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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
..
..
..