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(Math113)[2010](s)midterm~912^_10453.pdf
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Midterm
24 March, 2010
Name: ID: L1A or L1B or L2A, L2B, or L3A, L3B
Directions: Do all four problems. You must show your work and justify your answers in order to receive credit. You have 1 hour, from 7pm to 8pm. No calculator will be allowed.
Problem 1 2 3 4 Total:
Scores
1. (20 pts.) Decide the following statements are true or false. Justify your answer.
(1) The following set of vectors are linearly independent. (10pts)
v1 =
,
v2 =
. .
.
,
v3 =
. .
.
,
v4 =
1
3
5
2
1
.3
0
0
0
7
3
2
9
3
0
.1
Solution: The statement above is false. The linear dependence rela-tion for v1, v2, v3 and v4 is: 0 . v1 +0 . v2 +1 . v3 +0 . v4 =0 Therefore v1,v2,v3,v4 are linearly dependent with weights (0, 0, 1, 0).
Decide the following statements are true or false. Justify your answer.
(2) If det A .to Rn
= 0, then the map from Rn de.ned by mapping x to Ax is one to one. (10pts)
Solution: The statement above is true. If the determinant of a matrix A is nonzero, A is a invertible matrix. By the inverse matrix theory, the map from x to Ax is one to one.
2. (25 pts.) Let BT denote the transpose of the matrix B. Compute the determinant of the matrix det(ABT ), where
A =
..
100
111
..
and B =
..
132
123
..
132
011
Solution: We have
det(ABT ) = det(A) det(BT ). and
det BT = det B Here by using the cofactor expansion across the .rst row for A.
.
11
det(A)=1 .
=2 . 3= .1
32
Using the cofactor expansion across the third row for B, then
.
1
2
.
1
3
det(B)= .1 .
+1 .
13 12
= .(3 . 2) + (2 . 3) = .2
Therefore,
det(ABT )=(.1) . (.2) = 2
3. (30 pts.) Find the inverse of matrix A by the row operations, where
1 .33 .1
0001
A =
. .
.
2 .67 0
1 .47 1
. .
.
Solution: We perform row operation on [ A | I5 ].
. .
.
. .
.
=.
. .
.
1
.33 .1
1000
0001
0100
2
.6
70
0010
.33 .1
1
1000
r1
.47 .1
1
0010
r4
[A|I5]=
.6
2
70
0001
r3
1 .47 1
0001
=.
..
. ..
.
=.
0001
0100
r2
. .
.
1 .33 .1
1 000
r1
r2 . r1
r3 . 2r1
.14 2
.1
0
001
010
.2
0012
0001
0 100
r4
1 .330
1 100
r1 + r4
r2 . 2r4
r3 . 2r4
.140
.1 .2
0
01
.2 .2
0 0 10
10
0 0 01
0 100
r4
. .
.
. .
.
.300
r1 . 3r3
r2 . 4r3
r3
.100
=.
0 1 00
r4
. .
.
1 0 00
.14 .11 9 .3
0 .100
76 .41
0 0 10
.2 .21 0
r1 . 3r2
r2
r3
=.
0 0 01
0 100
r4
=.
. .
.
. .
.
1000
.14 .11 9 .3
0100
.7 .64 .1
0010
r1
.r2
r3
r4
Therefore, A.1 is equal to
. .
.
.14 .11 9 .3
.7 .64 .1
.2 .21 0
0 100
. .
.
A.1
=
. .
.
4. (25 pts.) Consider the following linear system:
x1 + x2 . x3 . 2x4 =1 x1 +2x2 +2x3 . x4 =0 .2x1 . 3x2 + kx3 +3x4 = .1
(i)
Find the value of k such that the linear system is con