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Transcript
Section 5.3 Matrices And Systems of Equations Systems of Equations in Two Variables Matrices A rectangular array of numbers is called a matrix (plural, matrices). Example: 4 2 3 1 5 4 The matrix shown above is an augmented matrix because it contains not only the coefficients but also the constant terms. 4 2 The matrix 1 5 is called the coefficient matrix. 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. 1 2 3 4 5 6 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. Gaussian Elimination with Matrices Row-Equivalent Operations 1. Interchange any two rows. 2. Multiply each entry in a row by the same nonzero constant. 3. Add a nonzero multiple of one row to another row. Example Solve the following system: 2x y z 8 x 3 y 2 z 1 4x 5 z 23 . Example continued First, we write the augmented matrix, writing 0 for the missing y-term in the last equation. 2x y z 8 x 3 y 2 z 1 4x 2 1 1 8 1 3 2 1 4 0 5 23 5 z 23 Our goal is to find a row-equivalent matrix of the form 1 0 0 a 1 0 b d 1 c e f Example continued 2 1 1 8 1 3 2 1 4 0 5 23 1 3 2 1 2 1 1 8 4 0 5 23 New row 1 = row 2 New row 2 = row 1 Example continued 1 3 2 1 2 1 1 8 4 0 5 23 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. Row 1 is unchanged 1 3 2 1 0 5 5 10 New row 2= 2(row 1) + row 2 0 12 13 27 New row 3= 4(row 1)+row3 Example continued 1 3 2 1 0 5 5 10 0 12 13 27 We multiply the second row by 1/5 to get a 1 in the second row, second column. 1 3 2 1 0 1 1 2 New row 2= 1 (row 2) 5 0 12 13 27 Example continued 1 3 2 1 0 1 1 2 0 12 13 27 We multiply the second row by 12 and add it to the third row. 1 3 2 1 0 1 1 2 0 0 1 3 New row 3= 12(row 2) + row 3 Example continued 1 3 2 1 0 1 1 2 0 0 1 3 Now, we can write the system of equations that corresponds to our last matrix. x 3 y 2 z 1 yz2 z 3 Example continued We back-substitute 3 for z in equation (2) and solve for y. yz2 y3 2 y 1 Example continued Next, we back-substitute 1 for y and 3 for z in equation (1) and solve for x. x 3 y 2 z 1 x 3(1) 2(3) 1 x 3 6 1 x 3 1 x2 The triple (2, 1, 3) checks in the original system of equations, so it is the solution. 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: Example Which of the following matrices are in row-echelon form? a) 1 6 7 5 0 1 3 4 0 0 1 8 b) 0 2 4 1 0 0 0 0 c) 1 2 7 6 0 0 0 0 0 1 4 2 d) 1 0 0 3.5 0 1 0 0.7 0 0 1 4.5 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. . Gauss-Jordan Elimination Example Example: Use Gauss-Jordan elimination to solve the system of equations from the previous example. 1 3 2 1 0 1 1 2 0 0 1 3 Gauss-Jordan Elimination continued 1 3 2 1 0 1 1 2 0 0 1 3 1 3 0 5 0 1 0 1 0 0 1 3 We continue to perform row-equivalent operations until we have a matrix in reduced row-echelon form. New row 1 = 2(row 3) + row 1 New row 2 = 1(row 3) + row 2 Gauss-Jordan Elimination continued 1 3 0 5 0 1 0 1 0 0 1 3 1 0 0 2 0 1 0 1 0 0 1 3 Next, we multiply the second row by 3 and add it to the first row. New row 1 = 3(row 2) + row 1 Gauss-Jordan Elimination continued Writing the system of equations that corresponds to this matrix, we have 1 0 0 2 0 1 0 1 0 0 1 3 2 1 x y z 3 We can actually read the solution, (2, 1, 3), directly from the last column of the reduced row-echelon matrix. Special Systems When a row consists entirely of 0’s, the equations are dependent and the system is equivalent. 1 0 4 6 0 1 4 8 0 0 0 0 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 1 0 4 6 0 1 4 8 0 0 0 9 the last row corresponds to the false equation 0 = 9, so we know the original system has no solution. Another Example Solve the system of equations using Gaussian elimination. x + 3y – 6z = 7 2x – y + 2z = 0 x + y + 2z = -1 Yet Another Example Solve the system of equations using Gaussian elimination. x + 2y = 1 2x + 4y = 3