  , because 6(2)3(3)4(4)13. This is not the only solution to the given linear equation, since x₁ 3,x₂ 1,dan x₃ 7 is another solution.

A system of equations that has no solutions is said to be inconsistent; if there is at least one solution of the system, it is called consistent. To illustrate the possibilities that can occur in solving systems of linear equations, consider a general system of two linear equations in the unknowns x and y:

  

 

 

zero) both not , (

zero) both not , (

2 2 2 2 2

1 1 1 1 1

b a c y b x a

b a c y b x a

The graphs of these equations are lines; call them l₁ and l₂. Since a point (x, y) lies on a line if and only if the numbers x and y satisfy the equation of the line, the solutions of the system of equations correspond to points of intersection of l₁ and l₂. There are three possibilities, illustrated in Figure 1:

The lines l₁ and l₂ may be parallel, in which case there is no intersection and consequently no solution to the system. The lines l₁ and l₂ may intersect at only one point, in which case the system has exactly one solution. The lines l₁ and l₂ may coincide, in which case there are infinitely many points of intersection and consequently infinitely many solutions to the system.

Although we have considered only two equations with two unknowns here, we will show later that the same three possibilities hold for arbitrary linear systems Every system of linear equations has no solutions, or has exactly one solution, or has infinitely many solutions.

More generally, a linear system is a system of m linear equations in n unknowns, and can be conveniently denoted by

      

.

2 2 1

1

2 2

2 22 1

21

1 1

2 12 1 11

m n mn m

m

n n

n n

b x a x

a x a

b x a x

a x a

b x a x

a x a

 

 

 

 

 

 



 

 

 

(2)

A solution to a linear system (2) is a sequence of nnumbers s₁,s₂,,s_n, which has the property that each equation in (2) is satisfied when

n n s x s x s

x₁  ₁, ₂  ₂,,  are subtituted in (2). The Solution of The Linear System

The basic method for solving a system of linear equations is to replace the given system by a new system that has the same solution set but is easier to solve. This new system is generally obtained in a series of steps by applying the following three types of operations to eliminate unknowns systematically:

The Linear Equation Operation 1. Interchange two equations.

2. Multiply an equation by a nonzero constant. 3. Add a multiple of one equation to another.

Example 2: Given the linear system:

    

  

  

  

0 5 6 3

1 3 4 2

9 2

z y x

z y x

z y x

To eliminate x from the second equation, we add (2) times the first equation to the second one to obtain

    

  

  

  

0 5

6 3

17 3

4

9 2

z y x

z y

z y x

To eliminate x from the second equation, we add (3) times the first equation to the third one to obtain

    

  

  

  

27 11

3

17 7

2

9 2

z y

z y

z y x

Multiply the second equation with 21 , to obtain

    

  

  

  

27 11

3

9 2

2 17 2

7

z y

z y

z y x

Added (3) times the second equation to the third one to obtain

    

  

  

  

2 3 2

1

2 17 2

7

9 2

z z y

z y x

Multiply the third equation with 2 , to obtain

    

   

 



3 9 2

2 17 2

7

z z y

z y

x

Added (1) times the second equation to the first one to obtain

    

   

 

3

2 17 2

7

2 35 2

11

z z y

z x

Added (11_{2 ) times the third equation to the first one and (} 2

7_{) times the third}

equation to the second one to obtain

    

  

3 2 1 z y x

We conclude that the solution of the linear system is x1,y2,z 3.

Exercises 1:

1. Which of the following are linear equations in x₁,x₂ andx₃? (a) x₁ 5x₂  2x₃ 1

(b)x₁ 3x₂ x₁x₃ 2 (c) x₁ 7x₂ 3x₃ (d)x₁2 x₂ 8x₃ 5 (e) 35 2 _{2 3} 4

1  x x 

x

(f) 13

3 2

1 7

3 1

2  

 x x

x 

2. Find the solution set of each of the following linear equations. (a)7x5y 3.

(b) 3x₁5x₂ 4x₃ 7

(c) 8x₁2x₂ 5x₃6x₄ 1 (d) 3v8w2xy4z 0.

