1. Lesson 8.1, page 782 Matrix Solutions to Linear Systems
Objective: To solve systems using matrix equations.

2. Review Solve this system of equations.
4x + 2y = -3
x + 5y = -4

3. Systems of Equations in Two Variables

4. Matrices
A rectangular array of numbers is called a matrix (plural, matrices).
The rows of a matrix are horizontal.
The columns of a matrix are vertical.
The matrix shown has 2 rows and 3 columns.
A matrix with m rows and n columns is said to be of order m  n.
When m = n the matrix is said to be square.
See Example 1, page 783.

5. Matrices Consider the system: 4x + 2y = -3 x + 5y = -4
Example:
The matrix shown above is an augmented matrix because it contains not only the coefficients but also the constant terms.
The matrix is called the coefficient matrix.

6. Row-Equivalent Operations, pg. 784
1) Interchange any two rows. (Ri Rj)
2) Multiply each entry in a row by the same nonzero constant. (kRi)
3) Add a nonzero multiple of one row to another row. (kRi + Rj)

7. See Example 2.
Check Point 2, page 785: Perform each indicated row operation on
(R1 R2)
¼R1
3R2 + R3 = new R3

8. Solving Systems using Gaussian Elimination with Matrices
Write the augmented matrix.
Use row operations to get the matrix in “row echelon” form:
Write the system of equations corresponding to the resulting matrix.
Use back-substitution to find the system’s solution.

9. Example – Watch and listen.
Solve the following system:

10. First, we write the augmented matrix, writing 0 for the missing y-term in the last equation.
Our goal is to find a row-equivalent matrix of the form

11. Example continued
R R2
We multiply the first row by 2 and add it to the second row.
We also multiply the first row by 4 and add it to the third row.

12. We multiply the second row by 1/5 to get a 1 in the second row, second column.

13. We multiply the second row by 12 and add it to the third row.
R3 = -12R2 + R3

14. Example continued
Now, we can write the system of equations that corresponds to the last matrix :

15. Example continued
We back-substitute 3 for z in equation (2) and solve for y.

16. Example continued
Next, we back-substitute 1 for y and 3 for z in equation (1) and solve for x.
The triple (2, 1, 3) checks in the original system of equations, so it is the solution.

17. See Example 3.
Check Point 3: Use matrices to solve the system 2x + y + 2z = 18
x – y + 2z = 9
x + 2y – z = 6.
Write the augmented matrix for the system.
Solve using Gauss Elimination or your calculator.
Check/verify your solution.

18. See Example 3.
2x + y + 2z = 18
x – y + 2z = 9
x + 2y – z = 6.

19. Row-Echelon Form, page 790
1. If a row does not consist entirely of 0’s, then the first nonzero element in the row is a 1 (called a leading 1).
2. For any two successive nonzero rows, the leading 1 in the lower row is farther to the right than the leading 1 in the higher row.
3. All the rows consisting entirely of 0’s are at the bottom of the matrix.
If a fourth property is also satisfied, a matrix is said to be in reduced row-echelon form:
4. Each column that contains a leading 1 has 0’s everywhere else.

20. Example -- Which of the following matrices are in row-echelon form?
a) b)
c) d)
Matrices (a) and (d) satisfy the row-echelon criteria. In (b) the first nonzero element is not 1. In (c), the row consisting entirely of 0’s is not at the bottom of the matrix.

21. Gauss-Jordan Elimination
We perform row-equivalent operations on a matrix to obtain a row-equivalent matrix in row-echelon form. We continue to apply these operations until we have a matrix in reduced row-echelon form.

22. Gauss-Jordan Elimination Example
Example: Use Gauss-Jordan elimination to solve the system of equations from the previous example; we had obtained the matrix

23. Next, we multiply the second row by 3 and add it to the first row.
Gauss-Jordan Elimination continued -- We continue to perform row-equivalent operations until we have a matrix in reduced row-echelon form.
Next, we multiply the second row by 3 and add it to the first row.

24. Gauss-Jordan Elimination continued
Writing the system of equations that corresponds to this matrix, we have
We can actually read the solution, (2, 1, 3), directly from the last column of the reduced row-echelon matrix.

25. Lesson 8.2 Special Systems
When a row consists entirely of 0’s, the equations are dependent and the system is equivalent, meaning you have a system that is colinear.
Answer: INFINITELY MANY SOLUTIONS

26. Lesson 8.2 Special Systems
When we obtain a row whose only nonzero entry occurs in the last column, we have an inconsistent system of equations. For example, in the matrix
the last row corresponds to the false equation
0 = 9, so we know the original system has no solution.

27. Check Point 2, page 797 Solve the system: x – 2y – z = 5