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Linear Algebra With Applications
Found in: Page 325
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

find an eigenbasis for the given matrice and diagonalize:


The eigenbasis for the given matrice is 0000000014.

See the step by step solution

Step by Step Solution

Step 1: Solving the given matrices

We solve:

det(A-λl)=0 1-λ2324-λ6369-λ=0 1-λ4-λ9-λ+36+36-361-λ-94-λ-49-λ=0 -λ3+14λ2=0 λ2-λ+14λ=0 λ1,2=0,λ3=14

Step 2: Solving by different values of λ

For λ=0, we solve:

[123246369]x1y1z1=000 x1+2y1+3z1=0

We can see all the eigenvector λ=0 for are spanned by vectors:


For λ=14, we solve:

[-13232-10636-5]x3y3z3=000 -13x3+2y3+3z3=0,2x3-10y3+6z3+0,3x3+6y3-5z3=0

We can choose an eigenvector:


Now, v1,v2,v3 is an eigenbasis for R3, therefore the diagonalization of A in the eigenbasis is


Hence the final answer is 0000000014.

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