Exam 2 Solution

# 3 10 08 2 132 1 8 0 1 8 0 0 1 4 8 1 that is

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Unformatted text preview: Find the coordinate vector of p (i) Consider p ,p , 2 . relative to B. 3 for standard basis C = 1, , p 101 10 1 10 1 101 Then 0 2 2 ~ 0 2 2~0 1 1 ~ 0 1 1 and the columns form a basis for 330 03 3 01 1 002 the IMT. is isomorphic to , so the corresponding polynomials form a basis for . (ii) Working in coordinates relative to the standard basis again, b b b 1 101 022 330 2 10 3~0 2 1 03 1 2 3 0 1 2 0 3~ 5 0 1 1 1 1 100 ~0 1 0 001 So p 23/12 19/12 1/12 2 101 3 011 2~ 5 002 3 23/12 19/12 1/12 p by p, so 2 101 3 011 2~ 1 001 6 2 3 2~ 1 12 Part IV. True/false questions [2pts each]: Write “True” or “False” in the blank. No justification is needed. TRUE 1. If and are similar matrices, then det = det . , then there is a 3 x 3 matrix A such that H = Col A. TRUE 2. If H is a subspace of FALSE 3. If A is m x n and rank A = m, then the linear transformation TRUE 4. If 0, then det is an n x n matrix and x x is one-to-one. 0. FALSE 5. An n x n matrix with n linearly independent eigenvectors is invertible. FALSE 6. A plane in is a two-dimensional subspace of TRUE 7. For a 5 x 7 matrix A, dim Nul A . 2. FALSE 8. Row operations on a matrix can change the null space of the matrix. TRUE 9. The matrix 5 0 0 0 0 3 2 0 0 0 4 1 6 0 0 7 0 0 3 0 8 2 1 is diagonalizable. 4 1 FALSE 10. If A is a 3 x 3 matrix, then det 3A = 3 det A....
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## This note was uploaded on 01/17/2014 for the course ESE 309 taught by Professor Staff during the Fall '08 term at Washington University in St. Louis.

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