Write matrices and identify their orders. Perform elementary row operations on matrices. Use matrices and Gaussian elimination to solve systems of linear equations. Use matrices and Gauss-Jordan elimination to solve systems of linear equations. What You Should Learn

Slide 4:

Matrices

Matrices:

Matrices In this section, we will study a streamlined technique for solving systems of linear equations. This technique involves the use of a rectangular array of real numbers called a matrix. The plural of matrix is matrices .

Matrices:

Matrices The entry in the i th row and j th column is denoted by the double subscript notation a ij . For instance, a 23 refers to the entry in the second row, third column. A matrix having m rows and n columns is said to be of order m n . If m = n , the matrix is square of order m m (or n n ). For a square matrix, the entries a 11 , a 22 , a 33 , . . . are the main diagonal entries.

Example 1 – Order of Matrices:

Example 1 – Order of Matrices Determine the order of each matrix. a. b. c. d. Solution: a. This matrix has one row and one column. The order of the matrix is 1 1. b. This matrix has one row and four columns. The order of the matrix is 1 4.

Example 1 – Solution:

Example 1 – Solution c. This matrix has two rows and two columns. The order of the matrix is 2 2. d. This matrix has three rows and two columns. The order of the matrix is 3 2. cont’d

Matrices:

Matrices A matrix that has only one row is called a row matrix, and a matrix that has only one column is called a column matrix. A matrix derived from a system of linear equations (each written in standard form with the constant term on the right) is the augmented matrix of the system. Moreover, the matrix derived from the coefficients of the system (but not including the constant terms) is the coefficient matrix of the system.

Matrices:

Matrices System : x – 4 y + 3 z = 5 – x + 3 y – z = –3 2 x – 4 z = 6 Augmented Matrix :

Matrices:

Matrices Coefficient Matrix : Note the use of 0 for the missing coefficient of the y -variable in the third equation, and also note the fourth column of constant terms in the augmented matrix. When forming either the coefficient matrix or the augmented matrix of a system, you should begin by vertically aligning the variables in the equations and using zeros for the coefficients of the missing variables.

Slide 12:

Elementary Row Operations

Elementary Row Operations:

Elementary Row Operations We have studied three operations that can be used on a system of linear equations to produce an equivalent system. 1. Interchange two equations. 2. Multiply an equation by a nonzero constant. 3. Add a multiple of an equation to another equation. In matrix terminology, these three operations correspond to elementary row operations.

Elementary Row Operations:

Elementary Row Operations An elementary row operation on an augmented matrix of a given system of linear equations produces a new augmented matrix corresponding to a new (but equivalent) system of linear equations. Two matrices are row-equivalent if one can be obtained from the other by a sequence of elementary row operations.

Elementary Row Operations:

Elementary Row Operations Although elementary row operations are simple to perform, they involve a lot of arithmetic. Because it is easy to make a mistake, you should get in the habit of noting the elementary row operations performed in each step so that you can go back and check your work.

Example 3 – Elementary Row Operations:

Example 3 – Elementary Row Operations a. Interchange the first and second rows of the original matrix. Original Matrix New Row-Equivalent Matrix

Example 3 – Elementary Row Operations:

Example 3 – Elementary Row Operations b. Multiply the first row of the original matrix by Original Matrix New Row-Equivalent Matrix cont’d

Example 3 – Elementary Row Operations:

Example 3 – Elementary Row Operations c. Add –2 times the first row of the original matrix to the third row. Original Matrix New Row-Equivalent Matrix Note that the elementary row operation is written beside the row that is changed . cont’d

Elementary Row Operations:

Elementary Row Operations The term echelon refers to the stair-step pattern formed by the nonzero elements of the matrix. To be in row echelon form, a matrix must have the following properties.

Elementary Row Operations:

Elementary Row Operations It is worth noting that the row-echelon form of a matrix is not unique. That is, two different sequences of elementary row operations may yield different row-echelon forms. However, the reduced row-echelon form of a given matrix is unique.

Slide 21:

Gaussian Elimination with Back-Substitution

Gaussian Elimination with Back-Substitution:

Gaussian Elimination with Back-Substitution Gaussian elimination with back-substitution works well for solving systems of linear equations by hand or with a computer. For this algorithm, the order in which the elementary row operations are performed is important. You should operate from left to right by columns, using elementary row operations to obtain zeros in all entries directly below the leading 1’s.

Example 6 – Gaussian Elimination with Back-Substitution:

Example 6 – Gaussian Elimination with Back-Substitution Solve the system y + z – 2 w = –3 x + 2 y – z = 2 2 x + 4 y + z – 3 w = – 2 x – 4 y – 7 z – w = –19 Solution: Write augmented matrix. .

Example 6 – Solution:

Example 6 – Solution cont’d Interchange R 1 and R 2 so first column has leading 1 in upper left corner. Perform operations on R 3 and R 4 so first column has zeros below its leading 1.

Example 6 – Solution:

Example 6 – Solution cont’d Perform operations on R 4 so second column has zeros below its leading 1. Perform operations on R 3 and R 4 so third and fourth columns have leading 1’s.

Example 6 – Solution:

Example 6 – Solution The matrix is now in row-echelon form, and the corresponding system is x + 2 y – z = 2 y + z – 2 w = –3 z – w = –2 w = 3 Using back-substitution, you can determine that the solution is x = –1, y = 2, z = 1, and w = 3. cont’d .

Gaussian Elimination with Back-Substitution:

Gaussian Elimination with Back-Substitution The procedure for using Gaussian elimination with back-substitution is summarized below.

Gaussian Elimination with Back-Substitution:

Gaussian Elimination with Back-Substitution When solving a system of linear equations, remember that it is possible for the system to have no solution. If, in the elimination process, you obtain a row of all zeros except for the last entry, it is unnecessary to continue the elimination process. You can simply conclude that the system has no solution, or is inconsistent .

Slide 29:

Gauss-Jordan Elimination

Gauss-Jordan Elimination:

Gauss-Jordan Elimination With Gaussian elimination, elementary row operations are applied to a matrix to obtain a (row-equivalent) row-echelon form of the matrix. A second method of elimination, called Gauss-Jordan elimination, continues the reduction process until a reduced row-echelon form is obtained. This procedure is demonstrated in Example 8.

Example 8 – Gauss-Jordan Elimination:

Example 8 – Gauss-Jordan Elimination Use Gauss-Jordan elimination to solve the system x – 2 y + 3 z = 9 – x + 3 y = –4 2 x – 5 y + 5 z = 17 Solution: The row-echelon form of the linear system can be obtained using the Gaussian elimination. .

Example 8 – Solution:

Example 8 – Solution Now, apply elementary row operations until you obtain zeros above each of the leading 1’s, as follows. Perform operations on R 1 so second column has a zero above its leading 1. Perform operations on R 1 and R 2 so third column has zeros above its leading 1. cont’d

Example 8 – Solution:

Example 8 – Solution The matrix is now in reduced row-echelon form. Converting back to a system of linear equations, you have x = 1 y = –1. z = 2 Now you can simply read the solution, x = 1, y = –1, and z = 2 which can be written as the ordered triple(1, –1, 2). cont’d

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