lecture_11 - MA 265 LECTURE NOTES: FRIDAY, FEBRUARY 1...

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Unformatted text preview: MA 265 LECTURE NOTES: FRIDAY, FEBRUARY 1 Finding Inverses of Matrices (contd) 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: A I n = a 11 a 12 a 1 n 1 0 a 21 a 22 a 2 n 0 1 . . . . . . . . . . . . . . . . . . . . . . . . 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: I n E = 1 0 e 11 e 12 e 1 n 0 1 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: A I 3 = 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:matrix....
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lecture_11 - MA 265 LECTURE NOTES: FRIDAY, FEBRUARY 1...

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