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Newberger Math 247 Spring 02
Solutions for Quiz 1.3 and 1.4
Let
b
=
2

1
6
u
1
=
1

2
0
u
2
=
0
1
2
u
3
=
5

6
8
.
1.
(12 points) Show that
b
is in
Span
{
u
1
,
u
2
,
u
3
}
. In other words
show that
b
is a linear combination of
u
1
,
u
2
and
u
3
.
We want to show we can ﬁnd numbers
x
1
,
x
2
and
x
3
, such that
x
1
u
1
+
x
2
u
2
+
x
3
u
3
=
b
.
To see this we reduce the following augmented matrix to echelon
form:
1 0
5
2

2 1

6
1
0 2
8
6
∼
1 0 5
2
0 1 4
3
0 2 8
6
∼
1 0 5
2
0 1 4
3
0 0 0
0
.
This system is consistent, so
b
is in
Span
{
u
1
,
u
2
,
u
3
}
.
2.
(4 points) Name three real numbers,
x
1
,
x
2
and
x
3
, such that
b
=
x
1
u
1
+
x
2
u
2
+
x
3
u
3
.
In (1), we showed the numbers
x
1
,
x
2
and
x
3
exist; in this problem
we are asked to ﬁnd such numbers. To solve for
x
1
,
x
2
and
x
3
, put
the matrix from (1) in reduced echelon form (as it turns out, in this
example it is already in reduced echelon form!
1 0 5
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This note was uploaded on 01/12/2010 for the course MATH 247 taught by Professor F,newberger during the Spring '03 term at Stanford.
 Spring '03
 F,newberger
 Math

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