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Unformatted text preview: Math 20F Linear Algebra Lecture 2: 1.2 RowEchelon form. Last time we saw that we could reduce an n × n system by row operations into an equivalent system in triangular form, ( a ij = 0 for i > j .) For example: x 1 2 x 2 + x 3 = 0 x 2 4 x 3 = 4 x 3 = 3 1 2 1 fl fl 1 4 fl fl 4 1 fl fl 3 If the diagonal entries are nonzero this can be solved by backsubstitution. If not, we might get no solutions, for example: x 1 2 x 2 + x 3 = 0 x 2 4 x 3 = 4 0 = 3 1 2 1 fl fl 1 4 fl fl 4 fl fl 3 Or we might get infinitely many solutions which happens for this 2 × 3 system in triangular form: (Notice that x 2 can be chosen freely.) x 1 2 x 2 + x 3 = 0 x 3 = 3 • 1 2 1 fl fl 1 fl fl 3 ‚ Now consider the system x 2 + x 3 = 0 x 2 x 3 = 0 x 3 = 1 0 1 1 fl fl 0 1 1 fl fl 0 0 1 fl fl 1 This matrix is in triangular form. However, it is not completely obvious without more checking that this system is inconsistent. Today we introduce special sorts of triangular form which are the most useful for solving equations. A matrix is said to be in Row Echelon Form (“steplike” or “staircase” form) if: Each leading entry (i.e. left most nonzero entry) of a row is in a column to the right of the leading entry of the row above it. For example, 1 * * * * 0 1 * * * 0 0 0 1 * 0 0 0 0 1 0 0 0 0 0 It is easy to determine if a system with augmented matrix in rowechelon form is consistent, and if so to solve it by backsubstitution.consistent, and if so to solve it by backsubstitution....
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This note was uploaded on 04/29/2008 for the course MATH 20F taught by Professor Buss during the Spring '03 term at UCSD.
 Spring '03
 BUSS
 Linear Algebra, Algebra

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