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(MATH113)midterm2000F_v2_sol.pdf
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Math 113, L4, Soltions of Midterm Eam, Fall 2000

Date: 30 October 2000 Time: 7:00p.m.| 8:00p.m. Venue: LTC

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).
nm
(1) Any linear transformation T : R ! R is not one-to-one if m . n.
p
True ( ). False ( ).
~
(2) Let A be an n . n matrix without zero rows and columns. If the equation A~x . b has

n
~
a solution for a nonzero vector b 2 R , then the solution is unique.
p
True ( ). False ( ). Counter Example:
. . .. ....
112 21
~
A. ;b.;~x.; :

112 01
~
(3) Let A be an m . n matrix. If the equation A~x . b has a unique solution for a nonzero

m~
~
vector b 2 R , then the equation A~x . 0 has only the trivial solution.
p
True ( ). False ( ).
n
~
(4) If a subset f~v; :::; ; ~vg of R is linearly dep endent and ~v6. 0, then there ex-
1p1
ists an index j 2 f2; :::; pg such that f~v; :::; ~vg is linearly indep endent and ~v2

1j.1j
Spanf~v; :::; ~vg.
1j.1
p
True ( ). False ( ).
3
(5) Any plane in R is spanned by two vectors, that is, Span f~v; ~vg.
12
p
True ( ). False ( ). Counter Example:
823 9

x
..

45
y j x;y 2R :
:;

1
(6) Iftwo n.n matricesA6.0 andB.60, then AB6. 0.
p
True ( ). False ( ). Counter Example:
.. ..
10 00
A. ;B.:

00 01
1

nn

(7) If T is a linear transformation from R to R with the standard matrix A, then T is

n
onto R if and only if the columns of A are linearly indep endent.
p
True ( ). False ( ).
~
(8) Let A be an n . n matrix. If the equation A~x . b has a solution for a given nonzero

~
vector b, then jAj 6. 0.
p
True ( ). False ( ). Counter Example
.... ..
112 1
~
A. ;b. ;~x. :

112 1
(9) For any two matrices A and B, jABj . jBAj if both AB and BA are de.ned.
p
True ( ). False ( ). Counter Example:
..
1
A. ; B.[1 1]:

1
..
..
11
..
jABj. .0; jBAj. det [2].2:

..
11
(10) Find the determinants of the following matrices (1 point for each)
23
23
34 5 67
23
0001
67
15 1000 89101112
6 767
0020
4 56 767
A.02 100;B. 105431;C. :

6 745
0300
45
00 4 105431
4000
23 4 57
Solution.
jAj.8; jBj.0; jCj. 24:

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

(20pts)
8
x.x+x. 1
.
234

x+2x.x+x. 1

1 234
:
2x+x+x.x..1

1 234
Solution. The augmented matrix is
2 32 3

01.111 12.111

4 54 5

12.1 1 1.01.1 1 1

21 1.1.1 21 1.1.1

2 323

1 2.1 1 1 10 1.1.1

4 545

.0 1.1 1 1.01.1 1 1:

0.33.3.3 00000

2

So the system is equivalent to

.

x+x.x..1
1 34

x.x+x. 1
234
Take xand xas free variables. Letting x. x. 0, we get a special solution

34 34
3

2

.1

1

664

775

:

0
0
The corresp onding homogeneous system is

.

x+x.x.0
1 3