mat1503 exam solutions
TRANSCRIPT
-
8/10/2019 MAT1503 Exam Solutions
1/93
May/June 2011 Examination Paper
Question 1
Consider the following system of linear equations
x1 + x2 x3 = 23x1 + 2x2 x3 = 3x1 x2 + 2x3 = 1
Write down the augmented matrix, reduce the augmented matrix to generalizedrow echelon form and then determine the solution of system
solution :
Augmented Matrix:
1 1 1 23 2 1 31 1 2 1
Apply the elimination row operations
3R1+ R2 > R2
R1+ R3 > R3
we get
1 1 1 20 1 2 3
0 0 1 1
The matrix is in generalized row echolen form so the solution is
x3 = 1,
x2+ 2x3= 3,
x1+ x2 x3 = 2
1
-
8/10/2019 MAT1503 Exam Solutions
2/93
thus x2= 3 + 2x3 = 5,
x1 = x2+ x3+ 2 = 2
Question 2
Consider system as in Question 1
2.1 Write down matrices A and b
2.2 What are the respective sizes of matrices A, xand b
2.3 Determine the inverse A1 of matrix A
solution:2.1
A=
1 1 13 2 11 1 2
b=
2
31
2.2 Dimension of matrix A:3 3
Dimension of matrixb:3 1
Dimension of matrixx:3 1
2.3
1 1 13 2 11 1 2
1 0 00 1 00 0 1
3R1+ R2 > R2
R1+ R3 > R3
2
-
8/10/2019 MAT1503 Exam Solutions
3/93
1 1 1
0
1 20 0 1
1 0 0
3 1 01 0 1
R2 > R2
1 1 10 1 20 0 1
1 0 03 1 01 0 1
2R3+ R2 > R2R3+ R1 > R1
1 1 00 1 00 0 1
2 0 15 1 21 0 1
R2+ R1 > R1
1 0 00 1 00 0 1
3 1 15 1 21 0 1
Thus the inverse
A1 =3 1 15 1 21 0 1
3
-
8/10/2019 MAT1503 Exam Solutions
4/93
May/June 2011 Examination Paper
Question 3
Consider the system
Ax= b
Multiply both sides by the inverse ofA
A1 (Ax) = A1b
Apply associative property for multiplication for matrices, we get
A1A
x= A1b
But AA1 = Iwe obtain
Ix= A1b
and so using the inverse calculated in Question 2 we get
x= A1b=
3 1 15 1 2
1 0 1
231
=
25
1
1
-
8/10/2019 MAT1503 Exam Solutions
5/93
May/June 2011 Examination Paper
Question 4If
C=
1 1 13 0 11 1 2
4.1 Evaluate the det(C)
4.2 Does the homogeneous system Cx= 0 have no solution, only one solution ,or infinitely many solutions ? Give a reason for your answer.
4.3 Calculate the following:
a) det(CC1)
b) det(3C)
c) det(CCT)
d) det(2C1)
solution:4.1 det(C)
= 1
0 11 2
1
3 11 2
+ (1)
3 01 1
= 1(1) 1(5) + (1)(3) = 3
4.2 Consider the homogeneous system Cx= 0.
Since the det(C) is nonzero, Chas an inverse C1.
Multiplying both sides of system by the inverse we get
C1
Cx= C10
x= C10 = 0
1
-
8/10/2019 MAT1503 Exam Solutions
6/93
that is x= 0 is the trivial solution (only solution)
4.3 a) det(CC1)
= det(C)det(C1)
Since det(C1) = 1
det(C)we get
det(CC1)
= det(C
)
1
det(C)
= 3 13
= 1
b) det(3C) = 33 det(C) = 27(3) = 81
c) det(CCT) = det(C)det(CT) = (det(C))2
= (3)2 = 9
since det(C) = det(CT)
d) det(2C1) = 23 det(C1) = 23 1
det(C) =
8
3
2
-
8/10/2019 MAT1503 Exam Solutions
7/93
May/June 2011 Examination Paper
Question 5
5.1 a) Identify those matrices that are elementary matrices
b) Give a definition of an elementary matrix
5.2 a) Is the inverse of an elementary matrix also an elementary matrix
b) Determine the inverse matrices of the elementary matrices that you wrote in5.1 a)
solution:
5.1 a) elementary matrices: Operation performed on identity
B 3R2 > R2
C R3 < > R1
F 2R1+ R2 > R2
b) An n n matrix is called an elementary if it can be obtained from then n identity matrix In by performing a single elementary row operation.
