Prove that if A and B are similar, then |A| |B|. Is the converse true? Illustrate the results using the matrices %3D [10 0 A =|0 3 0 [oo -2 -1 1 1] -2 1 1, -2 1 2 5 -2 -2 1 -1 B = -6 4 P = 2 0 -1 10 -5 -7 0 -1 where B = P-'AP. STEP 1: First note that A and B are similar. 1 -1 -- 1 0 0 3 -1 1 1 p-lAP = 2 0 -1 -2 1 1 0 -1 0 0 -2 -2 1 2 - E 1 -1 2 0 -1 0 -1 STEP 2: Take the determinant of B by expanding along the first row. |B| = 5 -6 + 2 6| 10 -7 + 2(-18) – 2
Prove that if A and B are similar, then |A| |B|. Is the converse true? Illustrate the results using the matrices %3D [10 0 A =|0 3 0 [oo -2 -1 1 1] -2 1 1, -2 1 2 5 -2 -2 1 -1 B = -6 4 P = 2 0 -1 10 -5 -7 0 -1 where B = P-'AP. STEP 1: First note that A and B are similar. 1 -1 -- 1 0 0 3 -1 1 1 p-lAP = 2 0 -1 -2 1 1 0 -1 0 0 -2 -2 1 2 - E 1 -1 2 0 -1 0 -1 STEP 2: Take the determinant of B by expanding along the first row. |B| = 5 -6 + 2 6| 10 -7 + 2(-18) – 2
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter6: Linear Systems
Section6.3: Matrix Algebra
Problem 79E
Related questions
Question
![|B|. Is the converse true? Illustrate the results using the matrices
1 -1
Prove that if A and B are similar, then |A|
[1 0
A = 0 3
-1 1 1
-2 1 1, p-:
-2 1 2]
5 -2 -2
B
-6
4
P =
2
0 -1
0 0 -2
10 -5 -7
0 -1
where B = P-1AP.
STEP 1: First note that A and B are similar.
1 -1
1 0
-1 1 1
p-1AP =
2
0 -1
0 3
-2 1 1
0 -1
1
0 0 -2
-2 1 2
1 -1
2
0 -1
0 -1
STEP 2: Take the determinant of B by expanding along the first row.
|B| = 5
|-6
+ 2
6 |
- 2
10 -7
-5
+ 2(-18)
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f30608e-bf31-4a49-92e6-e964952b6a83%2F067e68a1-e4d6-434d-9bd3-7aadd8ef5cb8%2Fj66jmnh_processed.png&w=3840&q=75)
Transcribed Image Text:|B|. Is the converse true? Illustrate the results using the matrices
1 -1
Prove that if A and B are similar, then |A|
[1 0
A = 0 3
-1 1 1
-2 1 1, p-:
-2 1 2]
5 -2 -2
B
-6
4
P =
2
0 -1
0 0 -2
10 -5 -7
0 -1
where B = P-1AP.
STEP 1: First note that A and B are similar.
1 -1
1 0
-1 1 1
p-1AP =
2
0 -1
0 3
-2 1 1
0 -1
1
0 0 -2
-2 1 2
1 -1
2
0 -1
0 -1
STEP 2: Take the determinant of B by expanding along the first row.
|B| = 5
|-6
+ 2
6 |
- 2
10 -7
-5
+ 2(-18)
2
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