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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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Transcribed Image Text:Sol:-
Given that
Now
e
S = {(-3, 2)}
(d) s= {(1,3,0), (4,1,2), (-2,5,-2)}
2
©
с
SO
5= {(-3,2)} is not a basis of 1R²
because, we know
dim (IR²) = 2,
Now if S is albo basis of IR2
then dim (IR²) = Number of element ins
= 1
which is
a contradiction.
S is not a
basis of IR2
d)
S =
{(1,3,0), (4,1,2), (-2,5,-2)}
I
4
3 1
let A =
-2
5
2 -2
det (A) = 1x (-2-10) -4 (-6) -2 (6)
= -12-24-12
= - 48 0
Hence s is a linearly dep independent set
containg 3 vectors of IR³
we know a famous result if A GIR" be a
linearly independent set containg n vecters
of Irn then
Rn
then A is basis of Rh
By
Therefore
(Ⓒ)
this results is a
basis of IR³
S is a basis
s={(-3,2)} is not a basis of 1R2
(d) S =
S
= {(1, 3,0), (4,1,2), (-2, 5, -2)} is a
basis of IR³
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