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(MATH113)midterm2000F_sol.pdf
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Math 113, L5, Soltions of Midterm Eam, Fall 2000

Date: 30 October 2000 Time: 3:55p.m.| 4:55p.m. Venue: RM2465

Name

Student Number
Tutorial Section
Score

1. Answer the following questions. Provide a counter example for each false question (3

points for each true question and 5 points for each false question).
mn
(1) Any linear transformation T : R ! R is not onto if m . n.
p
True ( ). False ( ).
~

(2) Let A be a matrix. The equation A~x . 0 has only the trivial solution if A has a pivot

position on each row.
p
True ( ). False ( ). Counter Example:
..
1
A.[1 1]; ~x. :

.1
(3) If the sets f~v; ~vg; f~v; ~vg and f~v; ~vg are linearly indep endent, then the set f~v; ~v; ~vg

121323123
are linearly indep endent.
p
True ( ). False ( ). Counter Example:
23 23 23
110
454545 ~

~v. .1 ;~v. 0 ;~v. 1 ;~v+~v+~v.0:

1 2 3 123
0 .1 .1
T
(4) Let A be n . n matrix. The matrix A is invertible if and only if A is invertible.
p
True ( ). False ( ).
(5) Let A and B be two n . n matrices. If jABj .60, then both A and B are invertible.

p
True ( ). False ( ).
3
(6) The span of any vector in R is a line.
p
True ( ). False ( ). Counter Example:
~~
Spanf0g . f0g; a point:

(7) For two matrices A and B, if AB . I, then A is invertible.
p
True ( ). False ( ). Counter Example:
23
..
10
100
45
A. ;B. 01:

010
00
A and B are not square matrices.
~
(8) If A is an n . n matrix and the equation A~x . 0 has nontrivial solution, then jAj . 0.

p
True ( ). False ( ).
(9)For two n.n matricesA andB,jAj.0 andjBj.0 implyjA+Bj. 0.
p
True ( ). False ( ). Counter Example:
.. ..
10 00
A. ; B. ; jAj.jBj.0; jA+Bj.1:

00 01
2000
(10) Find A if
..
10
A . : (3pts)
.2 1
Solution. By induction, we can prove
..
10
n
A. :

.2n 1
Thus
..
10
2000
A. :

.4000 1
2. Find the general solution in vector form of the following system of linear equations

(20pts)
8
x+x.1
.
34
x.x+x.3

12 3
:
x.x+2x+x.4

12 34
Solution. The augmented matrix is
2 32 3

0 0111 1.1103

4 54 5

1.1103 . 0 0111

1.1214 1.1214

232 3

1.1103 1.10.12

454 5

.0 0111.0 01 11:

0 0111 0 00 00

So the system is equivalent to
.
x.x.x.2

12 4
x+x.1
34
Take xand xas free variables. Letting x. x. 0, we get a special solution

24 24
3

2

2

0

664

775

:

1
0
The corresp onding homogeneous system is

.

x.x.x.0
12 4
x+x.0
34
Letting x. 1 and x. 0, we obtain a solution vector

24
3

2

1

1

664

775

:

0
0
Letting x. 0 and x. 1, we obtain another solution vector

24
3

2

1

0

664

775

:

.1
1
Thus the general solution of the homogeneous system is

3

2

3

2

11

x

2

664

1

775

+x

4

664

0

775

:

0 .1
01
Therefore, the general solution of the original system is

3

2

3

2

3

2

1 12

x

2