Math 20F
First Midterm Solutions
January 27,2003
The two versions are nearly the same and problems 1 and 3 have been interchanged.
The solutions here are for version A with notes on changes for B.
1. (#3 in version B) (a)
AB
=
BA
=
I
. I mentioned in class that
AB
=
I
or
BA
=
I
is
sufficient, so either “
AB
=
I
” and “
BA
=
I
” are also acceptable.
(b) transpose
2. There are many ways to convert a matrix to row echelon form. I’ll choose one way.
Rn means row n. For version A:
1
0

1
0
0
0
0
1
1
1
0

1
2
2
0
3
fl
fl
fl
fl
fl
fl
fl
1
0
0
0
0
0
0
1
→
1
0

1
0
1
1
0

1
0
0
0
1
0
0
0
5
fl
fl
fl
fl
fl
fl
fl
1
0
0
0
0
0
0
1
add (

2)
×
(R3) to R4
then switch R2 and R3
→
1
0

1
0
0
1
1

1
0
0
0
1
0
0
0
0
fl
fl
fl
fl
fl
fl
fl
1
0

1
0
0
0
0
1
subtract R1 from R2
add (

5)
×
(R3) to R4
Thus (a) has solutions; e.g., (1

1 0 0)
T
; however (b) does not have a solution because
the last row of the row echelon form augmented matrix is inconsistent.
For version B:
1
1

1
0
2
2
3
0
1
0
0

1
0
0
1
0
fl
fl
fl
fl
fl
fl
fl
0
0
1
0
0
1
0
0
→
1
1

1
0
0
0
5
0
0

1
1

1
0
0
1
0
fl
fl
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 Spring '03
 BUSS
 Math, Jaguar Racing, Row echelon form, AAT

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