Monthly Archives: July 2011

Linear Algebra and Its Applications, Review Exercise 1.19

Review exercise 1.19. Solve the following systems of equations using elimination and back substitution:    and    Answer: We start with the system and subtract 1 times the first equation from the second and 3 times the first equation from … Continue reading

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Linear Algebra and Its Applications, Review Exercise 1.18

Review exercise 1.18. Given the following matrix (a) Factor into the form (if ). (b) Show that exists and has the same form as . Answer: (a) We start elimination by subtracting times the second row from the third row … Continue reading

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Linear Algebra and Its Applications, Review Exercise 1.17

Review exercise 1.17. Factor the following two symmetric matrices into the form . Answer: We start elimination for the first matrix by subtracting 2 times the first row from the second row (): and then subtract 2 times the second … Continue reading

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Linear Algebra and Its Applications, Review Exercise 1.16

Review exercise 1.16. Given the following system of equations: what must be for the system to have no solution? One solution? An infinite number of solutions? Answer: If then the system reduces to for which and is (obviously) a solution. … Continue reading

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Linear Algebra and Its Applications, Review Exercise 1.15

Review exercise 1.15. For the following by matrix and its inverse what is the value of ? Answer: Since we must have Multiplying both sides by we obtain or so that . NOTE: This continues a series of posts containing … Continue reading

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Linear Algebra and Its Applications, Review Exercise 1.14

Review exercise 1.14. For each of the following find a 3 by 3 matrix such that for any matrix (a) (b) (c) The first row of is the last row of and the last row of is the first row … Continue reading

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Linear Algebra and Its Applications, Review Exercise 1.13

Review exercise 1.13. Given the following: use the triangular systems and to find a solution to Answer: We have Since and we have also. Since we have so that We then have Since and we have . We then have … Continue reading

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