1. Solution of System of Linear Equations Using Matrix
Ms. Sharmila A K RC Team No :5
Lecture -1
              SOLUTION OF SYSTEM OF LINEAR ALGEBRAIC EQUATIONS USING MATRIX by SHARMILA A K ,TEAM 5,RC-1229 is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License. Based on a work at
You are free to use, distribute and modify it, including for commercial purposes, provided you acknowledge the source and Share-alike 
To view a copy of this license, visit  send a letter to Creative Commons, PO Box 1866, Mountain View, CA 94042, USA

2. Prerequisites Types of Matrices, Algebra of Matrices Rank of a Matrix
Elementary Transformations - Gaussian elimination

3. Learning Objectives Students will be able
To understand how to write a System of Linear Algebraic Equations in the matrix equation form.
To Identify Homogeneous and Non Homogeneous equations
To identify whether the given system of equation is consistent or not
To understand when the system of equation has unique solution or multiple
solutions
Cont’d

4. Learning Objectives Students will be able
To use Gaussian elimination method in Matrix to find the consistency
6. To solve a large system of algebraic linear equations using Matrix equation
(Gaussian elimination method)
7. To model and solve real-world problems.
To develop the analytical ability to apply these concepts to the real world
problems

5. Rank of a Matrix Definition 1:
By performing a sequence of elementary transformations every non-
zero matrix can be reduced to one of the following form.
Where Ir is unit matrix of order r , then r is called the rank of the matrix

6. Examples

7. Rank of the matrix Definition 2:
A matrix that has undergone Gaussian elimination is said to be in row Echelon form if
1) All zero rows are at the bottom of the matrix.
2) The leading entry in any non zero row is 1.
3) All entries in the column below leading entry 1 are zero.
In Echelon form of the matrix number of non zero rows is the rank of the matrix.

8. Examples

9. System of Linear Equations
This can be written as A X = B , Where

10. X is called column matrix of unknown
A is called coefficient matrix
X is called column matrix of unknown
B is called column matrix of constants
If we adjoin the matrix B to A the resulting matrix [A B] is called Augmented Matrix

11. TYPES OF LINEAR EQUATIONS
Homogeneous equations
Non Homogeneous Equations
In the Matrix equation A X = B ,
If B = O the system is called Homogeneous
If B ≠ O The system is said to be Non Homogeneous

12. Example homogeneous and Non homogenous equation
1) 2x + y + 2z = ) 3x + 5y +8z =0
x + y + 3z = x + 6y + z = 0
4x + 3y + 5z = x + z = 0
3) 2x + 6y = – ) x + 7y – 3z=5
6x + 20y – 6z = – x + z = 6
3y – 18z = – x + 6y – 2z= 1
1 and 2 are homogeneous equations and 3 and 4 are nonhomogeneous equations

13. Consistency
If the system of equation has solution it is said to be consistent And If the system of equation does not have solution it is said to be inconsistent Theorem ( Rouche’s ) : If the matrix A and the Augmented matrix [AB] have the same rank , the system of equation is consistent

14. Trivial / Unique Solution
A X = B
B = O
r = n
Trivial / Unique Solution
r < n
Infinite Solutions
B ≠ 0
r > n
No Solution
Unique Solution
r is rank of [AB] , n is number of unknowns

15. In case of Homogeneous system of equations the system is always consistent because
is always a solution , and is known as trivial solution
That is when r = n
If r < n the system will have infinite solutions

16. In both Homogeneous and Non homogeneous system of linear algebraic equations
If m < n , where m is number of equations and n is number of unknowns,
The system will not have unique nontrivial solution

17. THANK YOU

18. Solution of System of Linear Equations Using Matrix
Ms. Sharmila A K RC Team No :5
Lecture -2
              SOLUTION OF SYSTEM OF LINEAR ALGEBRAIC EQUATIONS USING MATRIX by SHARMILA A K ,TEAM 5,RC-1229 is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License. Based on a work at
You are free to use, distribute and modify it, including for commercial purposes, provided you acknowledge the source and Share-alike 
To view a copy of this license, visit  send a letter to Creative Commons, PO Box 1866, Mountain View, CA 94042, USA

19. Example - homogeneous equation ( trivial solution )
Solve 2x + y + 2z = 0 x + y + 3z = 0 4x + 3y + 5z = 0 Solution : Above equation can be written as AX = O ( Here B = O ) Where A = X = O =
Cont’d

