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Solution for prelim 1
Math 294 spring, 2005
1a.
This is a hw problem, see hw solution for problem (17) in section 2.2.
1b.
Since A is a reflection, the inverse of A
is itself, that is,
1
AA
−
=
(Recall HW problem
41 in section 2.3).
2a.
This is also a hw problem, see hw solution for problem (19) in section 2.3.
2b.
It is not invertible.
To see this, note that if A is invertible if and only if the
RREF A is the identity matrix.
However, if two columns of A are identical, then,
the RREF A cannot be the identity matrix since the two columns of A that are
identical will be transformed, after row reduction, to two columns that are identical.
(Recall that the columns of an identity matrix are all different).
3
Let
1
2
3
4
x
x
x
x
x
=
G
be any vector that is perpendicular to
12
11
02
,
13
04
vv
==
GG
.
The
conditions
0
xv xv
•=•=
GG GG
is equivalent to the linear systems
1234
0
2340
xx
xxxx
+
=
+
++
=
The homogeneous solution of this system is, using row operations,
10100
12340
02240
01120
A
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This note was uploaded on 04/04/2008 for the course MATH 2940 taught by Professor Hui during the Spring '05 term at Cornell University (Engineering School).
 Spring '05
 HUI
 Math, Linear Algebra, Algebra

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