X Document1 - Word AutoSave evie tucker Off ET Mailings Review Share File Design References View Help Search Home Insert Draw Layout Comments Find Cut 11A A Aa A E2T Calibri (Body) AаBЬСcDc AaBЬСcDc AaBbC AаBbСcL A ав Replace Copy Paste Dictate No Spa.... Heading T Normal Heading 2 Title В IUab- х, х* А Select Format Painter Clipboard Paragraph Styles Editing Voice Font N 1 3 1 1 C Col(A) 1. Let A= 2 1 2 1 a) Find an orthonormal basis for C. D. Focus Page 1 of 1 0 words 100% 8:29 PM a Type here to search W 25 12/2/2019 !Tili

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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AаBЬСcDc AaBЬСcDc AaBbC AаBbСcL A ав
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В IUab- х, х* А
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N
1
3
1
1
C Col(A)
1. Let A=
2
1
2
1
a) Find an orthonormal basis for C.
D. Focus
Page 1 of 1
0 words
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8:29 PM
a
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W
25
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!Tili
Transcribed Image Text:X Document1 - Word AutoSave evie tucker Off ET Mailings Review Share File Design References View Help Search Home Insert Draw Layout Comments Find Cut 11A A Aa A E2T Calibri (Body) AаBЬСcDc AaBЬСcDc AaBbC AаBbСcL A ав Replace Copy Paste Dictate No Spa.... Heading T Normal Heading 2 Title В IUab- х, х* А Select Format Painter Clipboard Paragraph Styles Editing Voice Font N 1 3 1 1 C Col(A) 1. Let A= 2 1 2 1 a) Find an orthonormal basis for C. D. Focus Page 1 of 1 0 words 100% 8:29 PM a Type here to search W 25 12/2/2019 !Tili
Expert Solution
Step 1

Covert the matrix into echelon form.

1
3
5
-1
-3
1
A =
2.
1
5
2
8
R, -R - R, R->R, - R
1
R2RR
1
1.
2
2
-3
O
2.
3
3
5
2
3
6
2
-3
2
RR R R, ->R, - R,
1
0 2
0 0
6
0 0 -6
0
0
Row1 and Row2 contains the lea ding entries, hence first and secondcolumn ofthe original
matrix forms the basis for column space of A
00
n o
en
en
O
O
n
O O
O
Step 2

Basis for Col(A).

1
3
-1
-3
Col (A)=0
2
1
5
1
5
Step 3

Find the orthogonal basis by Gram Schmidt Orthogonalisation process.

Let (1-1,0,1,1
v= (3,-3,2,5,5)
be the orthogonal basis. Then by Gram Schmidt's process
Let Wi and W
(1,-1,0,1,1)
-Wi
(Mi.Wa
((3,-3,2,5,5).(1.-1,0,1,1)
(1-1,0,1,1),(1-1,0,1,1)
(3,-3,2,5,5)
-(3,-3,2,5,5)-4(1-1,0,1,1)
= (-1,1,2,1,1)
Hence,1,-1,0,1,1),(-1,1,2,1,1)} is the orthogonal basis.
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