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(MATH113)[2011](s)midterm~3938^_87308.pdf
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Math113 Introduction to Linear Algebra
Spring 2011
C Midterm Examination (All Sections) C

Name:
Student ID:

Lecture Section:

.
There are FOUR questions in this midterm examination.

.
Answer all the questions.

.
You may write on both sides of the paper if necessary.

.
You may use a HKEA approved calculator. Calculators with symbolic calculus functions are not allowed.

.
The full mark is 100.


1. Let T : R3 . R3 be the linear transformation de.ned by
T

. ..

. ..

=

. ..

2x1 . x2 +2x3
4x1 + x2 + x3
.6x1 +2x2 . 5x3
. ..

.

x1
x2
x3
(a) Find the set of vectors that maps to zero vector under T . . lie in the image of T ? If yes, .nd the set of vectors that maps
..
. ..

.1

(b) Does the vector v =

1
2

to v. [25 points]
Solution:
(a) The standard matrix de.ned by the transformation T is :
A =

. ..

. ..

2 .12

411

.62 .5

..
Row reduction on [A 0] yields:
.
..
.
The set of vectors that maps to the zero vector under T is
. ..

. ..

. ..

..
.
.2
... ..
. ,x3 any real number
. ..

2 .1 20

2 .1 20

2 .1 20

20 10

4 1 10

03 .30

01 .10

01 .10

.62 .50

0 .1 10

0 0 00

00 00

. ..

1

x3
.

2...
.
..
(b) Row reduction on [A v] yields:
.
..
....
.
Take x3 to be .2, a special solution to Ax = v is
. + x3 . ,x3 any real number
. ..

. ..

. ..

. ..

. ..

2 .12 .1

2 .12 .1

2 .12 .1

20 10

01 .11

00 00

4111

03 .33

01 .11

.62 .52

0 .11 .1

..
0000
..
. ..

1

.1

.2

. ..

. ..

1

1

Therefore the set of vectors
maps to the vector

.1

.2

.2

v under T .

2. (a) Compute the determinant of the following matrix A:
..

2 .61 1
5 .3 .2 .2

..

A =

.

00 0 2
31 .27

.

(b) Whats the determinant of the matrix 2A?
[25 points]
Solution:
(a) Using the cofactor expansion on the third row, we get
2 .61 1

.
2 .61

5 .3 .2 .2
det(A)= = .25 .3 .2
000 2

31 .2

31 .27

The determinant of the submatrix

2 .61

.3 .2

1 .2

+6

5 .2

3 .2

5 .3

31

= 16 . 24+14 = 6

.
=2

5 .3 .2

+1

31 .2
Therefore, det(A)= .12.

(b) Each row of the matrix A is multiplied by 2 to get 2A, det 2A =24 . (.12) = .192.
3. (a) Find the inverse of matrix A if the inverse exists:
A =

. ..

. ..

.

1 .11


2 .53

3 .43

(b) If e2
=

. ..

., .nd the solution to the matrix equation Ax = e2.
0
[25 points]
..
0

1

Solution:
Performing row operation on the matrix [ A | I3 ]
. ..

. ..

. ..

. ..

1 .1 1 1 0 0
2 .5 3 0 1 0
3 .4 3 0 0 1

1 .1 1 1 0 0
0 .3 1 .2 1 0
0 .1 0 .3 0 1

r4.3r1
.


r2..r3
.


scalings

.2r1+r2
. ..

..

. .

3r2+r3
. ..

... ..
., A.1e2 is the second
1 .11
100
30 .1
1 .11
10 0
010
30