1 1 0 0 0 -1 100 0 0 00 1 0 00 0 1 00 0 0 10 0 0 0 (E1):col2col2+col₁ (E2):col3 cols-col₁ Bker (8¹) 1 0 1 1 0 1 0 0 0 0 00 ---6 V2 0 Exercise Check that V2, V3, V4, V5 € ker(8¹) by the definition of kernel. V3 The key is now that the vectors of Q under the zero "columns" of 8¹. Q will actually form a basis for ker(8¹). So if we call v₁ through v5 the five columns of 8¹ - Q, the four vectors in our case that form a basis for the kernel are: 0 -1 0 0 0 1 0 0 0 1 0 0 0 1 V4 00 00 1 V5 8¹. Q Q = IE₂ E₁) 0

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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Question

Linear Algebra is the topic

1
1
0
0
0
-1 100
0 0 00
1 0 00
0
1 00
0
0 10
0
0 0
(E1):col2col2+col₁
(E2):col3 cols-col₁
Bker (8¹)
1
0
1 1
0 1
0 0
0 0
00
---6
V2
0
Exercise Check that V2, V3, V4, V5 € ker(8¹) by the definition of kernel.
V3
The key is now that the vectors of Q under the zero "columns" of 8¹. Q will actually form a basis for ker(8¹). So if we call
v₁ through v5 the five columns of 8¹ - Q, the four vectors in our case that form a basis for the kernel are:
0
-1
0
0 0
1
0
0
0 1 0
0 0 1
V4
00
00
1
V5
8¹. Q
Q = IE₂ E₁)
0
Transcribed Image Text:1 1 0 0 0 -1 100 0 0 00 1 0 00 0 1 00 0 0 10 0 0 0 (E1):col2col2+col₁ (E2):col3 cols-col₁ Bker (8¹) 1 0 1 1 0 1 0 0 0 0 00 ---6 V2 0 Exercise Check that V2, V3, V4, V5 € ker(8¹) by the definition of kernel. V3 The key is now that the vectors of Q under the zero "columns" of 8¹. Q will actually form a basis for ker(8¹). So if we call v₁ through v5 the five columns of 8¹ - Q, the four vectors in our case that form a basis for the kernel are: 0 -1 0 0 0 1 0 0 0 1 0 0 0 1 V4 00 00 1 V5 8¹. Q Q = IE₂ E₁) 0
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