3. Find the solution of given the linear system. (a)

    

  

   

  

10 7 4 3

1 2 3

8 2

z y x

z y x

z y x

(b)

    

  

  

 

0 2

6

7 4 3

1 2

2

z y x

z y x

z x

4. (a) Find a linear equation in the variables x and y that has the general solution t

(b)Show that xt,

2 5 2 1



 t

y is also the general solution of the equation in part (a).

5. Consider the system of equations

    

  

 

  

c z y x

b z x

a z y x

3 2

2

Show that for this system to be consistent, the constant a, b, and c must satisfy c = a + b.

6. Show that if the linear equations x₁ kx₂ c and x₁ lx₂ d have the same solution set, then the equations are identical.

7. For which value(s) of the constant k does the system 

 

 

 

k y x

y x

2 2

3

have no solutions? Exactly one solution? Infinitely many solutions? Explain your reasoning.

8. Consider the system of equations

    

 

 

 

m fy ex

l dy cx

k by ax

Indicate what we can say about the relative positions of the lines axbyk, l

dy

cx  , and ex fymwhen (a) the system has no solutions.

(b) the system has exactly one solution. (c) the system has infinitely many solutions.

9. If the system of equations in Exercise 8 is consistent, explain why at least one equation can be discarded from the system without altering the solution set. 10. If k = l = m = 0 in Exercise 8, explain why the system must be consistent.

What can be said about the point of intersection of the three lines if the system has exactly one solution?

YOGYAKARTA STATE UNIVERSITY

MATHEMATICS AND NATURAL SCIENCES FACULTY MATHEMATICS EDUCATION STUDY PROGRAM Topic: Matrices

Definition 1. An mn matrix A is a rectangular of mn real (or complex) numbers arranged in m horizontal rows and n vertical columns :

   

 

   

  

mn m

m

n n

a a

a

a a

a

a a

a A

    

 

2 1

2 22

21

1 12

11

. (3)

The ith row of A,



a_i1 a_i2  a_in



, (1  i  m); the jth column of A,     

 

    

 

mj j j

a a a



2 1

,

(1 j n).

If m = n, we say that A is a square matrix of order n, and that the numbers nn

a , , a ,

a_{11 22}  form the main diagonal A. We refer to the number a_ij, which is in the ith row and jth column of A, and (3) often write as A[a_ij]. Sum of all main diagonal is called trace A (tr A).

Example 3: _

  

   

1 0 1

3 2 1

A is an 23 matrix with

. a , a , a ,

a₁₂2 ₁₃ 3 ₂₂0 ₂₃ 1 _     

 

3 2

4 1

B is an 22 matrix, with

). 2 ( ) 3 ( 1 )

(A      tr

    

      

2 1 1

    

    

 

2 1 3

1 0 2

0 1 1

D is an 33 matrix, with tr(A)3. E 

 

3 is an 11 matrix.



1 0 2





F is an 13 matrix or row matrix.

The Equality of Two Matrices

Definition 2. Two mn matrices A[a_ij] and B[b_ij] are said to be equal if ij

ij b

a  , 1im,1 j n, that is, if corresponding elements are equal.

Example 4: The matrices

    

    

 

 

5 4 0

4 3 2

1 2 1

A and

    

    

 

z y

x w B

4 4 2

2 1

are equal if w1,x3,y0 and

5  z .

Matrix Addition

Definition 3. If A = [aij] and B = [bij] are mn matrices, then A+B is a mn matrix C =[cij], defined by cij = aij + bij (1im,1 jn) and the difference between A and B is D = [dij] noted as A(1)B AB, defined by

ij ij ij a b d   .

Example 5:

1. Given _

  

 

  

3 1 2

4 2 1

A and _

  



 



1 3 1

4 2 0

B .

Then _

  

      

 

 

 

      

4 2 3

0 0 1 1

3 3 1 1 2

) 4 ( 4 2 2 0 1 B

A .

