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(MATH113)[2011](s)makeUpMidterm~3938^_39464.pdf
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Math113 Introduction to Linear Algebra
Spring 2011
C Make up Examination for Midterm(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 v1
=


. ..

.1., v2

..
=


. ..

1
., v3

..
. ..

..
.1
..
1

1


Let T : R3 R3 be the linear transformation such that
1

1

=

0

0

1

1


2

T (v1)=
. ..


.3. ,
..
T (v2)=
. ..

. ,
.4
..

T (v3)=
. ..

. ..


.

1


.4

2


1

(a)
Find the standard matrix A such that T (x)= Ax.

(b)
Is the transformation x T (x) one to one? Is T onto? State your reason.


[25 points]
Solution: Let e1
=


. ..

., e2
0

..
. ..

. , e3
0

..
=


. ..

1

..
1

0

0

. ..

. be the standard basis of R3
.

Then we have

0

1

0

=

. ..


1

1

T (e1)= (T(v1)+ T(v2)) =
.1

2


1
0

2

. ..

. ..

1
T (e2)= (.T(v1)+ T(v2)) =
2

.3

. ..

.1

5
.5

. ..

T (e3)= T(v2) . T(v3)) =
Therefore, the standard matrix of T is equal to

. ..

. ..

10 .1

.12 5
.1 .3 .5

A =[T (e1), T(e2), T(e3)] =
The matrix A is row equivalent to the matrix

. ..

..
10 .1
..

There are two pivots for A. Therefore,

02 4

00 0


T is not one to one and not onto by the invertible matrix theorem.

2. Let

A =

. ..

123

.1 .3 .1

3 h 1


. ..

(a)
Compute the determinant of A.

(b)
For what value of h, A is invertible?


[25 points] Solution: The determinant of A is equal to 20 . 2h. Since A is invertible if and only if det(A) = 0. When .
h .
= 10, A is not invertible.
3. (a) Find the inverse of A if the inverse exist:
A =


. ..

1 11

.101

0 11


. ..

.

(b) Whats the inverse of A2?
[25 points]
Solution: The inverse of A is equal to
A.1
=

. ..

10 .1

.1 .12

11 .1


.

Since A . A.1 = I, A . A . A.1 . A.1 = I. The inverse of A2 is equal to:
A.2 = A.1 . A.1
=

. ..


0 .10

23 .3

.1 .2 .2

.


4. Let v1
=


. ..

1
., v2

..
=


. ..

1
., v3

..
=


. ..

., and v4
.1

..
=


. ..

..
.1

..
0

1


Let A =[v1, v2, v3, v4].
0

1

1

2

1

0

3

. ..

..
.1
..

(a) Let v =

Find the solutions for the matrix equation Ax = v.

4

..
b3

[25 points] Solution: Performing row operation on the matrix [A v], the echelon form of the matrix is
..
. ..

b1

(b) Let b =

Determine the condition on b1,b2,b3 such that b lies in Span{v1, v2, v3, v4}.
b2
. ..

. ..

10 .1 .1 .1
0112 4
0000 0


The set of solution is

..

..

..