Dheeraj Kataria, profile picture
Uploaded byDheeraj Kataria
PPTX, PDF2,567 views

Matrix presentation By DHEERAJ KATARIA

1) A matrix is a rectangular array of numbers arranged in rows and columns. The dimensions of a matrix are specified by the number of rows and columns. 2) The inverse of a square matrix A exists if and only if the determinant of A is not equal to 0. The inverse of A, denoted A^-1, is the matrix that satisfies AA^-1 = A^-1A = I, where I is the identity matrix. 3) For two matrices A and B to be inverses, their product must result in the identity matrix regardless of order, i.e. AB = BA = I. This shows that one matrix undoes the effect of the other.

Embed presentation

Downloaded 59 times
1 / 113
2 / 113
3 / 113
Most read
4 / 113
5 / 113
6 / 113
7 / 113
8 / 113
9 / 113
10 / 113
Most read
11 / 113
12 / 113
13 / 113
14 / 113
15 / 113
Most read
16 / 113
17 / 113
18 / 113
19 / 113
20 / 113
21 / 113
22 / 113
23 / 113
24 / 113
25 / 113
26 / 113
27 / 113
28 / 113
29 / 113
30 / 113
31 / 113
32 / 113
33 / 113
34 / 113
35 / 113
36 / 113
37 / 113
38 / 113
39 / 113
40 / 113
41 / 113
42 / 113
43 / 113
44 / 113
45 / 113
46 / 113
47 / 113
48 / 113
49 / 113
50 / 113
51 / 113
52 / 113
53 / 113
54 / 113
55 / 113
56 / 113
57 / 113
58 / 113
59 / 113
60 / 113
61 / 113
62 / 113
63 / 113
64 / 113
65 / 113
66 / 113
67 / 113
68 / 113
69 / 113
70 / 113
71 / 113
72 / 113
73 / 113
74 / 113
75 / 113
76 / 113
77 / 113
78 / 113
79 / 113
80 / 113
81 / 113
82 / 113
83 / 113
84 / 113
85 / 113
86 / 113
87 / 113
88 / 113
89 / 113
90 / 113
91 / 113
92 / 113
93 / 113
94 / 113
95 / 113
96 / 113
97 / 113
98 / 113
99 / 113
100 / 113
101 / 113
102 / 113
103 / 113
104 / 113
105 / 113
106 / 113
107 / 113
108 / 113
109 / 113
110 / 113
111 / 113
112 / 113
113 / 113
PREPARED BY:- DHEERAJ KATARIA(007)
Matrix - a rectangular array of variables or constants in horizontal rows(m) and vertical columns (n) enclosed in brackets. Element - each value in a matrix; either a number or a constant. Dimension - number of rows by number of columns of a matrix. **A matrix is named by its dimensions.
11 12 13 14 21 22 23 24 31 32 33 34 mn mn mn mn a a a a a a a a a a a a a a a a          Row 1 Row 2 Row 3 Row m Column 1 Column 2 Column 3 Column 4
Examples: Find the dimensions of each matrix. 1. A = 2 1 0 5 4 8           2. B = 1 2 3 4             0 5 3 1 3. C = 2 0 9 6       Dimensions: 3x2 Dimensions: 4x1 Dimensions: 2x4
Different types of Matrices • Column Matrix - a matrix with only one column. • Row Matrix - a matrix with only one row. • Square Matrix - a matrix that has the same number of rows and columns.
row nmrows                  mnmmm n n n aaaa a a a aaa aaa aaa A      321 3 2 1 333231 232221 131211 A matrix is a rectangular array of numbers. We subscript entries to tell their location in the array Matrices are identified by their size. Matrices and Rows "" "" columnthj rowthiaij
A matrix of m rows and n columns is called a matrix with dimensions m x n. 2 3 4 1.) 1 1 2           3 8 9 2.) 2 5 6 7 8          10 3.) 7       4.) 3 4 2 X 3 3 X 3 2 X 1 1 X 2
To add matrices, we add the corresponding elements. They must have the same dimensions. 5 0 6 3 4 1 2 3 A B               A + B 5 6 0 3 4 2 1 3           1 3 6 4       
To subtract matrices, we subtract the corresponding elements. The matrices must have the same dimensions. 1 2 1 1 3.) 2 0 1 3 3 1 2 3                     1 1 2 ( 1) 2 1 0 3 3 2 1 3                0 3 3 3 5 4           
ADDITIVE INVERSE OF A MATRIX: 1 0 2 3 1 5 A       1 0 2 3 1 5 A         
Equal Matrices - two matrices that have the same dimensions and each element of one matrix is equal to the corresponding element of the other matrix. *The definition of equal matrices can be used to find values when elements of the matrices are algebraic expressions.
1. 2x 2x  3y       y 12       * Since the matrices are equal, the corresponding elements are equal! * Form two linear equations. 2x  y 2x  3y  12 * Solve the system using substitution. Examples: Find the values for x and y 2x  y y  3y 12 4y 12 y  3 2x  3 x  3 2
* Write as linear equations.2. 3x  y x  2y       x  3 y 2       7y  7 y  1 2x  y  3 2x  6y  4 2x  1  3 2x  2 x 1 Now check your answer * Combine like terms. * Solve using elimination. 3x  y  x  3 x  2y  y  2 x  2y  y  2 1  2 1   1 2 1  2  1 1  1 2x  y  3 x  3y  2
Scalar Multiplication: 1 2 3 1 2 3 4 5 6 k           We multiply each # inside our matrix by k. 1 2 3 1 2 3 4 5 6 k k k k k k k k k           
                            812 026 2 14 58 13 2 y x               812 026 528 1143 2 y x
              812 026 56 043 2 y x               812 026 21012 086 y x Scalar Multiplication:
6x+8=26 6x=18 x=3 10-2y=8 -2y=-2 y=1
Associative Property of Addition (A+B)+C = A+(B+C) Commutative Property of Addition A+B = B+A Distributive Property of Addition and Subtraction S(A+B) = SA+SB S(A-B) = SA-SB NOTE: Multiplication is not included!!!
THE IDENTITY MATRIX IS A SQUARE MATRIX WITH ONE DOWN THE DIAGONALS In a 2 X 2 In a 3 X 3       10 01           100 010 001
FINDING THE INVERSE OF A MATRIX
THE MULTIPLICATIVE IDENTITY The multiplicative identity for real numbers is the number 1. The property is: In terms of matrices we need a matrix that can be multiplied by a matrix (A) and give a product which is the same matrix (A). If a is a real number, then a x 1 = 1 x a = a.
THE INVERSE OF A MATRIX (A-1) For an n  n matrix A, there may be a B such that AB = I = BA. The inverse is analogous to a reciprocal A matrix which has an inverse is nonsingular. A matrix which does not have an inverse is singular. An inverse exists only if 0A
PROPERTIES OF INVERSE MATRICES      111 -- ABAB      '11 - AA'     AA  11-
THE IDENTITY MATRIX FOR MULTIPLICATION Let A be a square matrix with n rows and n columns. Let I be a matrix with the same dimensions and with 1’s on the main diagonal and 0’s elsewhere. Then AI = IA = A
THE MULTIPLICATIVE IDENTITY                1406 7410 2973 9470 B              1000 0100 0010 0001 I Give the multiplicative identity for matrix B. This identity matrix is I4.
THE MULTIPLICATIVE INVERSE                             10 01 )3(1)1(2)2(1)1(2 )3(1)1(3)2(1)1(3 32 11 12 13 For every nonzero real number a, there is a real number 1/a such that a(1/a) = 1. In terms of matrices, the product of a square matrix and its inverse is I.
