Now, see what happens to the matrix 0 I 27 A3 0 0 if we multiply it on the left by L2 1 of 01 0 E₁₁ =1 12 Lo 00 We get 0 1 300 0 1 2= R(A) L2 1 0 0 1 0 0 1 2 1 0 0 3 0 0 Lo o 1JL2 1 0 Similarly, AE₁2 = C₁₂(A). 1 0 0 Again, consider E, (2)A = 0 1 0 0 1 2 3 0 0 Lo o 2] L2 1 0] [012] =3 0 0=R, (2)(A) L4 2 0 Similarly, AE,(2)=C₂(2)(A) 1 0 5 Finally, E, (5)A=0 1 0 Lo 0 1 -R₂ (5)(A) 0 1 2] 3 0 0 2 1 0 0 1 2 1 0 5 0 1 0 0 0 1 But, AE(5)= 3 0 0 L2 1 0 C₁15X(A) 10 6 2 3 0 0 21 0 [012] 3 0 15 2 1 10 How is this matrix R₁3 (5)A?? How is this matrix C31(5)A??
Now, see what happens to the matrix 0 I 27 A3 0 0 if we multiply it on the left by L2 1 of 01 0 E₁₁ =1 12 Lo 00 We get 0 1 300 0 1 2= R(A) L2 1 0 0 1 0 0 1 2 1 0 0 3 0 0 Lo o 1JL2 1 0 Similarly, AE₁2 = C₁₂(A). 1 0 0 Again, consider E, (2)A = 0 1 0 0 1 2 3 0 0 Lo o 2] L2 1 0] [012] =3 0 0=R, (2)(A) L4 2 0 Similarly, AE,(2)=C₂(2)(A) 1 0 5 Finally, E, (5)A=0 1 0 Lo 0 1 -R₂ (5)(A) 0 1 2] 3 0 0 2 1 0 0 1 2 1 0 5 0 1 0 0 0 1 But, AE(5)= 3 0 0 L2 1 0 C₁15X(A) 10 6 2 3 0 0 21 0 [012] 3 0 15 2 1 10 How is this matrix R₁3 (5)A?? How is this matrix C31(5)A??
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Now, see what happens to the matrix
го | 2
A3 0 0 if we multiply it on the left by
2 1 0
E
12
Го і
=|1 00 We get
0
0 1
300
3 000 1 2= R₁₂(A)
L2 1 0
[0 1 0 0 1 2
01
1 0 0 3 0
Lo
IJL2 1 0J
O
o
Similarly, AE₁₂ = C₁₂(A).
1 0 0 0 1 2
Again, consider E₂ (2)A = 0 1 0
Ę
Lo o
[012]
= 3 0 0=R, (2)(A)
[4 2 0]
01 03 00
Similarly, AE,(2)=C₂(2)(A)
0 1 2
But, AE(5)= 3 0 0
L2 1 0
=C(5XA)
[1 0 50 1 2]
Finally, E, (5)A = 0 1 0 3 0 0 =
Lo 0 12 1 0J
-R₁ (5)(A)
2] L2 1 0
1 05
0 1 0
0 0 1
=
1062
300
210]
012
3 0 15
2 1 10
How is this matrix R₁3 (5)A??
13
How is this matrix C31(5)A??](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2Ffc5a36c0-464f-4429-8915-e4526d04b36b%2Fbjepk2_processed.png&w=3840&q=75)
Transcribed Image Text:Now, see what happens to the matrix
го | 2
A3 0 0 if we multiply it on the left by
2 1 0
E
12
Го і
=|1 00 We get
0
0 1
300
3 000 1 2= R₁₂(A)
L2 1 0
[0 1 0 0 1 2
01
1 0 0 3 0
Lo
IJL2 1 0J
O
o
Similarly, AE₁₂ = C₁₂(A).
1 0 0 0 1 2
Again, consider E₂ (2)A = 0 1 0
Ę
Lo o
[012]
= 3 0 0=R, (2)(A)
[4 2 0]
01 03 00
Similarly, AE,(2)=C₂(2)(A)
0 1 2
But, AE(5)= 3 0 0
L2 1 0
=C(5XA)
[1 0 50 1 2]
Finally, E, (5)A = 0 1 0 3 0 0 =
Lo 0 12 1 0J
-R₁ (5)(A)
2] L2 1 0
1 05
0 1 0
0 0 1
=
1062
300
210]
012
3 0 15
2 1 10
How is this matrix R₁3 (5)A??
13
How is this matrix C31(5)A??
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