REDUCTIONMETHOD052

REDUCTIONMETHOD052 - Chapter 6 MATRIX ALGEBRA Dr Raja...

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Chapter 6: MATRIX ALGEBRA Dr. Raja Mohammad Latif We will show how to reduce a matrix and to use matrix reduction to solve a linear system.

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Math 131-052. Dr. Raja Latif 2 6.4-6.5: Method of Reduction Performing any one of the following Row Operations on the augmented matrix of a system of linear equations produces the augmented matrix of an equivalent system. 1. Interchange any two rows. 2. Multiply each element of a row by a nonzero constant. 3. Replace a row by the sum of itself and a constant multiple of another row of the matrix. DEPENDENT AND INCONSISTENT SYSTEMS. The possible number of solutions of a system with more than two variables or equations is the same as for smaller systems. Such a system has exactly one solution (independent system); or in°nitely many solutions (dependent system). Both the equation method and the matrix method always produce the unique solution of an independent system. REDUCED ROW ECHELON FORM: A system of m Department of Mathematics, KFUPM.
Math 131-052. Dr. Raja Latif 3 linear equations containing n variables will have either no solution, one solution, or in°nitely many solutions. We can determine which of these possibilities occurs and, if solutions exist, °nd them, by performing row operations on the augmented matrix of the system until we arrive at the reduced row echelon form of the augmented matrix. Reduced Row Echelon Form of an Augmented Matrix: The °rst nonzero entry in each row is 1 and it has 0s above it and below it. The leftmost 1 in any row is to the right of the leftmost 1 in the row above. Any rows that contain all 0s to the left of the vertical bar appear at the bottom. THE GAUSS-JORDAN METHOD FOR SOLVING A SYSTEM OF LINEAR EQUATIONS (Reduced Row Echelon Form of an Augmented Matrix) 1. Arrange the equations with the variable terms in the Department of Mathematics, KFUPM.

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Math 131-052. Dr. Raja Latif 4 same order on the left of the equals sign and the constants on the right. 2. Write the augmented matrix that represents the system. 3. Perform row operations that place the entry 1 in row 1 , column 1 unchanged, while causing 0 s to appear below it in column 1 : i.e., Choose the leftmost nonzero column and use appropriate row operations to get a 1 at the top.) 4. Use multiples of the row containing the 1 from step 1 to get zeros in all remaining places in the column containing this 1 : 5. Perform row operations that place the entry 1 in row 2 ; column 2 ; but leave the entries in columns to the left unchanged. If it is impossible to place a 1 in row 2 ; column 2 ; then proceed to place a 1 in row 2 ; column 3 : Once a 1 is in place, perform row operations to place 0 s below it. Department of Mathematics, KFUPM.
Math 131-052. Dr. Raja Latif 5 (If any rows are obtained that contain only 0 s on the left side of the vertical bar, place such rows at the bottom of the matrix.) 6. Now repeat Step 4, placing a 1 in the next row, but one column to the right. Continue until the bottom row or the vertical bar is reached.

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