2. If _

  



 



1 2 4

5 3 2

A and _

  

 

  

2 5 3

3 1 2

   

 

  

   

 

  

     

3 3 1

8 4 0 2 1 5 2 3 4

3 5 1 3 2 2 B

A .

3. Given _

     

14 2

8 4

A , _

  



 



4 10

2 6

B and _

  

 

  

3 21

4 10

C . Then

   

      

 

  

    

21 9

-2 20 3

4 14 21 -10 2

4 -2 -8 10 6 4 C B

A .

Additon Matrix Properties

Given A, B, C, and D are mn matrices. (a) A + B = B + A.

(b) A + (B + C) = (A + B) + C.

(c) A + O = A. O matrix is called zero matrix or the addition identity.

(d) A + D = O. DA is called the inverse addition matrix or the negative of A.

Example 6:

1. Given .

0 0

0 0 O and 3 2

1 4

       

  



 



A Then

           

 

 

    

3 2

1 4 0 3 0 2

0 1 0 4 O

A = A.

2. Given .

2 5 4

4 3 2

   

 

 



A Then ,

2 5 4

4 3 2

   

 

    

A and we obtained

O A

A( ) .

Scalar Multiplication

Definition 4. If A=[aij] is mn matrix and r is a real number, then scalar multiple of A by r, rA, is the mn matrix B=[bij], where bij = raij (1im,1 jn).

Example 7: 1. If r3 and

              2 6 3 0 5 2 3 2 4

A , then

. 6 18 9 0 15 6 9 6 12 ) 2 )( 3 ( ) 6 )( 3 ( ) 3 )( 3 ( ) 0 )( 3 ( ) 5 )( 3 ( ) 2 )( 3 ( ) 3 )( 3 ( ) 2 )( 3 ( ) 4 )( 3 ( 2 6 3 0 5 2 3 2 4 3                                                       rA

2. Given ,

4 3 2 1        A , 1 2 3 4       

B and _

        4 2 3 1

C . Then

                           19 7 22 6 20 3 4 10 6 3 15 9 2 5 12 1 5

3B C

A .

3. Given , andany scalar

2

,

2 b a k

b a B a b b a A _              

 . Then

. 2 5 5 2 5 5 4 2 2 2 6 3 3 3 2 3 _                                   a b b a k ka kb kb ka ka kb kb ka ka kb kb ka kB kA

Scalar Multiplication Matrix Properties

If r and s are real number and A and B are matrices, then (a) r(sA) = (rs)A. (c) r(A + B) = rA + rB. (b) (r + s)A = rA + sA. (d) A(rB) = r(AB) = (rA)B.

Example 8: Given r 2, .

2 -0 4 1 1 -2 and , 1 0 2 3 2 1                    B

A Then

. 0 8 2 8 0 4 -1 4 (-2) ) ( and , 0 8 2 8 4 0 8 2 2 4 1 0 2 3 2 1 ) (                                             AB r rB A

Matrix Multiplication

Definition 5. If A = [aij] is an mpmatrix and B = [bij] is an pn matrix, then the product of A and B, denoted AB, is the mnmatrix C = [cij], defined by

           p k ik kj pj ip j i j i

ij a b a b a b a b i m j n

c

1 2

2 1

1 ... (1 ,1 ).

Example 9:

1. Given .

1 2 3 -4 5 2 -and 4 1 3 1 2 1                    B

A Then

. 16 6 2 4 ) 1 )( 4 ( ) 3 )( 1 ( ) 5 )( 3 ( ) 2 )( 4 ( ) 4 )( 1 ( ) 2 )( 3 ( ) 1 )( 1 ( ) 3 )( 2 ( ) 5 )( 1 ( ) 2 )( 1 ( ) 4 )( 2 ( ) 2 )( 1 (                              AB

2. Given .

2 2 -1 -3 4 1 and 2 1 0 1 2 4 3 2 1                         B

A If AB=C then the (3,2) entry of

AB is c32, which is R3(A) x C2(B). We now have





5.