20. AB =
Here r (A) = r( A O) (As the last column is zero)
Using Elementary transformation,
AB ~ R1↔R ~ R 2 → R 2 – 2R 1
R 3 → R 3 – 4R 1
~ R 2 → (–1) R 2
R 3 → R 3 – R 2
Cont’d

21. Here r (A) = r( A O) = 3, the number of unknowns n
So the system has only trivial solution
Solution is { x = y = z = 0 }
Explanation : If we take the matrix and rewrite as an equation again , we get
From the last row z = z = 0
From the second row y + 4 z = y = 0
From the first row x + y +3 z = x = 0

22. Example - homogeneous equation ( Infinite solution )
Solve 2x + y + 2z = 0 x + y + 3z = 0 4x + 3y + 8z = 0 Solution : Above equation can be written as AX = O ( Here B = O ) Where A = X = O =
Cont’d

23. Augmented Matrix [AB] = [AO] =
Here r (A) = r( A O) (As the last column is zero)
Using Elementary transformation,
AB ~ R1↔R ~ R 2 → R 2 – 2R 1
R 3 → R 3 – 4R 1
~ R 2 → (–1) x R R 3 → R 3 – R 2
Cont’d

24. Here r (A) = r (A O) = 2 , ( the rank r ≠ n, r < n ,the number of unknowns )
So the system has infinite solutions
Explanation : If we take the matrix and rewrite as an equation again , we get
From the second row y + 4 z = 0 , if we put z = k, an arbitrary constant
We get y = - 4 k,
From the first row x + y +3 z = 0 , we get x = k ( by substituting the values of y and z in terms of k )
Solution is { x = k, y = - 4k, z = k , k can be any constant }

25. Trivial / Unique Solution
A X = B
B = O
r = n
Trivial / Unique Solution
r < n
Infinite Solution
B ≠ 0
r > n
No Solution
Unique Solution
r is rank of [AB] , n is number of solutions

26. In case of Non Homogeneous system of equations the system may be consistent or inconsistent
If r > n the system of equation is inconsistent
If r = n , the system of equation is consistent and solution is unique
If r < n , the system of equation is consistent and has infinite solution

27. Example – Non homogeneous equation ( inconsistent )
Solve 2x + 6y = – 11 6x + 20y – 6z = – 3 6y – 18z = – 1 Solution : Above equation can be written as AX = B Where A = X = B =
Cont’d

28. Augmented Matrix [AB] =
Let us find r( A B)
Using Elementary transformation,
AB ~ R2→R 2– 3R 1
~ R 3 → R 3 – 3R 2
Cont’d

29. [AB] ~ and Matrix A =
r [AB] = 3 and r (A) = the system of equation is inconsistent
Explanation : If we take the matrix and rewrite as an equation again , we get
From the third row x + 0 y + 0 z = – 91 , which is absurd
So system of equation is inconsistent
Cont’d

30. THANK YOU

31. Solution of System of Linear Equations Using Matrix
Ms. Sharmila A K RC Team No :5
Lecture -3
              SOLUTION OF SYSTEM OF LINEAR ALGEBRAIC EQUATIONS USING MATRIX by SHARMILA A K ,TEAM 5,RC-1229 is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License. Based on a work at
You are free to use, distribute and modify it, including for commercial purposes, provided you acknowledge the source and Share-alike 
To view a copy of this license, visit  send a letter to Creative Commons, PO Box 1866, Mountain View, CA 94042, USA

32. Example – Non homogeneous equation (unique solution)
Solve x + y + z = 8 x – y + 2z = 6 3x + 5y –7z = 14 Solution : Above equation can be written as AX = B Where A = X = B =
Cont’d

33. Augmented Matrix [AB] =
Let us find r( A B) , we denote it as r
Using Elementary transformation,
AB ~ R2 → R2– R1
R3 → R3 – 3R1
~ R 3 → R 3 + R 2
Cont’d

34. [AB] ~ and Matrix A =
r [AB] = 3 and r (A) = the system of equation is consistent
Also r = n the system has unique solution
Explanation : If we take the matrix and rewrite as an equation again , we get
From the third row we get – 9 z = – z =
Cont’d

35. From the second row we have – 2y + z = – 2 , putting the value of z we get
– 2 y = – y =
From the first row we have x + y + z = 8 , putting the values of y and z we get
x = 5
Solution is { x = 5, y = , z = }
Cont’d

36. Example–Non homogeneous equation (infinite solutions)
Solve x – 3y – 8z = – 10 3x + y – 4z = 0 2x + 5y + 6z = 13 Solution : Above equation can be written as AX = B Where A = X = B =
Cont’d

