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Section 5.3 MatricesAnd Systems of Equations. Systems of Equations in Two Variables.

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Presentation on theme: "Section 5.3 MatricesAnd Systems of Equations. Systems of Equations in Two Variables."— Presentation transcript:

1 Section 5.3 MatricesAnd Systems of Equations

2 Systems of Equations in Two Variables

3 Matrices A rectangular array of numbers is called a matrix (plural, matrices). Example: 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.

4 Matrices continued 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.

5 Gaussian Elimination with Matrices Row-Equivalent Operations 1.Interchange any two rows. 2.Multiply each entry in a row by the same nonzero constant. nonzero constant. 3.Add a nonzero multiple of one row to another row. another row.

6 Example Solve the following system:.

7 Example continued 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

8 Example continued New row 1 = row 2 New row 2 = row 1

9 Example continued 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.

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

11 Example continued We multiply the second row by  12 and add it to the third row.

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

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

14 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.

15 Row-Echelon Form 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: 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.

16 Example Which of the following matrices are in row-echelon form? a)b) c)d)

17 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..

18 Gauss-Jordan Elimination Example Example: Use Gauss-Jordan elimination to solve the system of equations from the previous example.

19 Gauss-Jordan Elimination continued We continue to perform row-equivalent operations until we have a matrix in reduced row-echelon form.

20 Gauss-Jordan Elimination continued Next, we multiply the second row by 3 and add it to the first row.

21 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.

22 Special Systems When a row consists entirely of 0’s, the equations are dependent and the system is equivalent.

23 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 the last row corresponds to the false equation 0 = 9, so we know the original system has no solution. 0 = 9, so we know the original system has no solution.

24 Another Example Solve the system of equations using Gaussian elimination. x + 3y – 6z = 7 x + 3y – 6z = 7 2x – y + 2z = 0 x + y + 2z = -1 x + y + 2z = -1

25

26 Yet Another Example Solve the system of equations using Gaussian elimination. x + 2y = 1 x + 2y = 1 2x + 4y = 3


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