1. MATH 250 Linear Equations and Matrices

2. Topics Preliminaries Systems of Linear Equations Matrices
Algebraic Properties of Matrix Operations
Special Types of Matrices and Partitioned Matrices
Matrix Transformations

3. Systems of Linear Equations
System of m equations in n unknowns

4. Systems of Linear Equations
Comments
If a system has a solution, call it consistent
If a system doesn’t have a solution, call it inconsistent
If , the system is called homogeneous. A homogeneous system always has the trivial solution
If two systems have the same solution, then they are called equivalent. The solution strategy for linear systems is to transform the system through a series of equivalent systems until the solution is obvious

7. Matrices
Systems of Equations
Consider
Define
Express system as AX = B

8. Matrices Systems of Equations
Since the solution of the system involves the a and b values only, will often work with the augmented matrix

9. Systems of Linear Equations
Elementary Operations on Systems
Switch two equations
Multiply an equation by nonzero constant
Add multiple of one equation to another
The application of any combination of elementary
operations to a linear system yields a new linear system
that is equivalent to the first

10. Inverting a Matrix Usually not a good idea to compute x=A-1b
Inefficient
Prone to roundoff error
In fact, compute inverse using linear solver
Solve Axi=bi where bi are columns of identity, xi are columns of inverse
Many solvers can solve several R.H.S. at once

11. Solving Linear Systems Using Gaussian Elimination
Write the augmented matrix for the system
Use matrix row operations to simplify the matrix to one with 1s down the diagonal from upper left to lower right, and 0s below the 1s
Write the system of linear equations corresponding to the matrix in step 2, and use back-substitution to find the system’s solutions

14. Theorem 1: Equivalent systems and equivalent matrices:
If the augmented coefficient matrices of two linear systems are row equivalent, then the two systems have the same solution set.
Definition: Echelon Matrix
The matrix E is called an echelon matrix provided it has the following two properties:
1. Every row of E that consists entirely of zeros (if any) lies beneath every row that contains a nonzero element.
2. In each row of E that contains a nonzero element, the nonzero element lies strictly to the right of the first nonzero element in the preceding row (if there is a preceding row).

15. Example Use matrices to solve the system Solution25 = 2z + 3y x 19 y 31 3x
Solution
Step 1 Write the augmented matrix for the system.
Linear System
Augmented Matrix 25 = 2z + 3y x 19 y 31 3x 25 2 3 1 19 31

16. Example cont.
Solution
Step 2 Use matrix row operations to simplify the matrix to one with 1s down the diagonal from upper left to lower right, and 0s below the 1s. Our goal is to obtain a matrix of the form f 1 e d c b a .
Our first step in achieving this goal is to get 1 in the top position of the first column.
We want 1 in this position. 25 2 3 1 19 31
To get 1 in this position, we interchange rows 1 and 2. (We could also interchange rows 1 and 3 to attain our goal.) 25 2 3 1 31 19
This was row 2; now it’s row 1.
This was row 1; now it’s row 2.

17. We want 0 in these positions. We want 0 in this position.
Example cont.
Solution
Now we want to get 0s below the 1 in the first column. 25 2 3 1 31 19
We want 0 in these positions.
Let’s first get a 0 where there is now a 3. If we multiply the top row of numbers by –3 and add these products to the second row of numbers, we will get 0 in this position. The top row of numbers multiplied by –3 gives -3(1) or –3, -3(1) or –3, -3(2) or –6, -3(19) or –57.
Now add these products to the corresponding numbers in row 2. Notice that although we use row 1 to find the products, row 1 does not change. 25 2 3 1
31 + (-3)
2 + (-3)
1 + (-3)
3 + (-3) 19 25 2 3 1
-26 -4 -2 19 =
We want 0 in this position.

18. We want 1 in this position.
Example cont.
Solution
We are not yet done with the first column. If we multiply the top row of numbers by –1 and add these products to the third row of numbers, we will get 0 in this position. The top row of numbers multiplied by –1 gives -1(1) or –1, -1(1) or –1, -1(2) or –2, -1(19) or –19.
Now add these products to the corresponding numbers in row 3.
25+(-19)
2 + (-2)
3 + (-1)
1 + (-1)
-26 -4 -2 19 2 1 6 2
-26 -4 -2 19 1 =
We move on to the second column. We want 1 in the second row, second column. 6 2
-26 -4 -2 19 1
We want 1 in this position.

19. We want 0 in this position.
Example cont.
Solution
To get 1 in the desired position, we multiply –2 by its reciprocal, -1/2. Therefore, we multiply all the numbers in the second row by –1/2 to get 6 2
› (-26)
› (-4)
› (-2)
› (0) 19 1 6 2 13 1 19 =
We want 0 in this position.
We are not yet done with the second column. If we multiply the top row of numbers by –2 and add these products to the third row of numbers, we will get 0 in this position. The second row of numbers multiplied by –2 gives -2(0) or 0, -2(1) or –2, -2(2) or –4, -2(13) or –26.
Now add these products to the corresponding numbers in row 3.
6+(-26)
0 + (-4)
2 + (-2)
0 + 0 13 2 1 19
-20 -4 13 2 1 19 =

20. We want 1 in this position.
Example cont.
Solution
We move on to the third column. We want 1 in the third row, third column.
-20 -4 13 2 1 19
We want 1 in this position.
To get 1 in the desired position, we multiply –4 by its reciprocal, -1/4. Therefore, we multiply all the numbers in the third row by –1/4 to get
-1/4(-20)
-1/4(-4)
-1/4(0) 13 2 1 19 5 1 13 2 19 =
We now have the desired matrix with 1s down the diagonal and 0s below the 1s.
Step 3 Write the system of linear equations corresponding to the matrix in step 2, and use back-substitution to find the system’s solution. The system represented by the matrix in step 2 is

21. Example cont.
Solution 25 2 3 1 19 31 5 = z 13 2z + y 19 x
We immediately see that the value for z is 5. To find y, we back-substitute 5 for z in the second equation.
y + 2z = 13 Equation 2
y + 2(5) = 13 Substitute 5 for x.
y = 3 Solve for y.
Finally, back-substitute 3 for y and 5 for z in the first equation:
x + y + 2z = 19 Equation 1
x (5) = 19 Substitute 3 for y and 5 for x.
x + 13 = 19 Multiply and add.
x = 6 Subtract 13 from both sides.
The solution set for the original system is {(6, 3, 5)}.

24. Gaussian Elimination Definition A matrix is in echelon form if
Any rows consisting entirely of zeros are grouped at the bottom of the matrix.
The first nonzero element of each row is 1. This element is called a leading 1.
The leading 1 of each row after the first is positioned to the right of the leading 1 of the previous row.
(This implies that all the elements below a leading 1 are zero.)

45. Example 1
Solving the following system of linear equations using the method of Gaussian elimination.
Solution
Starting with the augmented matrix, create zeros below the pivot in the first column.
At this stage, we create a zero only below the pivot.
Echelon form
We have arrived at the echelon form.

46. The corresponding system of equation is
We get
Substituting x4 = 2 and x3 = -5 into the first equation,
Let x2 = r. The system has many solutions. The solutions are

47. Example 2
Solving the following system of linear equations using the method of Gaussian elimination, performing back substitution using matrices.
Solution
We arrive at the echelon form as in the previous example.
Echelon form
This marks the end of the forward elimination of variables from equations. We now commence the back substitution using matrices.

48. 2 3
This matrix is the reduced echelon form of the original augmented matrix. The corresponding system of equations is
Let x2 = r. We get same solution as previously,