THE INVERSE OF A MATRIX Let A be a square matrix with n rows and n columns. If there is an n x n matrix B such that AB = I and BA = I, then A and B are inverses of one another. The inverse of matrix A is denoted by A-1.
THE INVERSE OF A MATRIX                23 35 53 32 BandA To show that matrices are inverses of one another, show that the multiplication of the matrices is commutative and results in the identity matrix. Show that A and B are inverses.
THE INVERSE OF A MATRIX                                10 01 )2(5)3(3)3(5)5(3 )2(3)3(2)3(3)5(2 23 35 53 32 AB and 
THE INVERSE OF A MATRIX                                10 01 )5(2)3(3)3(2)2(3 )5)(3()3(5)3)(3()2(5 53 32 23 35 BA
FINDING THE INVERSE OF A MATRIX - METHOD 1              dc ba BandALet 53 21                   10 01 53 21 dc ba Use the equation AB = I. Write and solve the equation:
INVERSES – METHOD 1, CONT.                   10 01 53 21 dc ba               10 01 5353 22 dbca dbca 1235 153 02 053 12            dandbcanda db db ca ca
INVERSES – METHOD 1, CONT.         13 25                              10 01 )1(5)2(3)3(5)5(3 )1(2)2(1)3(2)5(1 13 25 53 21 So the inverse of A = We can check this by multiplying A x A-1
Find the multiplicative inverse of:        43 21 A                     2 1 2 3 12 13 24 2 11 A 2)2(3)4(1 43 21  )( || 11 Aadj A A 
We can check to see if we are correct by multiplying. Remember that AA-1 = I                               10 01 )2/1(4)1(3)2/3(4)2(3 )2/1(2)1(1)2/3(2)2(1 2 1 2 3 12 43 21
ELEMENTARY ROW OPERATIONS 1. Interchange the order in which the equations are listed. 2. Multiply any equation by a nonzero number. 3. Replace any equation with itself added to a multiple of another equation.
RANK r is said to be rank of a matrix if it possess these two conditions:- 1). It contains a non-zero minor of order r. 2).All minor of order (r+1) vanishes. Rank Theorem Let A be the coefficient matrix of a system of linear equations with n variables. If the system is consistent, then Number of free variable = n – rank(A)
RANK OF MATRIX Equal to the dimension of the largest square sub-matrix of A that has a non-zero determinant. Example: has rank 3
RANK OF MATRIX (CONT’D) Alternative definition: the maximum number of linearly independent columns (or rows) of A. Therefore, rank is not 4 !Example:
RANK AND SINGULAR MATRICES
ECHELON FORM A rectangular matrix is in echelon form if it has the following properties: 1. All nonzero rows are above any rows of all zeroes. 2. Each leading entry of a row is in a column to the right of the leading entry of the row above it.
ECHELON FORM
ECHELON FORM
EXAMPLE
EXAMPLE
EXAMPLE
ECHELON FORM A rectangular matrix is in row reduced echelon form if it has the following properties: 1. It is in echelon form. 2. All entries in a column above and below a leading entry are zero. 3. Each leading entry is a 1, the only nonzero entry in its column.
1 0 0 0 0 −2 0 1 0 0 0 3 0 0 1 0 0 1 0 0 0 1 0 4 0 0 0 0 1 2 REDUCED ROW ECHELON FORM
REDUCED ROW ECHELON FORM
HOMOGENEOUS SYSTEM
AUGMENTED MATRIX
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 2 2 1 1 2 0 1 1 0 1 1 2 3 1 R R R          0 1 1 1 1 2 0 2 3 1 2 1             Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 1 2 0 0 1 1 1 2 2 1 3            1 3 3 2 2 4 0 2 2 1 3 3 2 0 0 3 R R R          Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 3 3 2 2 4 0 2 2 1 3 3 2 0 0 3 R R R          0 0 1 1 2 0 0 1 1 3 1 3            Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 1 2 0 0 1 1 1 0 0 3 3            2 2 0 1 1 1 1 R R   Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            2 2 0 1 1 1 1 R R   0 1 1 1 1 1 2 0 0 0 3 3            Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 1 2 0 0 1 1 1 0 0 3 3           2 1 1 0 1 1 1 1 1 0 1 1 1 2 0 R R R     Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            2 1 1 0 1 1 1 1 1 0 1 1 1 2 0 R R R     0 1 1 1 0 0 3 3 1 0 1 1          Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 0 1 1 0 1 1 1 0 0 3 3           3 3 1 3 0 0 1 1 R R  Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            3 3 1 3 0 0 1 1 R R 1 0 1 1 0 1 11 0 0 1 1          Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 0 1 1 0 1 11 0 0 1 1          3 2 2 0 0 1 1 0 1 1 1 1 0 0 2 R R R   Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            3 2 2 0 0 1 1 0 1 1 1 1 0 0 2 R R R   1 0 1 1 0 0 1 1 0 1 0 2          Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 0 1 1 0 1 0 2 0 0 1 1          3 1 1 0 1 0 0 1 0 0 1 1 0 1 1 R R R      Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            3 1 1 0 1 0 0 1 0 0 1 1 0 1 1 R R R      0 1 0 2 0 0 1 1 1 0 0 0         Gauss-Jordan Method
Solve the system of equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 0 0 0 0 1 0 2 0 0 1 1          (0, 2, 1) Gauss-Jordan Method
The homogeneous equation Ax = 0 has a nontrivial solution if and only if the equation has at least one free variable. Basic variables: The variables corresponding to pivot columns            00000 31100 02010 1x 2x 3x 4x Free variables: the others          freeis 3 2 freeis 4 43 42 1 x xx xx x 
A SYSTEM OF LINEAR EQUATIONS
TWO EQUATIONS, TWO UNKNOWNS: LINES IN A PLANE
THREE POSSIBLE TYPES OF SOLUTIONS 1. No solution
THREE POSSIBLE TYPES OF SOLUTIONS 1. A unique solution
THREE POSSIBLE TYPES OF SOLUTIONS 1. Infinitely many solutions
THREE EQUATIONS,THREE UNKNOWNS: PLANES IN SPACE
Definition of Homogeneous A system of linear equations is said to be homogeneous if it can be written in the form Ax = 0, where A is an matrix and 0 is the zero vector in Rm. nm Example:         023 034 0452 321 321 321 xxx xxx xxx Note: Every homogeneous linear system is consistent. i.e. The homogeneous system Ax = 0 has at least one solution, namely the trivial solution, x = 0.
The homogeneous equation Ax = 0 has a nontrivial solution if and only if the equation has at least one free variable. Basic variables: The variables corresponding to pivot columns            00000 31100 02010 1x 2x 3x 4x Free variables: he others          freeis 3 2 freeis 4 43 42 1 x xx xx x 