2 1 4 2 1 0 ) ( ) ( ₂

3 

             

C B A

R

3. Given , then and are,

6 12 If . y 4 2 and 1 1 2 3 1 y x AB B x A _                          6. and -2 have we , 6 then 12 3 4 2 Then . 6 12 4 4 3 4 2                         y x y y x y y x AB

4. Given .

1 0 1 2 and 3 1 2 1                B

A Then .

3 1 7 1 but 2 2 3 2                 BA AB

We conclude AB  BA.

5. Given .

1 2 3 4 3 2 -and 5 1 4 3 2 1                    B

. 7 17 7 2 3 5 1 4 3 2 1 ) ( 2                              C B A

6. Given _

             2 1 22 21 12 11 , x x x a a a a

A . Then _

        2 22 1 21 2 12 1 11 x a x a x a x a Ax . 7. If                   j i h g f e B d c b a

A , , then AB undefined, because A and B ukurannya

tidak sesuai.

Matrix Multiplication Properties If A, B, and C have the suitable size, then

(a) A(BC) = (AB)C. (b)A(B + C) = AB + AC. (c)(A + B)C = AC + BC.

Example 10:

1. Given .

0 1 2 3 0 0 0 3 -2 2 0 1 and , 3 1 0 3 2 2 2 0 0 1 1 -2 , 4 3 2 3 2 5                                

 B C

A

Then _

                         8 30 12 56 16 43 3 3 9 6 4 8 7 3 0 4 3 2 3 2 5 ) (BC A

and .

8 30 12 56 16 43 0 1 2 3 0 0 0 3 2 2 0 1 6 8 8 16 13 6 1 19 ) ( _                              C AB

2. Given .

2 -2 0 1 2 1 -and , 1 -3 2 2 0 1 , 2 1 3 3 2 2                             

 B C

Then _   

 

  

    

    

    

 

  

2 7

1 21 3

5 2 3

2 0 2 1 3

3 2 2 ) (B C A

and .

2 7

1 21 2

0 2 6 4 7

1 15

   

 

  

           

 

  AC AB

3. Given .

2 0 5

3 2 4

   

   

A Then .

2 0 5

3 2 4 2 0 5

3 2 4 1 0

0 1

2A A

I _

  

       

         

And .

2 0 5

3 2 4 1 0 0

0 1 0

0 0 1 2 0 5

3 2 4

3 

  

        

        

    AI

Notes :

1. If AB = AC, then uncertain B = C.

2. If AB = O, then uncertain A = O or B = O. Example 11:

1. If _

  

   

     

3 2

-6 -4 and

4 2

2 1

B

A , then .

0 0

0 0

      

AB So A and B are not nol matrix but AB = O.

2. If ,

2 3

1 2 , 4 2

2 1

            

 B

A ,then

1 5

7 2

and _

  

 

   C

. but , 10 16

5 8

C B AC

AB _ 

  

   

The Transpose of Matrix

Definition 6. If A = [aij] is an mn matrix, then AT=

 

aijT , where ji T ij a

a  , 1  i  m, 1  j  n, is called the transpose of A.

Example 12:

    

    

   

  

 

  

2 3

5 2

0 4 then

, 2 5 0

3 2

4 _T

A A

    

    

  

    

    

  

3 2 4

4 1 2

0 3 6 then

, 3 4 0

2 1 3

4 2 6

T B B

   

 

   

   

    

  

3 2 4

2 3 5 then

, 3 2

2 3

4 5

T C C





    

       



1 5 3 then

, 1 5

3 _DT

D



2 1 3



then , 3 1 2

  

   

     

 T

E E

The Transpose Properties

If r is scalar number and A and B are matrices, then (a) (AT)T = A. (c) (AB)T = BTAT. (b) (A + B)T = AT + BT. (d) (rA)T = rAT.

Example 13: Given .

1 -3

2 2

1 0 and

3 1 2

2 3 1

    

      

  

 



 B

A

. 3 5

7 12 3 2

1 3

2 1 1 2 1

3 2 0 and

3 5

7 12

Then _

  

 

 

    

    

    

 

 

   

 



 T T

T

A B (AB)

Example 14: The matrices are symmetric,

    

      

   

     

1 0 0

0 1 0

0 0 1 and

6 5 3

5 4 2

3 2 1

3

I

A .