37. Augmented Matrix [AB] =
Let us find r( A B)
Using Elementary transformation,
AB ~ R2 → R2 – 3R1
R3 → R3 – 2R1
R2 → R2 / 10
~ R3 → R3 /11
Cont’d

38. ~ R3 → R3 – R2
r [AB] = 2 and r (A) = the system of equation is consistent
r < n , so the system has infinite solution
Explanation : If we take the matrix and rewrite as an
equation again , we get
Cont’d

39. From the second row y + 2 z = 3 , if we put z = t, an arbitrary constant
We get y = 3 – 2t ,
From the first row x – 3y – 8 z = – 10 , we get x = –1 + 2t ( by substituting the values of y and z in terms of k )
Solution is { x = – 1 +2t, y = 3 – 2t, z = t , t can be any constant }

40. Example – Non homogeneous equation (infinite solutions and more than three variables)
Solve x1 – x2 + x3– x4 + x5 = 1 2x1 – x2 + 3x3 + 4x5 = 2 3x1 – 2x2 + 2x3 + x4 + x5 = 1 x1 + x3 + 2x4 + x5 = 0 Solution : Above equation can be written as AX = B
Cont’d

41. where A = X = B =
Augmented Matrix [AB]=
Cont’d

42. R2 → R2 – 2R1
[AB] ~ R3 →R3 – 3R1
R4→ R4 – R1
~ R3 ↔ R 4
Cont’d

43. ~ R3 → R3 – R2
R4 → R4 – R3
~ R4 → R4 – R3
r [AB] = 3 = r (A) the system of equation is consistent and it has infinite solution Cont’d

44. the rank r ≠ n, r < n ,the number of unknowns
So the system has infinite solutions
Explanation : If we take the matrix
and rewrite as an equation again , we get
From the third row – x3 + x4 – 2x5 = – 1
Put x4 = a and x5 = b , a and b are any arbitrary constants
We obtain x3 = 1 + a – 2b Cont’d

45. From the second row x2 + x3 + 2 x4 + 2x5 = 0
Putting x4 = a , x5 = b and x3 = 1 + a – 2b , we obtain
x2 = – 1 – 3a
Similarly from the first row x1 – x2 + x3– x4 + x5 = 1 , we obtain
x1 = – 1 – 3a + b
Solution is {x1 = – 1 – 3a + b , x2 = – 1 – 3a , x3 = 1 + a – 2b , x4 = a , x5 = b where a and b can be any arbitrary constants }

46. Real Life Word Problem
Chand Novelty wishes to produce three types of toys: types A, B and C. To manufacture a type A toy requires 2 minutes on machine I, 1 minute on machine II, and 2 minutes on machine III. A type B toy requires 1 minute on machine I, 3 minutes on machine II, and 1 minute on machine III. A type C toy requires 1 minute on machine I, 2 minutes each on machine II and machine III. There are 3 hours available on machine I, 5 hours available on machine II, and 4 hours available on machine III for processing the order. How many toys of each type should Chand Novelty make in order to use all of the available time?
Type A
Type B
Type C
Time Available
Machine 1 2 1
180
Machine 2 3
300
Machine 3
240

47. solution
Type A
Type B
Type C
Time Available
Machine 1 2 1
180
Machine 2 3
300
Machine 3
240
Let x, y and z denote the respective number of type A, B and C toys to be made. Total time that machine I is used is given by 2x + y + z = 180 ( minutes) Total time that machine II is used x + 3y + 2z = 300 And for machine III 2x + y + 2z = 240

48. Problems for Practice and Self assessment
1) x + 2y + 3z = ) x – 2y + 3z = 0
2x + 3y + z = x + 5y + 6z = 0
4x + 5y + 4z = 0
x + y – 4 z =0
3) 2x – 3y + 7z = ) 4x – 2y + 6z = 8
3x + y – 3z = x + y – 3z = –1
2x + 19y – 47z = x – 3y + 9z = 21
Answers
1. Trivial Solution { x = 3k, y = 0 and z = k, k is an arbitrary constant}
3. Inconsistent 4. { x = 1, y= 3k – 2 , z = k, k is an arbitrary constant}

49. Problems for Practice and Self assessment
5) x + y + z = ) x1 + x2 – x3 + x4 = 0
x + 2y + 3z = x1 – x2 + 2x3– x4 = 0
x + 4y + 9z = x1 + x2 + x4 = 0
Answers
5. {x =2, y=1, z=0 }
6. { x1 = – a/2, x2 = (3a/2) – b, x3 = a and x4 =b where a and b are arbitrary constants}

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