Example 3: Describe all solutions for         486 1423 7453 321 321 321 xxx xxx xxx Solutions of Nonhomogeneous Systems i.e. Describe all solutions of where  Ax  b               816 423 453 A and  b  7 1 4           Geometrically, what does the solution set represent?
Homogeneous         086 0423 0453 321 321 321 xxx xxx xxx Nonhomogeneous         486 1423 7453 321 321 321 xxx xxx xxx                   1 0 -4 3 0 0 1 0 0 0 0 0 0                   1 0 -4 3 -1 0 1 0 2 0 0 0 0           freex x xx 3 2 31 0 3 4           freex x xx 3 2 31 2 1 3 4                      1 0 3/4 3 3 2 1 x x x x                                 1 0 3/4 0 2 1 3 3 2 1 x x x x
THEOREM
CRAMER’S RULE Example: Solve the system: 3x - 2y = 10 4x + y = 6 10 2 6 1 22 2 3 2 11 4 1 x      3 10 4 6 22 2 3 2 11 4 1 y       The solution is (2, -2)
CRAMER’S RULE ●Example: Solve the system 3x - 2y + z = 9 x + 2y - 2z = -5 x + y - 4z = -2 x  9 2 1 5 2 2 2 1 4 3 2 1 1 2 2 1 1 4  23 23  1 y  3 9 1 1 5 2 1 2 4 3 2 1 1 2 2 1 1 4  69 23  3
CRAMER’S RULE Example, continued: 3x - 2y + z = 9 x + 2y - 2z = -5 x + y - 4z = -2 z  3 2 9 1 2 5 1 1 2 3 2 1 1 2 2 1 1 4  0 23  0 The solution is (1, -3, 0)
GAUSSIAN ELIMINATION To solve a system of equations using Gaussian elimination with matrices, we use the same rules as before. 1. Interchange any two rows. 2. Multiply each entry in a row by the same nonzero constant. 3. Add a nonzero multiple of one row to another row.
A method to solve simultaneous linear equations of the form [A][X]=[C] Two steps 1. Forward Elimination 2. Back Substitution
FORWARD ELIMINATION                                 735.0 21.96 8.106 7.000 56.18.40 1525 3 2 1 x x x                                2.279 2.177 8.106 112144 1864 1525 3 2 1 x x x The goal of forward elimination is to transform the coefficient matrix into an upper triangular matrix
FORWARD ELIMINATION A set of n equations and n unknowns 11313212111 ... bxaxaxaxa nn  22323222121 ... bxaxaxaxa nn  nnnnnnn bxaxaxaxa  ...332211 . . . . . . (n-1) steps of forward elimination
Step 1 For Equation 2, divide Equation 1 by and multiply by . )...( 11313212111 11 21 bxaxaxaxa a a nn       1 11 21 1 11 21 212 11 21 121 ... b a a xa a a xa a a xa nn  11a 21a FORWARD ELIMINATION
FORWARD ELIMINATION 1 11 21 1 11 21 212 11 21 121 ... b a a xa a a xa a a xa nn  22323222121 ... bxaxaxaxa nn  1 11 21 21 11 21 2212 11 21 22 ... b a a bxa a a axa a a a nnn              ' 2 ' 22 ' 22 ... bxaxa nn  Subtract the result from Equation 2. − _________________________________________________ or
Repeat this procedure for the remaining equations to reduce the set of equations as 11313212111 ... bxaxaxaxa nn  ' 2 ' 23 ' 232 ' 22 ... bxaxaxa nn  ' 3 ' 33 ' 332 ' 32 ... bxaxaxa nn  '' 3 ' 32 ' 2 ... nnnnnn bxaxaxa  . . . . . . . . . End of Step 1 FORWARD ELIMINATION
Step 2 Repeat the same procedure for the 3rd term of Equation 3. 11313212111 ... bxaxaxaxa nn  ' 2 ' 23 ' 232 ' 22 ... bxaxaxa nn  " 3 " 33 " 33 ... bxaxa nn  "" 3 " 3 ... nnnnn bxaxa  . . . . . . End of Step 2 FORWARD ELIMINATION
At the end of (n-1) Forward Elimination steps, the system of equations will look like ' 2 ' 23 ' 232 ' 22 ... bxaxaxa nn  " 3 " 33 " 33 ... bxaxa nn     11   n nn n nn bxa . . . . . . 11313212111 ... bxaxaxaxa nn  End of Step (n-1) FORWARD ELIMINATION
MATRIX FORM AT END OF FORWARD ELIMINATION                                                   )(n- n " ' n )(n nn " n " ' n '' n b b b b x x x x a aa aaa aaaa 1 3 2 1 3 2 1 1 333 22322 1131211 0000 00 0    
BACK SUBSTITUTION                                 735.0 21.96 8.106 7.000 56.18.40 1525 3 2 1 x x x Solve each equation starting from the last equation Example of a system of 3 equations
BACK SUBSTITUTION STARTING EQNS ' 2 ' 23 ' 232 ' 22 ... bxaxaxa nn  " 3 " 3 " 33 ... bxaxa nn     11   n nn n nn bxa . . . . . . 11313212111 ... bxaxaxaxa nn 
Start with the last equation because it has only one unknown )1( )1(    n nn n n n a b x BACK SUBSTITUTION
          1,...,1for ... 1 1 ,2 1 2,1 1 1, 1            ni a xaxaxab x i ii n i nii i iii i ii i i i       1,...,1for1 1 11       ni a xab x i ii n ij j i ij i i i )1( )1(    n nn n n n a b x BACK SUBSTITUTION
FORWARD ELIMINATION
NUMBER OF STEPS OF FORWARD ELIMINATION Number of steps of forward elimination is (n1)(31)2
Divide Equation 1 by 25 and multiply it by 64, . .    408.27356.28.126456.28.1061525        208.9656.18.40 408.27356.28.1264 177.21864                2.279112144 2.1771864 8.1061525               2.279112144 208.9656.18.40 8.1061525    56.2 25 64  Subtract the result from Equation 2 Substitute new equation for Equation 2 FORWARD ELIMINATION
FORWARD ELIMINATION: STEP 1 (CONT.) .    168.61576.58.2814476.58.1061525              2.279112144 208.9656.18.40 8.1061525         968.33576.48.160 168.61576.58.28144 279.2112144                  968.33576.48.160 208.9656.18.40 8.1061525    Divide Equation 1 by 25 and multiply it by 144, .76.5 25 144  Subtract the result from Equation 3 Substitute new equation for Equation 3
FORWARD ELIMINATION: STEP 2    728.33646.58.1605.3208.9656.18.40               968.33576.48.160 208.9656.18.40 8.1061525         .7607.000 728.33646.516.80 335.96876.416.80                 76.07.000 208.9656.18.40 8.1061525    Divide Equation 2 by −4.8 and multiply it by −16.8, .5.3 8.4 8.16    Subtract the result from Equation 3 Substitute new equation for Equation 3
BACK SUBSTITUTION
BACK SUBSTITUTION                                            76.0 208.96 8.106 7.000 56.18.40 1525 7.07.000 2.9656.18.40 8.1061525 3 2 1 a a a    08571.1 7.0 76.0 76.07.0 3 3 3    a a a Solving for a3
BACK SUBSTITUTION (CONT.) Solving for a2 690519. 4.8 1.085711.5696.208 8.4 56.1208.96 208.9656.18.4 2 2 3 2 32         a a a a aa                                 76.0 208.96 8.106 7.000 56.18.40 1525 3 2 1 a a a
BACK SUBSTITUTION (CONT.) Solving for a1 290472.0 25 08571.16905.1958.106 25 58.106 8.106525 32 1 321       aa a aaa                                 76.0 2.96 8.106 7.000 56.18.40 1525 3 2 1 a a a
GAUSSIAN ELIMINATION SOLUTION                                2279 2177 8106 112144 1864 1525 3 2 1 . . . a a a                      08571.1 6905.19 290472.0 3 2 1 a a a
THANK YOU

