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Math 113, L5, Solutions of Midterm Exam, Fall 2000
Date:

30 October 2000

Time:

3:55p
...
| 4:55p
...


Venue:

RM2465

Name
Student Number
Tutorial Section
Score
1
...
Provide a counter example for each false question (3
points for each true question and 5 points for each false question)
...

p
True ( )
...

(2) Let A be a matrix
...

p
True ( )
...
Counter Example:
1
A = 1 1]; ~ = ?1 :
x
(3) If the sets f~1;~2 g; f~1;~3 g and f~2;~3 g are linearly independent, then the set f~1;~2 ;~3g
v v v v
v v
v v v
are linearly independent
...
False ( )
...
The matrix A is invertible if and only if AT is invertible
...
False ( )
...
If jAB j 6= 0, then both A and B are invertible
...
False ( )
...

p
True ( )
...
Counter Example:
Spanf~ g = f~ g; a point:
0
0
(7) For two matrices A and B , if AB = I , then A is invertible
...
False ( )
...

(8) If A is an n n matrix and the equation A~ = ~ has nontrivial solution, then jAj = 0
...
False ( )
...

p
True ( )
...
Counter Example:
A = 1 0 ; B = 0 0 ; jAj = jB j = 0; jA + B j = 1:
0 0
0 1
(10) Find A2000 if
Solution
...
Find the general solution in vector form of the following system of linear equations
(20pts)
8
x3 +x4 = 1
<
x1 ?x2 +x3
= 3
:
x1 ?x2 +2x3 +x4 = 4
Solution
...
Letting x2 = x4 = 0, we get a special solution
2
3
2
6 0 7
6
7:
4 1 5
0
The corresponding homogeneous system is
x1 ?x2

?x4 = 0

x3 +x4 = 0
Letting x2 = 1 and x4 = 0, we obtain a solution vector
2
3
1
6 1 7
6
7:
4 0 5
0
Letting x2 = 0 and x4 = 1, we obtain another solution vector
3
2
1
6
0 7:
7
6
4 ?1 5
1
Thus the general solution of the homogeneous system is
3
3
2
2
1
1
7
7
6
6
x2 6 1 7 + x4 6 ?0 7 :
4
4 0 5
15
1
0
Therefore, the general solution of the original system is
3
2
3
3
2
2
2
1
1
6
7
6
07 6 7
x2 6 1 7 + x4 6 ?1 7 + 6 0 7 :
5
4 1 5
4
4 0 5
0
1
0
3
...


By elementary row operations, we have
2
3
2
3
1 ?1 0 1 0 0
1 ?1 0 1 0 0
A I3 ] = 4 2 ?1 1 0 1 0 5 4 0 1 1 ?2 1 0 5
2 0 1 0 0 1
0 2 1 ?2 0 1
3

2
4

Thus

1 0 1 ?1 1
0 1 1 ?2 1
0 0 ?1 2 ?2
2
1 0
4 0 1
0 0

0
1 0 0 1 ?1 1
0 5 4 0 1 0 0 ?1 1
1
0 0 ?1 2 ?2 1
3
0 1 ?1 1
0 0 ?1 1 5
1 ?2 2 ?1
3

2

3
5

1 ?1 1
?1 = 4 0 ?1 1 5 :
A
?2 2 ?1
2

3

4
...


By elementary row operations, we have
4 ?2 1 ?3
8 2 1 ?3
4 1 1 ?2
8 ?1 1 ?3

4 ?2 1 ?3
6 ?1 3
= 0 6 ?1 3 = 4 3 0 1
0 3 0 1
3 ?1 3
0 3 ?1 3
6 ?1 3
3
= 4 3 0 1 = 4 ?3 1 = 12:
0
?3 0 0

4



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