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Unformatted text preview: Solutions to quiz # 5 (October 7)
1. Writing X =
13
31
a + 3c
b + 3d
⇐⇒
3a + c
3b + d a
c
a
c
=
=
=
= b
, we obtain
d
b
ab
13
a + 3c
=
⇐⇒
d
cd
31
3a + c
a + 3b
3a + b
b=c
⇐⇒
c + 3d
a = d.
3c + d b + 3d
a + 3b 3a + b
=
3b + d
c + 3d 3c + d Answer:
X= ab
,
ba where a and b can be any numbers.
2. To ﬁnd a basis of the kernel, we solve the system of linear equations: x1 x2 x3 x4
12 0
3 0
120 3 0
2
0
3  0
1 −→ 0 0 1 −1  0 −→ 0 0 1 −1  0 ,
1
2
1
20
0 0 −1 1  0
000 0 0
1
2 −1 4  0
from which the kernel consists of the vectors −2
−3
x1 = −2t2 − 3t1
t2
1 0 x2 =
, + t2 = t1 0
1
x3 =
t1
0
1
x4 =
t1
where t1 and t2 can be any numbers. Hence vectors −2
−3
1
0 and 0
1
0
1
constitute a basis of the kernel.
To ﬁnd a basis of the image, we pick the columns 1
0 1 and 1 1
−1
of the matrix corresponding to the leading variables x1 and x3 . −2 −3 01
4
Answer: The kernel is a plane in R with the basis and the image is
,
0
1 0
1 0
1
a plane in R3 with the basis 1 , 1 . 1
−1
1 ...
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This note was uploaded on 12/26/2011 for the course MATH 214 taught by Professor Conger during the Fall '08 term at University of Michigan.
 Fall '08
 Conger
 Linear Algebra, Algebra

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