More Related Content

PDF
Introduction of matrices
byShakehand with Life
 
PPTX
Matrices and System of Linear Equations ppt
byDrazzer_Dhruv
 
PPTX
Matrices ppt
byaakashray33
 
PPT
systems of linear equations & matrices
byStudent
 
PPTX
System of linear equations
byDiler4
 
PPTX
system linear equations and matrices
byAditya Vaishampayan
 
PPTX
Cramers rule
bymstf mstf
 
PPT
matrices basic operation.ppt
byMahesh Kumar Lohano
 
Introduction of matrices
byShakehand with Life
 
Matrices and System of Linear Equations ppt
byDrazzer_Dhruv
 
Matrices ppt
byaakashray33
 
systems of linear equations & matrices
byStudent
 
System of linear equations
byDiler4
 
system linear equations and matrices
byAditya Vaishampayan
 
Cramers rule
bymstf mstf
 
matrices basic operation.ppt
byMahesh Kumar Lohano
 

What's hot

PDF
Matrices and determinants
byoscar
 
PPTX
Linear differential equation of second order
byShri Shankaracharya College, Bhilai,Junwani
 
PPT
System Of Linear Equations
bysaahil kshatriya
 
PPT
Matrix Algebra : Mathematics for Business
byKhan Tanjeel Ahmed
 
PDF
Matrix introduction and matrix operations.
byLearnbay Datascience
 
PDF
Linear algebra-Basis & Dimension
byManikanta satyala
 
PPTX
Matrices
bySanthanam Krishnan
 
PPT
Ppt on matrices and Determinants
byNirmalaSolapur
 
PPTX
Linear Algebra - Row Echelon Form and Reduced Row Echelon Form
byJaressen Kyle Salcedo
 
PPTX
Numerical solution of system of linear equations
byreach2arkaELECTRICAL
 
PPTX
Matrix inverse
bymstf mstf
 
PPTX
Presentation on inverse matrix
bySyed Ahmed Zaki
 
PDF
INVERSE OF MATRIX
byRajneesh Pachwaria
 
PDF
7.1 Systems of Linear Equations - Two Variables
bysmiller5
 
PDF
Matrix algebra
byFarzad Javidanrad
 
PPT
Introduction to Matrices
byholmsted
 
PPT
Solving systems of Linear Equations
byswartzje
 
PPTX
Gauss jordan and Guass elimination method
byMeet Nayak
 
PPTX
Introduction of matrix
byPankaj Das
 
PPTX
Matrix Algebra seminar ppt
bySwetalina Pradhan
 
Matrices and determinants
byoscar
 
Linear differential equation of second order
byShri Shankaracharya College, Bhilai,Junwani
 
System Of Linear Equations
bysaahil kshatriya
 
Matrix Algebra : Mathematics for Business
byKhan Tanjeel Ahmed
 
Matrix introduction and matrix operations.
byLearnbay Datascience
 
Linear algebra-Basis & Dimension
byManikanta satyala
 
Matrices
bySanthanam Krishnan
 
Ppt on matrices and Determinants
byNirmalaSolapur
 
Linear Algebra - Row Echelon Form and Reduced Row Echelon Form
byJaressen Kyle Salcedo
 
Numerical solution of system of linear equations
byreach2arkaELECTRICAL
 
Matrix inverse
bymstf mstf
 
Presentation on inverse matrix
bySyed Ahmed Zaki
 
INVERSE OF MATRIX
byRajneesh Pachwaria
 
7.1 Systems of Linear Equations - Two Variables
bysmiller5
 
Matrix algebra
byFarzad Javidanrad
 
Introduction to Matrices
byholmsted
 
Solving systems of Linear Equations
byswartzje
 
Gauss jordan and Guass elimination method
byMeet Nayak
 
Introduction of matrix
byPankaj Das
 
Matrix Algebra seminar ppt
bySwetalina Pradhan
 

Similar to Matrix presentation By DHEERAJ KATARIA

PPTX
Mathes for-social-Matrics-Updated ppt.pptx
bychuchu159198
 
PPTX
Matrix
byওমর ফারুক
 
DOCX
University of duhok
byRwan Kamal
 
DOCX
Matrices and its Applications to Solve Some Methods of Systems of Linear Equa...
byRwan Kamal
 
DOCX
Matrices and its Applications to Solve Some Methods of Systems of Linear Equa...
byAbdullaا Hajy
 
PPTX
Bba i-bm-u-2- matrix -
byRai University
 
PPT
Matrix and its applications by mohammad imran
byMohammad Imran
 
PDF
Engg maths k notes(4)
byRanjay Kumar
 
PPTX
matrix algebra
bykganu
 
PDF
Matrices & Determinants
byIshant Jain
 
PPTX
Introduction about matrix in Finite Mathematics.pptx
bymariceltoropilla1
 
PPT
Matrix algebra
byPrasenjitRathore
 
PPT
Matrices
bySatriaWiguna4
 
PDF
matrix
bySatyamTiwari491261
 
PDF
Matrices & Determinants
byBirinder Singh Gulati
 
PDF
math Combined Matrices Lecture for college
byAdeIrma366624
 
PPTX
Precalculus 09 Matrices.pptx
byKarloFritz
 
PPTX
Precalculus 09 Matrices.pptx
byPiyaliBhar1
 
PPT
Matrix
byST ZULAIHA NURHAJARURAHMAH
 
PPTX
Linear Algebra Presentation including basic of linear Algebra
byMUHAMMADUSMAN93058
 
