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lecture_11- Finding Inverses of Matrices Part 3 and Determinants Part 1

Lecture_11- Finding Inverses of Matrices Part 3 and Determinants Part 1

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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 0 · · · 0 a 21 a 22 · · · a 2 n 0 1 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . a n 1 a n 2 · · · a nn 0 0 · · · 1 . We perform a series of elementary row operations to place this matrix in reduced row echelon form: bracketleftbig I n E bracketrightbig = 1 0 · · · 0 e 11 e 12 · · · e 1 n 0 1 · · · 0 e 21 e 22 · · · e 2 n . . . . . . . . . . . . . . . . . . . . . . . . 0 0 · · · 1 e n 1 e n 2 · · · e nn . Example. Consider the following 3 × 3 matrix: A = 1 1 1 0 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 0 0 0 2 3 0 1 0 5 5 1 0 0 1 We proceed to perform Gauss-Jordan 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 0 0 0 2 3 0 1 0 0 0 4 5 0 1 Now divide the second row by 2 in order to find the “leading one” for that row: (1 / 2) r 2 r 2 : 1 1 1 1 0 0 0 1 3 2 0 1 2 0 0 0 4 5 0 1
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