Exercises 2:

1. Suppose that A, B, C, D, and E are matrices with the following sizes: )

5 4 ( 

A , B(45), C(52), D(42), E(54)

Determine which of the following matrix expressions are defined. For those that are defined, give the size of the resulting matrix.

(a)BA (b)AC + D (c)AE + B (d)AB + B (e)E(A + B) (f)E(AC) (g)ETA (h)(AT + E)D

2. Given r = -4 and .

5 1 2

1 2 4

, 3 1 2

2 3 1

   

  

 

   

 



 B

A

Show that : (a) (A + B)T = AT + BT (b) (rA)T = rAT. 3. Given

    

          

 

 

2 1

4 2

1 3 , 3 1 2

2 3 1

B

A . Show that (AB)T = BTAT.

4. Given

. 3 2

0 1 and

, 1 2 3

-3 1 2

2 1 1

2 3

-1 2

, 0 1 3

4 2 1

-3 1 2

, 2 1

4 3

1 2 , 5 2 3

2 1 2

   

 

  

   

    

  

   



 

     

          

    

  

   



 



F E

D C

B A

If possible compute :

(a) (AB)T (b) BTAT (c) ATBT (d) BBT

(e) BTB

(f) (3C - 2E)TB (g) AT(D + F) (h) BTC + A (i) (2E)AT (j) (BT + A)C.

5. Determine a constant k such that (kA)T(kA) = 1, where

    

    

  

1 1 2

A .

For no. 6 – 7, given

           

                  

          

         

  

3 2

5 4 , 1 2 3

4 1 0

5 4 2 ,

4 2

2 -3 ,

3 1 2

5 1 4

3 1 -3 ,

2 3

1 2

0 1 ,

4 1 2

3 2 1

F E

D C

B A

If possible compute :

(a). AT and (AT)T (c). (2D + 3F)T (e). 2AT + B (b). (C + E)T and CT + ET (d). D - DT (f). (3D - 2F)T If possible compute :

(a). (2A)T (c). (3BT - 2A)T (e). (A)Tand(A)T (b). (A - B)T (d). (3AT - 5BT)T (f). (C + E + FT)T

Types of Matrices A. Identity Matrix Example 15:

1. Upper Triangular Matrix

   

 

   

       

           

 

3 0 0 0

2 -0 0 0

2 1 3 0

1 1 2 -1 , 3 0 0

1 2 -0

3 -2 1 , 1 0

2 1

C B

2. Lower Triangular Matrix                                 3 2 -2 1 0 0 1 1 0 0 3 2 -0 0 0 1 , 3 1 3 -0 2 -2 0 0 1 , 1 2 0 1 C B A

3. Diagonal Matrix

                                3 0 0 0 0 0 0 0 0 0 3 0 0 0 0 1 , 3 0 0 0 2 -0 0 0 1 , 1 0 0 1 C B A

4. Scalar Matrix

                                 3 0 0 0 0 3 0 0 0 0 3 0 0 0 0 3 , 2 -0 0 0 2 -0 0 0 2 , 1 0 0 1 C B A

5. Identity Matrix

                               1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 , 1 0 0 0 1 0 0 0 1 , 1 0 0 1 C B A

B. Square Matrices Example 16: 1. If                                       1 0 0 0 1 0 0 0 1 then 1 0 1 0 1 1 3 2 6 and 4 2 1 3 3 1 3 2 1 BA AB B

A . A and

B are called commutative matrices.

2. If 0 0 and , 0 0 , 0 1 1 0                        i i C i i B

A then AB = -BA, AC = -CA,

and

BC = -CB. A, B, and C are called not commutative matrices . 3. A is a periodic matrix with period 2, because

                                                3 0 2 9 2 3 6 2 1 dan 3 4 4 9 10 9 6 6 5 maka 3 0 2 9 2 3 6 2 1 1 2 2 A A A .