Mathes for-social-Matrics-Updated ppt.pptx
bychuchu159198
 
Matrix
byওমর ফারুক
 
University of duhok
byRwan Kamal
 
Matrices and its Applications to Solve Some Methods of Systems of Linear Equa...
byRwan Kamal
 
Matrices and its Applications to Solve Some Methods of Systems of Linear Equa...
byAbdullaا Hajy
 
Bba i-bm-u-2- matrix -
byRai University
 
Matrix and its applications by mohammad imran
byMohammad Imran
 
Engg maths k notes(4)
byRanjay Kumar
 
matrix algebra
bykganu
 
Matrices & Determinants
byIshant Jain
 
Introduction about matrix in Finite Mathematics.pptx
bymariceltoropilla1
 
Matrix algebra
byPrasenjitRathore
 
Matrices
bySatriaWiguna4
 
matrix
bySatyamTiwari491261
 
Matrices & Determinants
byBirinder Singh Gulati
 
math Combined Matrices Lecture for college
byAdeIrma366624
 
Precalculus 09 Matrices.pptx
byKarloFritz
 
Precalculus 09 Matrices.pptx
byPiyaliBhar1
 
Matrix
byST ZULAIHA NURHAJARURAHMAH
 
Linear Algebra Presentation including basic of linear Algebra
byMUHAMMADUSMAN93058
 

More from Dheeraj Kataria

PPTX
Information Security- Threats and Attacks presentation by DHEERAJ KATARIA
byDheeraj Kataria
 
PPTX
Microprocessor Protected Mode Memory addressing By DHEERAJ KATARIA
byDheeraj Kataria
 
PPT
C++ functions presentation by DHEERAJ KATARIA
byDheeraj Kataria
 
PPTX
E facilities of Municipal Corporation of Delhi By DHEERAJ KATARIA
byDheeraj Kataria
 
PPTX
Types Of Recursion in C++, Data Stuctures by DHEERAJ KATARIA
byDheeraj Kataria
 
PPTX
Heritage and tourism in india by DHEERAJ KATARIA
byDheeraj Kataria
 
Information Security- Threats and Attacks presentation by DHEERAJ KATARIA
byDheeraj Kataria
 
Microprocessor Protected Mode Memory addressing By DHEERAJ KATARIA
byDheeraj Kataria
 
C++ functions presentation by DHEERAJ KATARIA
byDheeraj Kataria
 
E facilities of Municipal Corporation of Delhi By DHEERAJ KATARIA
byDheeraj Kataria
 
Types Of Recursion in C++, Data Stuctures by DHEERAJ KATARIA
byDheeraj Kataria
 
Heritage and tourism in india by DHEERAJ KATARIA
byDheeraj Kataria
 

Recently uploaded

PDF
PPT Daftar Bahan Kelas Bedah Kitab Matius Khotbah di Bukit
bySABDA
 
PDF
Tetracycline Class of Antibiotics: SAR, MOA and Uses
bymominzarinamdnajeeb
 
PDF
PHARMACOLOGY 1 ( BP 404 T) Unit 1 (Cology 1)
byChetan Mali
 
PPTX
ECP Demo ppt.pptx Visualizing and Identifying solid figures using concrete an...
byireneodechavez1
 
PPTX
How to Manage Checkout Policy & Customer Portal in Odoo 18 Website
byCeline George
 
PPTX
Lesson_1 Acceleration.pptx_Science 8-4th quarter
byAnnabelAlarcon
 
PPTX
A brief introduction to Minor vegetable crops.pptx
byUmeshTimilsina1
 
PPTX
Modeling Simple Equation using Bar Model.pptx
byshahanieabbat3
 
PDF
Types of Foot & Foot Modifications in Birds
byDr. Ramzan Virani
 
PPTX
MEMORY &FORGETTING. Shilpa Hotakar.Psychology pptx
bySHILPA HOTAKAR
 
PPTX
GRADE 8 - QUARTER 4 TYPES AND PARTS OF LETTER.pptx
byvillariasjazminercai
 
PDF
AI Is a Tool, Not a Voice: Radio, Trust, and the Human Factor
byN.A. Shah Ansari
 
PPTX
GRADE 8_WEEK 2_QUARTER 4_ENGLISH_MATATAG.pptx
byRoinelReyes1
 
PPTX
ENGLISH-7-Quarter-4-Week-3 Matatag .pptx
byKimberlyJadeCuyos
 
PDF
Chapter 05 Drugs Acting on the Central Nervous System: Anti-Depressant Drugs
bySandeshSul
 
PPTX
World cancer day 2026 4febuary united by unique
byDr Santosh Vishwakarma
 
PPTX
Problem and Solution (PowerPoint Game Template)
byacpaulite
 
PDF
Nursing care plan for Vomiting /B.Sc nsg
byDr. Sudhadevi Sadanandan
 
PPTX
Q4_PPT MUSIC & ARTS 7 week 1-2, matatag curriculum
byZEN
 
PPTX
How to Track & Manage My Time Section in Odoo 18 Time Off
byCeline George
 
PPT Daftar Bahan Kelas Bedah Kitab Matius Khotbah di Bukit
bySABDA
 
Tetracycline Class of Antibiotics: SAR, MOA and Uses
bymominzarinamdnajeeb
 
PHARMACOLOGY 1 ( BP 404 T) Unit 1 (Cology 1)
byChetan Mali
 
ECP Demo ppt.pptx Visualizing and Identifying solid figures using concrete an...
byireneodechavez1
 
How to Manage Checkout Policy & Customer Portal in Odoo 18 Website
byCeline George
 
Lesson_1 Acceleration.pptx_Science 8-4th quarter
byAnnabelAlarcon
 
A brief introduction to Minor vegetable crops.pptx
byUmeshTimilsina1
 
Modeling Simple Equation using Bar Model.pptx
byshahanieabbat3
 
Types of Foot & Foot Modifications in Birds
byDr. Ramzan Virani
 
MEMORY &FORGETTING. Shilpa Hotakar.Psychology pptx
bySHILPA HOTAKAR
 
GRADE 8 - QUARTER 4 TYPES AND PARTS OF LETTER.pptx
byvillariasjazminercai
 
AI Is a Tool, Not a Voice: Radio, Trust, and the Human Factor
byN.A. Shah Ansari
 
GRADE 8_WEEK 2_QUARTER 4_ENGLISH_MATATAG.pptx
byRoinelReyes1
 
ENGLISH-7-Quarter-4-Week-3 Matatag .pptx
byKimberlyJadeCuyos
 
Chapter 05 Drugs Acting on the Central Nervous System: Anti-Depressant Drugs
bySandeshSul
 
