(b) A = 3 is an eigenvalue of A. Find an eigenvector of A with this eigenvalue. (c) The matrix A is diagonalizable. Using your answers to parts (a) and (b) above, write down an invertible matrix P and a diagonal matrix D such that A = PDP-1. (d) Circle whichever of the following are equal to A100 (no justification necessary; this question may have multiple correct answers): (-1)100 4100 (-2)1007 |(-3)100 4100 |(-3)100 0100 3100 PD100 p-1; p100 p100 (P-1)100, (PDP-1)!00. 100
(b) A = 3 is an eigenvalue of A. Find an eigenvector of A with this eigenvalue. (c) The matrix A is diagonalizable. Using your answers to parts (a) and (b) above, write down an invertible matrix P and a diagonal matrix D such that A = PDP-1. (d) Circle whichever of the following are equal to A100 (no justification necessary; this question may have multiple correct answers): (-1)100 4100 (-2)1007 |(-3)100 4100 |(-3)100 0100 3100 PD100 p-1; p100 p100 (P-1)100, (PDP-1)!00. 100
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
100%
Please do part b), c) and d). Do not answer part a).
Transcribed Image Text:-1 4 -2
Consider the matrix A =
-3 4
-3 1
3
2
3 are eigenvectors of A. Find the corresponding eigenvalues.
(a) The vectors i = |1 and j:
%3D
(b) A = 3 is an eigenvalue of A. Find an eigenvector of A with this eigenvalue.
(c) The matrix A is diagonalizable. Using your answers to parts (a) and (b) above, write down an invertible
matrix P and a diagonal matrix D such that A = PDP-1.
(d) Circle whichever of the following are equal to A100 (no justification necessary; this question may have
multiple correct answers):
[(-1)100 4100
(-3)100 4100
(-3)100 1100
(-2)1007
0100
3100
PD100 p-1;
p100 D100 (p-1)100,
(PDP-1)100
![key
Andwer
For part (a), the eigenvalue for is d = 1 (since A = #) and the eigenvalue for j is A = 2 (since
Aj
25). For part (b), the eigenvectors of A with eigenvalue A = 3 (found by solving the equation
(A – 31) = 0) are all the nonzero multiples of 3/
[1/4]
so any one of these is a correct answer. For part
(c), one possible diagonalization of A is A = PDP-1 where
[1 2 1]
P = 1 3
1 3 4
[1 0 0]
D = 0 2 0
0 0 3
[1/4]
other correct answers are possible (here I scaled the 3-eigenvector 3/4 to clear fractions). Finally, for
1
part (d), only the second and fourth options (PD100 p-1 and (PDP¯1)100) are equal to A100.
1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa923b6f-81dd-482c-8885-6de6bc295751%2F0c2870b5-983c-45cf-9bbd-7c21b392719e%2Fjx9w9jd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:key
Andwer
For part (a), the eigenvalue for is d = 1 (since A = #) and the eigenvalue for j is A = 2 (since
Aj
25). For part (b), the eigenvectors of A with eigenvalue A = 3 (found by solving the equation
(A – 31) = 0) are all the nonzero multiples of 3/
[1/4]
so any one of these is a correct answer. For part
(c), one possible diagonalization of A is A = PDP-1 where
[1 2 1]
P = 1 3
1 3 4
[1 0 0]
D = 0 2 0
0 0 3
[1/4]
other correct answers are possible (here I scaled the 3-eigenvector 3/4 to clear fractions). Finally, for
1
part (d), only the second and fourth options (PD100 p-1 and (PDP¯1)100) are equal to A100.
1
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