4. A and B are idempotent matrices, cause

if thenB B B

    

    



    

   

    



   

5 3 1

5 3 1

5 3 1 then

5 3 1

5 3 1

5 3 1

2

.

5. A is called nilpotent matrix with index 2, because

if A A O

    

      

   

    

  

  

0 0 0

0 0 0

0 0 0 then

4 3 1

4 3 1

4 3 1

2

.

6. A and B are called involuntary matrices, because

if then 2 ₃

4 3 3

4 3 4

1 1 0

I A

A 

    

    

 



 , and then 2 ₃

3 4 4

1 0 1

3 3 4

I B

B 

    

    

  

 

 .

C. Hermitian Matrix Example 17: Given

0 1

3 2

1 2

1

3 2 1

and 0 3

2

2 1

3 2 1 1

    

    

 

  

  

    

    



 

  

i i i

i i

i B

i i

i i

i i

A then

1. A is a Hermitian matrix, because A A

i i

i i

i i

At t 

    

    

 



 

 and

0 3

2

2 1

3 2 1 1

2. B is a Skew Hermitian Matrix,

because B B

i

i i

i i

i

Bt t 

    

    



 

   

 and

0 1

3 2

1 2

1

3 2 1

3. iB is a Hermitian Matrix, because

iB B i i

i

i i

i i

iB t t 

    

    



 



    

 and( )

0 3

2

2 1

3 2 1 1

)

(

. then 0 3 2 2 1 3 2 1 1 and , 0 3 2 2 1 3 2 1 1 A (A) i i i i i i A i i i i i i A t t _                                 

5. B is a skew Hermitian matrix, because

B B i i i i i i B i i i i i i B t

t _

                                      

 then

0 1 3 2 1 2 1 3 2 1 and , 0 1 3 2 1 2 1 3 2 1

YOGYAKARTA STATE UNIVERSITY

MATHEMATICS AND NATURAL SCIENCES FACULTY MATHEMATICS EDUCATION STUDY PROGRAM Topic: Augmented Matrix

An arbitrary system of m linear equations in n unknowns can be written as,

       . 2 2 1 1 2 2 2 22 1 21 1 1 2 12 1 11 m n mn m m n n n n b x a x a x a b x a x a x a b x a x a x a                    (4)

where x₁, x₂, …,x_n are the unknowns and the a_ijand b_i, i1,2,,m, n

j 1,2,, denote constants.

This system can be abbreviated by writing in matrix term

             mn m m n n a a a a a a a a a A        2 1 2 22 21 1 12 11 ,              n x x x x  2 1 ,              m b b b b  2 1

or Ax b.

With A is a coefisien matrix and

            m mn m m n n b a a a b a a a b a a a         2 1 2 2 22 21 1 1 12 11

is called an augmented

Example 18: The augmented matrix for the system of 3 linear equations and 3 unknowns

    

. 3 2 2 3

7 4 3 2

5 2

  

  

  

z y x

z y x

z x

With

    

    

 



2 2 3

4 3 2

1 0 2

A ,

    

     

z y x

x , and

    

     

3 7 5

b , is

    

    

 

3 2 2 3

7 4 3 2

5 1 0 2

.

Example 19: The Matrix _   



 

5 2 0 3

4 3 1 2

is an augmented matrix of linear system

  

. 5 2 3

4 3 2

 

  

z x

z y x

Exercises 3:

1. Let see a linear system

      

. 5 3

3 4

3 2

2 3

2 3

7 2

 

 



  



 

z x

z y x

z y x

w x

(a) Find the coefisien matrix. (b) Find the augmented matrix. (c) Write the system on matrix term.

2. Write the linear system of this augmented matrix :

   

 

   

  

 

6 3 1 0 3

4 2 0 0 1

3 8 7 2 3

5 4 0 1 2

.

3. Write the linear system of this augmented matrix :

    

    

 

1 4 3 1

5 2 1 0

3 4 0 2

.

      

. 4 4

7 3

2 2

4 2 3

 

 

 

 



z x

z y

y x

z y

x

(a) Find the coefisien matrix. (b) Find the augmented matrix. (c) Write the linear system on matrix term.

