Math
2940
Solutions, Fall
2011
Section
1
.
3
3
) The equations are
y
1
=
B
1
,
y
1
+
y
2
=
B
2
,
y
1
+
y
2
+
y
3
=
B
3
so substituting, we see
y
2
=
B
2

B
1
and
y
3
=
B
3

y
1

y
2
=
B
3

B
1

(
B
2

B
1
) =
B
3

B
2
.
So
y
1
y
2
y
3
=
1
0
0

1
1
0
0

1
1
B
1
B
2
B
3
.
4
) We have to solve
x
1
w
1
+
x
2
w
2
+
x
3
w
3
= 0. The equations are
x
1
+ 4
x
2
+ 7
x
3
= 0
,
2
x
1
+ 5
x
2
+ 8
x
3
= 0
,
3
x
1
+ 6
x
2
+ 9
x
3
= 0
.
We eliminate the varible
x
1
from the last trwo equations by adding

2 and

3 times the
first equation to these two to get

3
x
2

6
x
3
= 0
,

6
x
2

12
x
3
= 0
.
These last two equations are multiples of one another, so the solutions to one are exactly the
solutions to the other. We’ll take
x
2
= 2 and
x
3
=

1. Plugging back into the first equation
gives
x
1
=

1.
The three vectors lie in a
plane
.
6
) Clearly the first two columns of
1
3
5
1
2
4
1
1
c
are independent.
We need to consider
x
1
1
1
1
+
x
2
3
2
1
=
5
4
c
.
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 '05
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 Linear Algebra, Algebra, Equations, X1, British B class submarine, B type proanthocyanidin, ﬁrst equation

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