# Week5 - h 6 = 0 and we will ﬁnd the inverse We must row...

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Solutions for Assignment 5 Applied Linear Algebra MATH 232 (Fall 2008) Section 2.3 Section 2.6

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Assignment 5 Applied Linear Algebra Math 232(Fall 2008) Additional question: For which values of h is the matrix 1 1 1 0 h 2 h 0 1 1 invertible? Find the inverse when it exists. A1. We create the 3 × 6 matrix 1 1 1 | 1 0 0 0 h 2 h | 0 1 0 0 1 1 | 0 0 1 . The ﬁrst column is OK. Let’s switch rows two and three to get a one as a pivot in the second row: 1 1 1 | 1 0 0 0 1 1 | 0 0 1 0 h 2 h | 0 1 0 . Now we take row 3 and subtract h times row 2 from it: 1 1 1 | 1 0 0 0 1 1 | 0 0 1 0 0 h | 0 1 - h . The matrix on the left is in echelon form. We have a pivot in every row and column if and only if h is nonzero. So the answer to the ﬁrst part is: the matrix is invertible for every h except for h = 0. For the remainder, we’ll assume
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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 /h-1 . Next we take row 3 and subtract it from both rows 2 and 1: 1 1 0 | 1-1 /h 1 0 1 0 |-1 /h 2 0 0 1 | 1 /h-1 . 1 To ﬁnish, let’s take row 2 and subtract it from row 1: 1 0 0 | 1-1 0 1 0 |-1 /h 2 0 0 1 | 1 /h-1 . This shows that the inverse (when h is nonzero) is the matrix 1-1-1 /h 2 1 /h-1 . 2...
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Week5 - h 6 = 0 and we will ﬁnd the inverse We must row...

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