World cancer day 2026 4febuary united by unique
byDr Santosh Vishwakarma
 
Problem and Solution (PowerPoint Game Template)
byacpaulite
 
Nursing care plan for Vomiting /B.Sc nsg
byDr. Sudhadevi Sadanandan
 
Q4_PPT MUSIC & ARTS 7 week 1-2, matatag curriculum
byZEN
 
How to Track & Manage My Time Section in Odoo 18 Time Off
byCeline George
 

Matrix presentation By DHEERAJ KATARIA

  • 1.
  • 2.
    Matrix - arectangular array of variables or constants in horizontal rows(m) and vertical columns (n) enclosed in brackets. Element - each value in a matrix; either a number or a constant. Dimension - number of rows by number of columns of a matrix. **A matrix is named by its dimensions.
  • 3.
    11 12 1314 21 22 23 24 31 32 33 34 mn mn mn mn a a a a a a a a a a a a a a a a          Row 1 Row 2 Row 3 Row m Column 1 Column 2 Column 3 Column 4
  • 4.
    Examples: Find thedimensions of each matrix. 1. A = 2 1 0 5 4 8           2. B = 1 2 3 4             0 5 3 1 3. C = 2 0 9 6       Dimensions: 3x2 Dimensions: 4x1 Dimensions: 2x4
  • 5.
    Different types ofMatrices • Column Matrix - a matrix with only one column. • Row Matrix - a matrix with only one row. • Square Matrix - a matrix that has the same number of rows and columns.
  • 6.
    row nmrows                  mnmmm n n n aaaa a a a aaa aaa aaa A      321 3 2 1 333231 232221 131211 A matrix isa rectangular array of numbers. We subscript entries to tell their location in the array Matrices are identified by their size. Matrices and Rows "" "" columnthj rowthiaij
  • 7.
    A matrix ofm rows and n columns is called a matrix with dimensions m x n. 2 3 4 1.) 1 1 2           3 8 9 2.) 2 5 6 7 8          10 3.) 7       4.) 3 4 2 X 3 3 X 3 2 X 1 1 X 2
  • 8.
    To add matrices,we add the corresponding elements. They must have the same dimensions. 5 0 6 3 4 1 2 3 A B               A + B 5 6 0 3 4 2 1 3           1 3 6 4       
  • 9.
    To subtract matrices,we subtract the corresponding elements. The matrices must have the same dimensions. 1 2 1 1 3.) 2 0 1 3 3 1 2 3                     1 1 2 ( 1) 2 1 0 3 3 2 1 3                0 3 3 3 5 4           
  • 10.
    ADDITIVE INVERSE OFA MATRIX: 1 0 2 3 1 5 A       1 0 2 3 1 5 A         
  • 11.
    Equal Matrices -two matrices that have the same dimensions and each element of one matrix is equal to the corresponding element of the other matrix. *The definition of equal matrices can be used to find values when elements of the matrices are algebraic expressions.
  • 12.
    1. 2x 2x  3y       y 12       *Since the matrices are equal, the corresponding elements are equal! * Form two linear equations. 2x  y 2x  3y  12 * Solve the system using substitution. Examples: Find the values for x and y 2x  y y  3y 12 4y 12 y  3 2x  3 x  3 2
  • 13.
    * Write aslinear equations.2. 3x  y x  2y       x  3 y 2       7y  7 y  1 2x  y  3 2x  6y  4 2x  1  3 2x  2 x 1 Now check your answer * Combine like terms. * Solve using elimination. 3x  y  x  3 x  2y  y  2 x  2y  y  2 1  2 1   1 2 1  2  1 1  1 2x  y  3 x  3y  2
  • 14.
    Scalar Multiplication: 1 2 3 12 3 4 5 6 k           We multiply each # inside our matrix by k. 1 2 3 1 2 3 4 5 6 k k k k k k k k k           
  • 15.
  • 16.
  • 17.
  • 18.
    Associative Property ofAddition (A+B)+C = A+(B+C) Commutative Property of Addition A+B = B+A Distributive Property of Addition and Subtraction S(A+B) = SA+SB S(A-B) = SA-SB NOTE: Multiplication is not included!!!
  • 19.
    THE IDENTITY MATRIXIS A SQUARE MATRIX WITH ONE DOWN THE DIAGONALS In a 2 X 2 In a 3 X 3       10 01           100 010 001
  • 21.
  • 22.
    THE MULTIPLICATIVE IDENTITY Themultiplicative identity for real numbers is the number 1. The property is: In terms of matrices we need a matrix that can be multiplied by a matrix (A) and give a product which is the same matrix (A). If a is a real number, then a x 1 = 1 x a = a.
  • 23.
    THE INVERSE OFA MATRIX (A-1) For an n  n matrix A, there may be a B such that AB = I = BA. The inverse is analogous to a reciprocal A matrix which has an inverse is nonsingular. A matrix which does not have an inverse is singular. An inverse exists only if 0A
  • 24.
    PROPERTIES OF INVERSEMATRICES      111 -- ABAB      '11 - AA'     AA  11-
  • 25.
    THE IDENTITY MATRIXFOR MULTIPLICATION Let A be a square matrix with n rows and n columns. Let I be a matrix with the same dimensions and with 1’s on the main diagonal and 0’s elsewhere. Then AI = IA = A
  • 26.
  • 27.
    THE MULTIPLICATIVE INVERSE                             10 01 )3(1)1(2)2(1)1(2 )3(1)1(3)2(1)1(3 32 11 12 13 Forevery nonzero real number a, there is a real number 1/a such that a(1/a) = 1. In terms of matrices, the product of a square matrix and its inverse is I.
  • 28.
    THE INVERSE OFA MATRIX Let A be a square matrix with n rows and n columns. If there is an n x n matrix B such that AB = I and BA = I, then A and B are inverses of one another. The inverse of matrix A is denoted by A-1.
  • 29.
    THE INVERSE OFA MATRIX                23 35 53 32 BandA To show that matrices are inverses of one another, show that the multiplication of the matrices is commutative and results in the identity matrix. Show that A and B are inverses.
  • 30.
    THE INVERSE OFA MATRIX                                10 01 )2(5)3(3)3(5)5(3 )2(3)3(2)3(3)5(2 23 35 53 32 AB and 
  • 31.
    THE INVERSE OFA MATRIX                                10 01 )5(2)3(3)3(2)2(3 )5)(3()3(5)3)(3()2(5 53 32 23 35 BA
  • 32.
    FINDING THE INVERSEOF A MATRIX - METHOD 1              dc ba BandALet 53 21                   10 01 53 21 dc ba Use the equation AB = I. Write and solve the equation:
  • 33.
