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Monthly Archives: March 2011
Linear Algebra and Its Applications, Exercise 1.6.9
Exercise 1.6.9. Given the singular matrix show that A has no inverse. If it did have an inverse then multiplying the third row of by the columns of A should give the third row of I. Explain why this is … Continue reading
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Linear Algebra and Its Applications, Exercise 1.6.8
Exercise 1.6.8. The matrix has no inverse. Demonstrate this by trying to solve the following: Answer: Multiplying the first row of the first marix by the first column of the second matrix gives However multiplying the second row of the … Continue reading
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Linear Algebra and Its Applications, Exercise 1.6.7
Exercise 1.6.7. Find three 2 by 2 matrices A such that and A is neither I nor I. Answer: We first note that the transpose of I is its own inverse: Note that this also follows from the result of … Continue reading
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Linear Algebra and Its Applications, Exercise 1.6.6
Exercise 1.6.6. Invert the following matrices using the GaussJordan method: Answer: For the first matrix GaussJordan elimination proceeds as follows: We first subtract 1 times the first row from the second row: This completes the process of forward elimination. We … Continue reading
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Linear Algebra and Its Applications, Exercise 1.6.5
Exercise 1.6.5. For a matrix A assume that is invertible and has inverse B. Prove that A is also invertible, with inverse AB. Answer: We have We then have We also have So AB is both a left and right … Continue reading
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Linear Algebra and Its Applications, Exercise 1.6.4
Exercise 1.6.4. (a) Given AB = AC, show that B = C if A is invertible. (b) Given find B and C such that AB = AC but . Answer: (a) If A is invertible then (b) If B is … Continue reading
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Linear Algebra and Its Applications, Exercise 1.6.3
Exercise 1.6.3. Given AB = C, express in terms of B and C. Similar, given PA = LU, express in terms of P, L, and U. Answer: Assume that both B and C are invertible (see below). We then have … Continue reading
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