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MATH1111/201011/Tutorial III Solution
1
Tutorial III Suggested Solution
1. Consider the vectors
a
= (1 1
±
1)
T
; b
= (2 0 1)
T
; c
= (
±
1 1
±
2)
T
; d
= (1 2 1)
T
:
(a) Reduce the matrix
±
a
b
c
d
²
in row echelon form.
(b) Could one decide by part (a) whether
a
;b
;c
;d
form a spanning set for
R
3
? How? What
else can you tell from the augmented matrix?
(c) Find a basis from
a
;b
;c
;d
for
R
3
.
(d) Do
a
;b
form a basis for
R
3
? If no, could you add suitable vector(s) other than
c
or
d
to get
a basis?
Ans
.
(a) The matrix is
0
@
1
2
±
1 1
1
0
1
2
±
1 1
±
2 1
1
A
. It can be reduced to the row echelon form
0
@
1 2
±
1
1
0 1
±
1 2
=
3
0 0
0
1
1
A
:
Remark. The reduced row echelon form is
0
@
1 2
±
1 0
0 1
±
1 0
0 0
0
1
1
A
:
(b) Yes.
a
;b
;c
;d
form a spanning set for
R
3
. Below are two ways to explain:
Explanation 1
. Let
y
=
0
@
y
1
y
2
y
3
1
A
2
R
3
. Consider
c
1
a
+
c
2
b
+
c
3
c
+
c
4
d
=
y
:
In matrix form, the augmented matrix is
0
@
1
2
±
1 1
1
0
1
2
±
1 1
±
2 1
y
1
y
2
y
3
1
A
;
which can be reduced to
0
@
1 2
±
1
1
0 1
±
1 2
=
3
0 0
0
1
²
²
²
1
A
:
(
²
denotes an entry whose value does not matter.) From the row echelon form, we see that
the system is always solvable for
c
1
;c
2
;c
3
;c
4
.
i.e.
R
3
= Span(
a
;b
;c
;d
).
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2
Explanation 2
. From the row echelon form we see that
a
;b
;d
are linearly independent, as
they are the vectors corresponding to the columns with leading 1’s. As dim
R
3
= 3, any
3 linearly independent vectors form a basis. (See Theorem 3.4.3 (I).) In particular,
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This note was uploaded on 05/04/2011 for the course MATH 1111 taught by Professor Dr,li during the Spring '10 term at HKU.
 Spring '10
 Dr,Li

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