(MATH2111)[2010](s)midterm~=gl0cn3^_35636.pdf downloaded by tnsitou from http://petergao.net/ustpastpaper/down.php?course=MATH2111&id=0 at 2022-12-05 11:23:53. Academic use within HKUST only.
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
Scores
1
2
3
4 Total:
(MATH2111)[2010](s)midterm~=gl0cn3^_35636.pdf downloaded by tnsitou from http://petergao.net/ustpastpaper/down.php?course=MATH2111&id=0 at 2022-12-05 11:23:53. Academic use within HKUST only.
1. (20 pts.) Decide the following statements are true or false. Justify your
answer.
(1) The following set of vectors are linearly independent. (10pts)



1
2
 3 
 1


v1 = 
 5  , v2 =  −3
9
3




0
7

 0 
 3
 , v3 =   , v4 = 

 0 
 2
0
−1




Solution: The statement above is false. The linear dependence relation 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).
(MATH2111)[2010](s)midterm~=gl0cn3^_35636.pdf downloaded by tnsitou from http://petergao.net/ustpastpaper/down.php?course=MATH2111&id=0 at 2022-12-05 11:23:53. Academic use within HKUST only.
Decide the following statements are true or false. Justify your answer.
(2) If det A 6= 0, then the map from Rn to Rn defined 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.
(MATH2111)[2010](s)midterm~=gl0cn3^_35636.pdf downloaded by tnsitou from http://petergao.net/ustpastpaper/down.php?course=MATH2111&id=0 at 2022-12-05 11:23:53. Academic use within HKUST only.
2. (25 pts.) Let B T denote the transpose of the matrix B. Compute the
determinant of the matrix det(AB T ), where




1 0 0
1 3 2
A =  1 1 1  and B =  1 2 3 
1 3 2
0 1 1
Solution: We have
det(AB T ) = det(A) det(B T ).
and
det B T = det B
Here by using the cofactor expansion across the first row for A.
¯
¯ 1 1
det(A) = 1 ∗ ¯¯
3 2
¯
¯
¯ = 2 − 3 = −1
¯
Using the cofactor expansion across the third row for B, then
¯
¯ 1 2
det(B) = −1 ∗ ¯¯
1 3
¯
¯
¯
¯
¯+1∗¯ 1 3
¯
¯ 1 2
¯
¯
¯ = −(3 − 2) + (2 − 3) = −2
¯
Therefore,
det(AB T ) = (−1) ∗ (−2) = 2
(MATH2111)[2010](s)midterm~=gl0cn3^_35636.pdf downloaded by tnsitou from http://petergao.net/ustpastpaper/down.php?course=MATH2111&id=0 at 2022-12-05 11:23:53. Academic use within HKUST only.
3. (30 pts.) Find the inverse of matrix A by the row operations, where


1 −3 3 −1
 0 0 0 1 

A=
 2 −6 7 0 
1 −4 7 1
Solution: We perform row operation on [ A | I5 ].

1 −3 3 −1 1 0 0
 0 0 0 1 0 1 0
[A|I5 ] = 
 2 −6 7 0 0 0 1
1 −4 7 1 0 0 0

1 −3 3
 0 −1 4
=⇒ 
 0 0 1
0 0 0

1 −3 3
 0 −1 4
=⇒ 
 0 0 1
0 0 0

1 −3 0
 0 −1 0
=⇒ 
 0 0 1
0 0 0

1 0 0 0
 0 −1 0 0
=⇒ 
 0 0 1 0
0 0 0 1

1 0 0
 0 1 0
=⇒ 
 0 0 1
0 0 0


1 −3 3 −1 1 0 0
0
 1 −4 7 −1 0 0 1
0 
 =⇒ 
 2 −6 7 0 0 0 0
0 
1
0 0 0 1 0 1 0

0
r1

1  r2 − r1
0  r3 − 2r1
0
r4

1 0 0
r1 + r4
0 1
0 −1 −2 0 1 
r
 2 − 2r4
0 −2 −2 1 0  r3 − 2r4
1 0
1 0 0
r4

0 7
7 −3 0
r1 − 3r3

0 7
6 −4 1  r2 − 4r3
0 −2 −2 1 0 
r3
1 0
1
0 0
r4

−14 −11 9 −3
r1 − 3r2
7
6 −4 1 
r2


−2 −2 1
0
r3
0
1
0
0
r4

r1
0 −14 −11 9 −3
−r
0 −7 −6 4 −1 
2

0 −2 −2 1 0  r3
r4
1 0 0
1 0
−1 1 0 0
2 −1 0 0
2 −2 0 1
0 1 0
1
Therefore, A−1 is equal to

A−1

−14 −11 9 −3
 −7 −6 4 −1 

=
 −2 −2 1 0 
0
1 0 0

r1
0
0 
 r4
1  r3
0
r2
(MATH2111)[2010](s)midterm~=gl0cn3^_35636.pdf downloaded by tnsitou from http://petergao.net/ustpastpaper/down.php?course=MATH2111&id=0 at 2022-12-05 11:23:53. Academic use within HKUST only.
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 consistent. (20
pts.)
(ii) When the system is consistent, point out the basic variables and
the free variables. (5 pts.)
Solution: The augmented matrix of the linear system is:


1
1 −1 −2 1
 1
2
2 −1 0 
−2 −3 k
3 −1
Performing the elementary row operation on the augmented matrix, we
get




1
1 −1 −2 1
1 1
−1 −2 1
R1
 1



2
2 −1 0
3
1 −1
R2 − R1
=⇒ 0 1
−2 −3 k
3 −1
0 −1 k − 2 −1 1
R3 + 2R1


1 1 −1 −2 1
R1


3
1 −1
R2
=⇒ 0 1
0 0 k+1 0
0
R3 + R2
For the system to be consistent, which means the linear system has at
least one solution, we should not have any row vector like
£
¤
0 0 0 0 b
in the echelon form of the augmented matrix, where b 6= 0.
The third row of the echelon form is:
£
¤
0 0 k+1 0 0
The constant coefficient is zero. Therefore, for any value of k ∈ R, the
linear system is consistent.
(2). When k + 1 = 0, namely k = −1, x1 , x2 are basic variable and
x3 , x4 are free variables.
When k 6= −1, then x1 , x2 , x3 are basic variables and x4 is the free
variable.
(MATH2111)[2010](s)midterm~=gl0cn3^_35636.pdf downloaded by tnsitou from http://petergao.net/ustpastpaper/down.php?course=MATH2111&id=0 at 2022-12-05 11:23:53. Academic use within HKUST only.
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