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Dawson College
Department of Mathematics
Final Examination
Linear Algebra
201-NYC-05-Science
Winter 2011
Student Name:โโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
Student I.D#:โโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
Teachers : K.Ameur, T.Fox, I.Gombos, N.Sabetghadam Haghighi, B.Szczepara
Date: May 16, 2011
Time: 9:30 AM to 12:30 PM
Instructions:
- Translation and regular dictionaries are permitted.
- Scientific non-programmable calculators are permitted.
- Print your name and ID in the provided space.
- This examination booklet must be returned intact.
This examination consists of 12 questions. Please ensure that you have a
complete examination before starting.
- (8 points) (a) Solve the following system using Gauss Jordan elimination.
x 1 + x 2 + 2x 3 โ x 4 = 4
3 x 2 โ x 3 + 4x 4 = 2
x 1 + 2x 2 โ 3 x 3 + 5x 4 = 0
2 x 1 + 2x 2 โ 3 x 3 + 5x 4 = 1
(b) Give a particular solution that statisfies x 3 =
3 2
Solution: (a) (5 points)
โR 1 + R 3
โ 2 R 1 + R 4
R 2 โ R 3
โ 3 R 2 + R 3
1 14
R 3
7 R 3 + R 4
โ 2 R 2 + R 1
5 R 3 + R 2
โR 2 + R 1
Hence (^) ๏ฃฑ
๏ฃด๏ฃด ๏ฃด๏ฃด ๏ฃด ๏ฃด๏ฃฒ
x 1 = 1
x 2 = โt + 1 t โ R
x 3 = t + 1
x 4 = t
(b) (3 points) x 3 =
3 2
โโ t + 1 =
3 2
โโ t =
1 2
, then
x 1 = 1
x 2 =
1 2
x 3 =
3 2
x 4 =
1 2
- (8 points) Let A =
, X =
x
y
z
, B =
(a) Find A
โ 1
(b) Use your answer in (a) to solve the system AX = B
Solution:
(a) (5 points ) ๏ฃฎ
Hence A
โ 1
(b) (3 points)
x
y
z
= A
โ 1 B =
- (8 points) Find a matrix A in each of the following:
(a)
A
T
(b)(I โ (2A)
โ 1 )
โ 1
๏ฃป A.
Solution:
(a) (4 points)
A
T
โ 1
Then A =
(b) (4 points) (I โ (2A)
โ 1 )
โ 1
๏ฃป A
I โ
A
โ 1 = A
โ 1
โ 1
A
โ 1
I =
A
โ 1
A
โ 1
A
โ 1
I +
A =
I +
3 2
- (10 points) Given the points A(3, 1 , โ2), B(5, 2 , โ1), C(4, 3 , โ3)
(a) Find the area of the triangle ABC.
(b) Find the angle at the vertex A.
(c) Find the equation of the plane that contains the triangle ABC.
Solution:
(a) (4 points)
AB = (2, 1 , 1)
AC = (1, 2 , โ1)
AB ร
AC =
~i ~j ~k
= โ 3 ~i + 3~j + 3~k
Area =
1 2 ||
AB ร
AC|| =
(b) (3 points) cos ฮธ =
AB ยท
AC
AB||||
AC||
โ ฮธ =
ฯ
(c) (3 points) Let N be the normal of this plane. N =
AB ร
AC = (โ 3 , 3 , 3)
โ3(x โ 3) + 3(y โ 1) + 3(z + 2) = 0
(x โ 3) โ (y โ 1) โ (z + 2) = 0
x โ y โ z โ 4 = 0
- (9 points) Consider the vectors U = (0, 1 , โ1), V = (1, 0 , 2), W = (1, 1 , 0) ( Equivalently
U = ~j โ ~k, V = ~i + 2~k , W = ~i + ~j).
(a) Find a vector of length 6 orthogonal to both U and V.
(b) Find the volume of the parallelepiped determined by 2U , V , and 3W.
(c) Find the orthogonal projection of W on V โ 2 U.
Solution:
(a) (3 points) Z =
6(U ร V )
||U ร V ||
U ร V =
~i ~j ~k
= 2~i โ ~j โ ~k
Z =
โ^6 6
(b) (3 points ) Volume =| 2 U ยท (V ร 3 W )| = 6|U ยท (V ร W )|
U ยท (V ร W ) =
Volume =
(b) (3 points) proj
W V โ 2 U =^
W ยท (V โ 2 U )
||V โ 2 U ||^2
V โ 2 U
V โ 2 U = (1, โ 2 , 4) W ยท (V โ 2 U ) = โ 1
proj
W V โ 2 U =^
- (9 points) Consider the line L 1 :
x = 1 + t
y = 2 + t
z = 1 โ t
, and the point A(2, 3 , โ6)
(a) Find the point on L 1 , which is closest to the point A.
