? 1 2 ? 2 1 3 11 1 4 a a pdp 1 3 1 11 4 2 0 0 1 3 1

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λ 1 = 2 , λ 2 3 11 1 4 A = = 1 =
λ 1 = 1 , λ 2 = 0 , λ 1 0 1 2 1 1 1 1 1 A = A = PDP 1 = = 1 2 1 0 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 2 1 0 1 1 1 1 1 1 1 2 1 0 1 1 1 1 1 1 0 0 0 0 0 0 0 1 2 3 1 1 2 1 1 1 1 1 2 0 1 1 1 3 4 2 3 = 1 =
12/4/17, 21(40 UW Common Math 308 Section 6.2 4. 1/1 points | Previous Answers HoltLinAlg2 6.2.009. Diagonalize A if possible. (Find P and D such that for the given matrix A . Enter your answer as one augmented matrix. If the matrix is not able to be diagonalized, enter DNE in any cell.) DNE DNE DNE DNE Solution or Explanation Since a basis for the eigenspace of is The
Page 2 of 4 6. 1/1 points | Previous Answers HoltLinAlg2 6.2.012.
multiplicity of the eigenvalue is two, which exceeds the dimension of its eigenspace. Thus is not diagonalizable. 5. 1/1 points | Previous Answers HoltLinAlg2 6.2.010. Diagonalize A if possible. (Find P and D such that for the given matrix A . Enter your answer as one augmented matrix. If the matrix is not able to be diagonalized, enter DNE in any cell.) = 1 1 0 0 1 0 0 -6 [1, 1, 0, 0; 1, 0, 0, -6] is is

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