    INVERSES – METHOD1, CONT.                   10 01 53 21 dc ba               10 01 5353 22 dbca dbca 1235 153 02 053 12            dandbcanda db db ca ca
  • 34.
    INVERSES – METHOD1, CONT.         13 25                              10 01 )1(5)2(3)3(5)5(3 )1(2)2(1)3(2)5(1 13 25 53 21 So the inverse of A = We can check this by multiplying A x A-1
  • 35.
    Find the multiplicativeinverse of:        43 21 A                     2 1 2 3 12 13 24 2 11 A 2)2(3)4(1 43 21  )( || 11 Aadj A A 
  • 36.
    We can checkto see if we are correct by multiplying. Remember that AA-1 = I                               10 01 )2/1(4)1(3)2/3(4)2(3 )2/1(2)1(1)2/3(2)2(1 2 1 2 3 12 43 21
  • 37.
    ELEMENTARY ROW OPERATIONS 1.Interchange the order in which the equations are listed. 2. Multiply any equation by a nonzero number. 3. Replace any equation with itself added to a multiple of another equation.
  • 39.
    RANK r is saidto be rank of a matrix if it possess these two conditions:- 1). It contains a non-zero minor of order r. 2).All minor of order (r+1) vanishes. Rank Theorem Let A be the coefficient matrix of a system of linear equations with n variables. If the system is consistent, then Number of free variable = n – rank(A)
  • 40.
    RANK OF MATRIX Equalto the dimension of the largest square sub-matrix of A that has a non-zero determinant. Example: has rank 3
  • 41.
    RANK OF MATRIX(CONT’D) Alternative definition: the maximum number of linearly independent columns (or rows) of A. Therefore, rank is not 4 !Example:
  • 42.
  • 44.
    ECHELON FORM A rectangularmatrix is in echelon form if it has the following properties: 1. All nonzero rows are above any rows of all zeroes. 2. Each leading entry of a row is in a column to the right of the leading entry of the row above it.
  • 45.
  • 46.
  • 47.
  • 48.
  • 49.
  • 50.
    ECHELON FORM A rectangularmatrix is in row reduced echelon form if it has the following properties: 1. It is in echelon form. 2. All entries in a column above and below a leading entry are zero. 3. Each leading entry is a 1, the only nonzero entry in its column.
  • 51.
    1 0 00 0 −2 0 1 0 0 0 3 0 0 1 0 0 1 0 0 0 1 0 4 0 0 0 0 1 2 REDUCED ROW ECHELON FORM
  • 52.
  • 54.
  • 55.
  • 56.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            Gauss-Jordan Method
  • 57.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 2 2 1 1 2 0 1 1 0 1 1 2 3 1 R R R          0 1 1 1 1 2 0 2 3 1 2 1             Gauss-Jordan Method
  • 58.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 1 2 0 0 1 1 1 2 2 1 3            1 3 3 2 2 4 0 2 2 1 3 3 2 0 0 3 R R R          Gauss-Jordan Method
  • 59.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 3 3 2 2 4 0 2 2 1 3 3 2 0 0 3 R R R          0 0 1 1 2 0 0 1 1 3 1 3            Gauss-Jordan Method
  • 60.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 1 2 0 0 1 1 1 0 0 3 3            2 2 0 1 1 1 1 R R   Gauss-Jordan Method
  • 61.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            2 2 0 1 1 1 1 R R   0 1 1 1 1 1 2 0 0 0 3 3            Gauss-Jordan Method
  • 62.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 1 2 0 0 1 1 1 0 0 3 3           2 1 1 0 1 1 1 1 1 0 1 1 1 2 0 R R R     Gauss-Jordan Method
  • 63.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            2 1 1 0 1 1 1 1 1 0 1 1 1 2 0 R R R     0 1 1 1 0 0 3 3 1 0 1 1          Gauss-Jordan Method
  • 64.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 0 1 1 0 1 1 1 0 0 3 3           3 3 1 3 0 0 1 1 R R  Gauss-Jordan Method
  • 65.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            3 3 1 3 0 0 1 1 R R 1 0 1 1 0 1 11 0 0 1 1          Gauss-Jordan Method
  • 66.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 0 1 1 0 1 11 0 0 1 1          3 2 2 0 0 1 1 0 1 1 1 1 0 0 2 R R R   Gauss-Jordan Method
  • 67.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            3 2 2 0 0 1 1 0 1 1 1 1 0 0 2 R R R   1 0 1 1 0 0 1 1 0 1 0 2          Gauss-Jordan Method
  • 68.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 0 1 1 0 1 0 2 0 0 1 1          3 1 1 0 1 0 0 1 0 0 1 1 0 1 1 R R R      Gauss-Jordan Method
  • 69.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            3 1 1 0 1 0 0 1 0 0 1 1 0 1 1 R R R      0 1 0 2 0 0 1 1 1 0 0 0         Gauss-Jordan Method
  • 70.
    Solve the systemof equations using Gauss-Jordan Method 2 0 2 3 1 2 2 3 x y z x y z x y z            1 0 0 0 0 1 0 2 0 0 1 1          (0, 2, 1) Gauss-Jordan Method
  • 71.
    The homogeneous equationAx = 0 has a nontrivial solution if and only if the equation has at least one free variable. Basic variables: The variables corresponding to pivot columns            00000 31100 02010 1x 2x 3x 4x Free variables: the others          freeis 3 2 freeis 4 43 42 1 x xx xx x 
  • 73.
    A SYSTEM OFLINEAR EQUATIONS
  • 74.
    TWO EQUATIONS, TWOUNKNOWNS: LINES IN A PLANE
  • 75.
    THREE POSSIBLE TYPESOF SOLUTIONS 1. No solution
  • 76.
    THREE POSSIBLE TYPESOF SOLUTIONS 1. A unique solution
  • 77.
    THREE POSSIBLE TYPESOF SOLUTIONS 1. Infinitely many solutions
  • 78.
  • 80.
    Definition of Homogeneous Asystem of linear equations is said to be homogeneous if it can be written in the form Ax = 0, where A is an matrix and 0 is the zero vector in Rm. nm Example:         023 034 0452 321 321 321 xxx xxx xxx Note: Every homogeneous linear system is consistent. i.e. The homogeneous system Ax = 0 has at least one solution, namely the trivial solution, x = 0.
  • 81.
    The homogeneous equationAx = 0 has a nontrivial solution if and only if the equation has at least one free variable. Basic variables: The variables corresponding to pivot columns            00000 31100 02010 1x 2x 3x 4x Free variables: he others          freeis 3 2 freeis 4 43 42 1 x xx xx x 