(b) Find the distance between the line L 1 , and the point A.
(c) Find the equation of the plane that contains the line L 1 , and the point B(1, 2 , 2).
Solution:
(a) (4 points) Let Q be the point on L 1 closest to A, Then
QA โฅ L 1
QA ยท V = (1, 1 , โ1) = 0
โ (1 โ t, 1 โ t, โ7 + t) ยท (1, 1 , โ1) = 0
โ 1 โ t + 1 โ t + 7 โ t = 0
โ 9 โ 3 t = 0
โ t = 3
โ Q(4, 5 , โ2)
(b) (2 points) D = ||
AQ|| =
22 + 2^2 + 4^2 = 2
(c) (3 points) Let C(1, 2 , 1) be a point in L 1. Let N be the normal of the plane that contains
L 1 , and the point B(1,2,3).
N =
CB ร V =
~i ~j ~k
= โ~i + ~j
โ(x โ 1) + (y โ 2) = 0
x โ y + 1 = 0
- (6 points) Show that the lines L 1 :
x = 1 + t
y = 2 + t
z = 1 โ t
, and L 2 :
x = s
y = โ1 + s
z = 2 + s
, are skew lines,
and find the distance between them.
Solution:
(3 points ) Let V 1 = (1, 1 , โ1) V 2 = (1, 1 , 1)
The lines L 1 and L 2 are not parallel since their direction vectors V 1 and V 2 are not scalar
multiple of each other, and they do not intersect since the system : ๏ฃฑ ๏ฃด๏ฃด ๏ฃด ๏ฃฒ
1 + t = s
2 + t = โ1 + s
1 โ t = 2 + s
t โ s = โ 1
t โ s = โ 3
โt โ s = 1
, is inconsistent. Hence L 1 and L 2 are skew lines.
(3 points ) Let P (1, 2 , 1) โ L 1 , Q(0, โ 1 , 2) โ L 2
D =
P Q ยท (V 1 ร V 2 )|
||V 1 ร V 2 ||
V 1 ร V 2 =
~i ~j ~k
= 2~i โ 2 ~j
P Q = (โ 1 , โ 3 , 1)
D =
22 + (โ2)^2
- (5 points) Find the basis and dimension of the solution space of the system
๏ฃฑ ๏ฃฒ
x 1 + 4x 2 โ 3 x 3 + x 4 โ 4 x 5 = 0
3 x 1 + 10x 2 โ x 3 โ 5 x 4 + 2x 5 = 0
Solution:
(2 points )Let W be the solution set.
โ 3 R 1 + R 2
1 2
R 2
โ 4 R 2 + R 1
x 1 = โ 13 t + 15s โ 24 r
x 2 = 4t โ 4 s + 7r
x 3 = t t, s, r โ R
x 4 = s
x 5 = r
(2 points)
(โ 13 t + 15s โ 24 r, 4 t โ 4 s + 7r, t, s, r) = (13t, 4 t, t, 0 , 0) + (15s, โ 4 s, 0 , s, 0) + (โ 24 r, 7 r, 0 , 0 , r)
= t(โ 13 , 4 , 1 , 0 , 0) + s(15, โ 4 , 0 , 1 , 0) + r(โ 24 , 7 , 0 , 0 , 1)
Then the vectors V 1 = (โ 13 , 4 , 1 , 0 , 0), V 2 = (15, โ 4 , 0 , 1 , 0), V 3 = (โ 24 , 7 , 0 , 0 , 1) span W.
(1 point ) k 1 V 1 + k 2 V 2 + k 3 V 3 = (0, 0 , 0 , 0 , 0)
โ (โ 13 k 1 + 15k 2 โ 24 k 3 , 4 k 1 โ 4 k 2 + 7k 3 , k 1 , k 2 , k 3 ) = (0, 0 , 0 , 0 , 0) โ k 1 = 0,k 2 = 0, k 3 = 0
โ The vectors V 1 , V 2 , V 3 are linearly independent. Hence they form a basis for the solution
space W , and dim(W ) = 3