5. Find the answer with elimination methods 

 

 

 

8 2

9 3

y x

y x

. 6. Find the answer (if any) of this linear system

    

  

  

  

5 13 4 2

2 2

z y x

z y x

z y x

.

7. Find a and b, so this linear system has

(i). exactly one solution (ii). no solution (iii). Infinitely many solution

(a).   

 

 

5 1 2 by ax

y x

(b).   

 

  

5 2

1 y ax

by x

(c).   

 

  

3 1 ay x

by x

. 8. Given the linear system

    

 

 

 

r fy ex

q dy cx

p by ax

When the lines of the 3 linear eqution makes (i). the system has exactly one solution, (ii). the system has no solution,

(iii). the system has infinitely many solution.

9. For number 8, if the equation systems consistent, then one of the equations can be eliminated without changing the solution set.

    

  

 

  

c z y x

b z x

a z y x

3 2

2

Suppose the system is consistent then a, b, and c has satisfy cab. 11.Prove: If the linear system x₁kx₂ candx₁lx₂ d has the same solution

2. Lower Triangular Matrix                                 3 2 -2 1 0 0 1 1 0 0 3 2 -0 0 0 1 , 3 1 3 -0 2 -2 0 0 1 , 1 2 0 1 C B A

3. Diagonal Matrix

                                3 0 0 0 0 0 0 0 0 0 3 0 0 0 0 1 , 3 0 0 0 2 -0 0 0 1 , 1 0 0 1 C B A

4. Scalar Matrix

                                 3 0 0 0 0 3 0 0 0 0 3 0 0 0 0 3 , 2 -0 0 0 2 -0 0 0 2 , 1 0 0 1 C B A

5. Identity Matrix

                               1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 , 1 0 0 0 1 0 0 0 1 , 1 0 0 1 C B A

B. Square Matrices Example 16: 1. If                                       1 0 0 0 1 0 0 0 1 then 1 0 1 0 1 1 3 2 6 and 4 2 1 3 3 1 3 2 1 BA AB B

A . A and

B are called commutative matrices.

2. If 0 0 and , 0 0 , 0 1 1 0                        i i C i i B

A then AB = -BA, AC = -CA,

and

BC = -CB. A, B, and C are called notcommutative matrices . 3. A is a periodicmatrix with period 2, because

                                                3 0 2 9 2 3 6 2 1 dan 3 4 4 9 10 9 6 6 5 maka 3 0 2 9 2 3 6 2 1 1 2 2 A A A .

4. A and B are idempotent matrices, cause

if thenB B B

    

    



    

   

    



   

5 3 1

5 3 1

5 3 1 then

5 3 1

5 3 1

5 3 1

2

.

5. A is called nilpotent matrix with index 2, because

if A A O

    

      

   

    

  

  

0 0 0

0 0 0

0 0 0 then

4 3 1

4 3 1

4 3 1

2

.

6. A and B are called involuntary matrices, because

if then 2 ₃

4 3 3

4 3 4

1 1 0

I A

A 

    

    

 



 , and then 2 ₃

3 4 4

1 0 1

3 3 4

I B

B 

    

    

  

 

 .

C. Hermitian Matrix Example 17: Given

0 1

3 2

1 2

1

3 2 1

and 0 3

2

2 1

3 2 1 1

    

    

 

  

  

    

    



 

  

i i i

i i

i B

i i

i i

i i

A then

1. A is aHermitian matrix, because A A

i i

i i

i i

At t 

    

    

 



 

 and

0 3

2

2 1

3 2 1 1

2. B is a SkewHermitian Matrix,

because B B

i

i i

i i

i

Bt t 

    

    



 

   

 and

0 1

3 2

1 2

1

3 2 1

3. iB is a Hermitian Matrix, because

iB B i i

i

i i

i i

iB t t 

    

    



 



    

 and( )

0 3

2

2 1

3 2 1 1

)

