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M313Lec07

# M313Lec07 - Math 313 Lecture#7 1.5 Elementary Matrices Part...

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§ 1.5: Elementary Matrices, Part II Finding the Inverse of an Invertible Matrix. In the last Theorem, we learned that A is invertible if and only if A is row equivalent to I . So when A is invertible, there are ﬁnitely many elementary matrices E 1 , E 2 , . . . , E k such that E k · · · E 2 E 1 A = I. Post multiplying (i.e., on the right) both sides by A - 1 gives a “formula” for the inverse of A : E k · · · E 2 E 1 = A - 1 . Practically, one row reduces the augmented matrix ( A | I ) to ( I | A - 1 ): ( A | I ) ( E 1 A | E 1 I ) ( E 2 E 1 A | E 2 E 1 I ) ⇔ · · · ⇔ ( E k · · · E 2 E 1 A | E k · · · E 2 E 1 ) = ( I | A - 1 ) . NOTE: if A is not invertible, then A cannot be row reduced to the identity matrix: there has to appear a row of zeros in the “ A ” part of ( A | I ) at some point in the row reduction. This is one way to detect a singular matrix. Example. Find, if possible, the inverse of A = 1 2 2 1 1 2 1 1 1 . Augment A with I and row reduce: ( A | I ) = 1 2 2 | 1 0 0 1 1 2 | 0 1 0 1 1 1 | 0 0 1 R 2 - R 1 R 2 R 3 - R 1 R 3 1 2 2 | 1 0 0 0 - 1 0 | - 1 1 0 0 - 1 - 1 | - 1 0 1 R 1 + 2 R 2 R 1 R 3 - R 2 R 3 1 0 2 | - 1 2 0 0 - 1 0 | - 1 1 0 0 0 - 1 | 0 - 1 1 R 1 + 2 R 3 R 1 1 0 0 | - 1 0 2 0 - 1 0 | - 1 1 0 0 0 - 1 | 0 - 1 1 - R 2 R 2 - R 3 R 3 1 0 0 | - 1 0 2 0 1 0 | 1 - 1 0 0 0 1 | 0 1 - 1

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M313Lec07 - Math 313 Lecture#7 1.5 Elementary Matrices Part...

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