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Unformatted text preview: h 6 = 0 and we will ﬁnd the inverse. We must row reduce to the identity, so let’s divide the third row by h . (This is OK, because we are assuming h is nonzero.) We get: 1 1 1  1 0 1 1  1 0 0 1  0 1 /h1 . Next we take row 3 and subtract it from both rows 2 and 1: 1 1 0  11 /h 1 0 1 0 1 /h 2 0 0 1  1 /h1 . 1 To ﬁnish, let’s take row 2 and subtract it from row 1: 1 0 0  11 0 1 0 1 /h 2 0 0 1  1 /h1 . This shows that the inverse (when h is nonzero) is the matrix 111 /h 2 1 /h1 . 2...
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 Fall '10
 Russel
 Linear Algebra, Algebra, Addition, Invertible matrix

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