(

. then

0 3

2

2 1

3 2 1 1 and

, 0 3

2

2 1

3 2 1 1

A (A) i

i

i i

i i

A i

i

i i

i i

A

t

t _

    

    



 

  

    

    

 

 

  

5. B is a skewHermitian matrix, because

B B i

i i

i i

i B

i i i

i i

i B

t

t _

    

    



 



      

   

    

 



  

  

 then

0 1

3 2

1 2

1

3 2 1

and , 0 1

3 2

1 2

1

3 2 1

YOGYAKARTA STATE UNIVERSITY

MATHEMATICS AND NATURAL SCIENCES FACULTY MATHEMATICS EDUCATION STUDY PROGRAM Topic: Augmented Matrix

An arbitrary system of m linear equations in n unknowns can be written as,

      

. 2

2 1

1

2 2

2 22 1

21

1 1

2 12 1

11

m n mn m

m

n n

n n

b x a x

a x a

b x a x

a x a

b x a x

a x a

 

 

 

 

 

 



 

 

 

(4)

where x₁, x₂, …,x_n are the unknowns and the a_ijand b_i, i1,2,,m, n

j 1,2,, denote constants.

This system can be abbreviated by writing in matrix term

   

 

   

  

mn m

m

n n

a a

a

a a

a

a a

a A

    

 

2 1

2 22

21

1 12

11

,

   

 

   

  

n x x x x



2 1

,

   

 

   

  

m b b b b



2 1

or Ax b.

With A is a coefisien matrix and

   

 

   

 

m mn m

m

n n

b a a

a

b a a

a

b a a

a



    

 

2 1

2 2 22

21

1 1 12

11

is called an augmented

Example 18: The augmented matrix for the system of 3 linear equations and 3 unknowns

    

. 3 2 2 3

7 4 3 2

5 2

  

  

  

z y x

z y x

z x

With

    

    

 



2 2 3

4 3 2

1 0 2

A ,

    

     

z y x

x , and

    

     

3 7 5

b , is

    

    

 

3 2 2 3

7 4 3 2

5 1 0 2

.

Example 19: The Matrix _   

  

5 2 0 3

4 3 1 2

is an augmented matrix of linear system

  

. 5 2 3

4 3 2

 

  

z x

z y x

Exercises 3:

1. Let see a linear system

      

. 5 3

3 4

3 2

2 3

2 3

7 2

 

 



  



 

z x

z y x

z y x

w x

(a) Find the coefisien matrix. (b) Find the augmented matrix. (c) Write the system on matrix term.

2. Write the linear system of this augmented matrix :

   

 

   

  

 

6 3 1 0 3

4 2 0 0 1

3 8 7 2 3

5 4 0 1 2

.

3. Write the linear system of this augmented matrix :

    

    

 

1 4 3 1

5 2 1 0

3 4 0 2

.

      

. 4 4

7 3

2 2

4 2 3

 

 

 

 



z x

z y

y x

z y

x

(a) Find the coefisien matrix. (b) Find the augmented matrix. (c) Write the linear system on matrix term.

5. Find the answer with elimination methods 

 

 

 

8 2

9 3

y x

y x

. 6. Find the answer (if any) of this linear system

    

  

  

  

5 13 4 2

2 2

z y x

z y x

z y x

.

7. Find a and b, so this linear system has

(i). exactly one solution (ii). no solution (iii). Infinitely many solution

(a).   

 

 

5 1 2 by ax

y x

(b).   

 

  

5 2

1 y ax

by x

(c).   

 

  

3 1 ay x

by x

. 8. Given the linear system

    

 

 

 

r fy ex

q dy cx

p by ax

When the lines of the 3 linear eqution makes (i). the system has exactly one solution, (ii). the system has no solution,

(iii). the system has infinitely many solution.

9. For number 8, if the equation systems consistent, then one of the equations can be eliminated without changing the solution set.

    

  

 

  

c z y x

b z x

a z y x

3 2

2

Suppose the system is consistent then a, b, and c has satisfy cab. 11.Prove: If the linear system x₁kx₂ candx₁lx₂ d has the same solution