  • 82.
    Example 3: Describeall solutions for         486 1423 7453 321 321 321 xxx xxx xxx Solutions of Nonhomogeneous Systems i.e. Describe all solutions of where  Ax  b               816 423 453 A and  b  7 1 4           Geometrically, what does the solution set represent?
  • 83.
    Homogeneous         086 0423 0453 321 321 321 xxx xxx xxx Nonhomogeneous         486 1423 7453 321 321 321 xxx xxx xxx                   1 0 -4 3 0 0 10 0 0 0 0 0                   1 0 -4 3 -1 0 1 0 2 0 0 0 0           freex x xx 3 2 31 0 3 4           freex x xx 3 2 31 2 1 3 4                      1 0 3/4 3 3 2 1 x x x x                                 1 0 3/4 0 2 1 3 3 2 1 x x x x
  • 84.
  • 85.
    CRAMER’S RULE Example: Solve thesystem: 3x - 2y = 10 4x + y = 6 10 2 6 1 22 2 3 2 11 4 1 x      3 10 4 6 22 2 3 2 11 4 1 y       The solution is (2, -2)
  • 86.
    CRAMER’S RULE ●Example: Solve thesystem 3x - 2y + z = 9 x + 2y - 2z = -5 x + y - 4z = -2 x  9 2 1 5 2 2 2 1 4 3 2 1 1 2 2 1 1 4  23 23  1 y  3 9 1 1 5 2 1 2 4 3 2 1 1 2 2 1 1 4  69 23  3
  • 87.
    CRAMER’S RULE Example, continued:3x - 2y + z = 9 x + 2y - 2z = -5 x + y - 4z = -2 z  3 2 9 1 2 5 1 1 2 3 2 1 1 2 2 1 1 4  0 23  0 The solution is (1, -3, 0)
  • 89.
    GAUSSIAN ELIMINATION To solvea system of equations using Gaussian elimination with matrices, we use the same rules as before. 1. Interchange any two rows. 2. Multiply each entry in a row by the same nonzero constant. 3. Add a nonzero multiple of one row to another row.
  • 90.
    A method tosolve simultaneous linear equations of the form [A][X]=[C] Two steps 1. Forward Elimination 2. Back Substitution
  • 91.
  • 92.
    FORWARD ELIMINATION A setof n equations and n unknowns 11313212111 ... bxaxaxaxa nn  22323222121 ... bxaxaxaxa nn  nnnnnnn bxaxaxaxa  ...332211 . . . . . . (n-1) steps of forward elimination
  • 93.
    Step 1 For Equation2, divide Equation 1 by and multiply by . )...( 11313212111 11 21 bxaxaxaxa a a nn       1 11 21 1 11 21 212 11 21 121 ... b a a xa a a xa a a xa nn  11a 21a FORWARD ELIMINATION
  • 94.
    FORWARD ELIMINATION 1 11 21 1 11 21 212 11 21 121 ...b a a xa a a xa a a xa nn  22323222121 ... bxaxaxaxa nn  1 11 21 21 11 21 2212 11 21 22 ... b a a bxa a a axa a a a nnn              ' 2 ' 22 ' 22 ... bxaxa nn  Subtract the result from Equation 2. − _________________________________________________ or
  • 95.
    Repeat this procedurefor the remaining equations to reduce the set of equations as 11313212111 ... bxaxaxaxa nn  ' 2 ' 23 ' 232 ' 22 ... bxaxaxa nn  ' 3 ' 33 ' 332 ' 32 ... bxaxaxa nn  '' 3 ' 32 ' 2 ... nnnnnn bxaxaxa  . . . . . . . . . End of Step 1 FORWARD ELIMINATION
  • 96.
    Step 2 Repeat thesame procedure for the 3rd term of Equation 3. 11313212111 ... bxaxaxaxa nn  ' 2 ' 23 ' 232 ' 22 ... bxaxaxa nn  " 3 " 33 " 33 ... bxaxa nn  "" 3 " 3 ... nnnnn bxaxa  . . . . . . End of Step 2 FORWARD ELIMINATION
  • 97.
    At the endof (n-1) Forward Elimination steps, the system of equations will look like ' 2 ' 23 ' 232 ' 22 ... bxaxaxa nn  " 3 " 33 " 33 ... bxaxa nn     11   n nn n nn bxa . . . . . . 11313212111 ... bxaxaxaxa nn  End of Step (n-1) FORWARD ELIMINATION
  • 98.
    MATRIX FORM ATEND OF FORWARD ELIMINATION                                                   )(n- n " ' n )(n nn " n " ' n '' n b b b b x x x x a aa aaa aaaa 1 3 2 1 3 2 1 1 333 22322 1131211 0000 00 0    
  • 99.
  • 100.
    BACK SUBSTITUTION STARTINGEQNS ' 2 ' 23 ' 232 ' 22 ... bxaxaxa nn  " 3 " 3 " 33 ... bxaxa nn     11   n nn n nn bxa . . . . . . 11313212111 ... bxaxaxaxa nn 
  • 101.
    Start with thelast equation because it has only one unknown )1( )1(    n nn n n n a b x BACK SUBSTITUTION
  • 102.
             1,...,1for ... 1 1 ,2 1 2,1 1 1, 1            ni a xaxaxab x i ii n i nii i iii i ii i i i       1,...,1for1 1 11       ni a xab x i ii n ij j i ij i i i )1( )1(    n nn n n n a b x BACK SUBSTITUTION
  • 103.
  • 104.
    NUMBER OF STEPSOF FORWARD ELIMINATION Number of steps of forward elimination is (n1)(31)2
  • 105.
    Divide Equation 1by 25 and multiply it by 64, . .    408.27356.28.126456.28.1061525        208.9656.18.40 408.27356.28.1264 177.21864                2.279112144 2.1771864 8.1061525               2.279112144 208.9656.18.40 8.1061525    56.2 25 64  Subtract the result from Equation 2 Substitute new equation for Equation 2 FORWARD ELIMINATION
  • 106.
    FORWARD ELIMINATION: STEP1 (CONT.) .    168.61576.58.2814476.58.1061525              2.279112144 208.9656.18.40 8.1061525         968.33576.48.160 168.61576.58.28144 279.2112144                  968.33576.48.160 208.9656.18.40 8.1061525    Divide Equation 1 by 25 and multiply it by 144, .76.5 25 144  Subtract the result from Equation 3 Substitute new equation for Equation 3
  • 107.
    FORWARD ELIMINATION: STEP2    728.33646.58.1605.3208.9656.18.40               968.33576.48.160 208.9656.18.40 8.1061525         .7607.000 728.33646.516.80 335.96876.416.80                 76.07.000 208.9656.18.40 8.1061525    Divide Equation 2 by −4.8 and multiply it by −16.8, .5.3 8.4 8.16    Subtract the result from Equation 3 Substitute new equation for Equation 3
  • 108.
  • 109.
  • 110.
    BACK SUBSTITUTION (CONT.) Solvingfor a2 690519. 4.8 1.085711.5696.208 8.4 56.1208.96 208.9656.18.4 2 2 3 2 32         a a a a aa                                 76.0 208.96 8.106 7.000 56.18.40 1525 3 2 1 a a a
  • 111.
    BACK SUBSTITUTION (CONT.) Solvingfor a1 290472.0 25 08571.16905.1958.106 25 58.106 8.106525 32 1 321       aa a aaa                                 76.0 2.96 8.106 7.000 56.18.40 1525 3 2 1 a a a
  • 112.
  • 113.