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Unformatted text preview: MA 265 LECTURE NOTES: FRIDAY, FEBRUARY 1 Finding Inverses of Matrices (cont’d) Computing Inverses. In the last lecture, we showed that an n × n matrix A is invertible if and only if A is row equivalent to the n × n identity matrix I n . In other words, say that there is a sequence of elementary row operations which brings A into its reduced row echelon form I n . These elementary row operations correspond to elementary matrices E i , so that I n = E A in terms of E = E 1 E 2 ··· E k . Then A 1 = E is the inverse of A . In a more practical sense, we may keep track of these elementary row operations by considering the n × 2 n augmented matrix: bracketleftbig A I n bracketrightbig = a 11 a 12 ··· a 1 n 1 ··· a 21 a 22 ··· a 2 n 1 ··· . . . . . . . . . . . . . . . . . . . . . . . . a n 1 a n 2 ··· a nn ··· 1 . We perform a series of elementary row operations to place this matrix in reduced row echelon form: bracketleftbig I n E bracketrightbig = 1 ··· e 11 e 12 ··· e 1 n 1 ··· e 21 e 22 ··· e 2 n . . . . . . . . . . . . . . . . . . . . . . . . ··· 1 e n 1 e n 2 ··· e nn . Example. Consider the following 3 × 3 matrix: A = 1 1 1 2 3 5 5 1 We show how to compute A 1 . We form the 3 × 6 augmented matrix: bracketleftbig A I 3 bracketrightbig = 1 1 1 1 2 3 1 5 5 1 1 We proceed to perform GaussJordan Reduction in order to compute the reduced row echelon form for this matrix. First, we eliminate the entries in the first column: ( − 5) r 1 + r 3 → r 3 : 1 1 1 1 2 3 1 − 4 − 5 1 Now divide the second row by 2 in order to find the “leading one” for that row:...
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This note was uploaded on 02/25/2010 for the course MA 00265 taught by Professor ... during the Spring '10 term at Purdue University Calumet.
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