5.2 a) Every elementary matrix is invertible and the inverse is also an elemen-tary matrix.
b)
B=
1 0 00 3 00 0 1
B1 =
1 0 00 1
3 0
0 0 1
C=
0 0 10 1 01 0 0
1
-
8/10/2019 MAT1503 Exam Solutions
8/93
C1 =
0 0 10 1 01 0 0
F =
1 0 02 1 00 0 1
F1 =
1 0 02 1 00 0 1
2
-
8/10/2019 MAT1503 Exam Solutions
9/93
May/June 2011 Examination Paper
Question 6
6.1 Let L1 be a line passing through the points A(3, 0, 2) and B(4, 3, 0) whileL2 is the line passing through the points B(4, 3, 0) and C(8, 1, 1).
a) Determine the parametric equations for lines
b) Determine whether L1 and L2 are mutually perpendicular
6.2 Does the point (8, 4,5) lie on the the plane 7x 3y+ 4z = 8
solution:
6.1a) L1:
direction vectorv = AB = OB OA= (4, 3, 0) (3, 0, 2) = (1, 3,2)
if (x,y,z) is the position vector of any point on the line and (4, 3, 0) is the po-sition vector of pointB then the parametric equation for L1:
x= 4 + t, y= 3 + 3t, z = 0 2t t R
L2 :
direction vector
v1= BC= (8, 1,1) (4, 3, 0) = (4, 2, 1)
Parametric equation for L2:x= 8 + 4t, y= 1 2t, z = 1 t, t R
6.2 Substituting x= 8, y= 4, z = 5 into equation of plane, and if equation ofplane is satisfied then the point lies on plane so
7(8) 3(4) + 4(5) = 56 12 20 = 24
which does not equal 8 so the point does not lie in the plane.
6.2 Two lines are perpendicular if dot product of their directional vectors is zero
1
-
8/10/2019 MAT1503 Exam Solutions
10/93
v
v1 = (4,
2,
1)
(1, 3,
2) = 4
6 + 2 = 0therefore the lines are perpendicular.
6.3 The lineL passes through the points P1(2, 4, 1) andP2(5, 0, 7). Determinethe point of intersection ofL and the xy -plane.
Equation of line, in parametric form is:
v= (5, 0, 7) (2, 4,1) = (3, 4, 8)
if (x,y,z) is the position vector of any point on line and (5, 0, 7) be the po-
sition vector of a point P2 then
L:
x= 5 + 3t, y= 4t, z= 7 + 8t, t R
All points in the xy-plane have the z = 0 coordinate equal to zero so setz = 7 + 8t= 0, then solving to find t , t= 7/8
Substituting this valve in the remaining equations we get the x and y coor-dinate
y= 4(7/8) = 7/2
x= 5 + 3(7/8) = 19/8
so the line intersects the xy-plane at point (7/2, 19/8, 0)
2
-
8/10/2019 MAT1503 Exam Solutions
11/93
May/June 2011 Examination Paper
Question 7
7.1 Determine an equation for the plane that passes through the point (1 , 1, 1)and is parallel to the plane x 3y2z4 = 0
solution:
For the equation of planex3y2z= 4 the normal to plane is n = (1,3,2)
Two planes are parallel if their normals are scalar multiples of each other
Takingr0 be the position vector of (1, 1, 1) and ifr is the position vector of anypoint in the plane (x,y,z) the the vector r r0 is a vector that is orthogonalto the normal n.
We getn(rr0) = 0
(1,3,2) (x1, y1, z1) = 0
(x1)3(y1)2(z1) = 0
x3y2z+ 4 = 0
7.2 Determine the volume of the parallelepiped determined by three vectorsu= (1, 3,1) , v = (1, 1, 2) and w= (3,1, 2).
Use the triple scalar product
u(vw) =
1 3 11 1 23 1 2
= 20
Volume of parallelepiped = |u(vw)|= 20
1
-
8/10/2019 MAT1503 Exam Solutions
12/93
7.3 Calculate the distance between the plane 2x3y+ 6z= 1 and the point(1,4, 3)
Let A be the point (1,4, 3) and B be any point on the plane B(x0, y0, z0)
let bbe the vector b = AB = (x0, y0, z0)(1,4, 3) = (x01, y0+ 4, z03).
The distance from A to the plane is equal to the absolute value of the scalarprojection ofb onto the normal vector n = (2,3, 6)
Distance
=b|cos |
= bn
|n|
= |(x01, y0+ 4, z03)(2,3, 6)|
22 + (3)2 + 62
= |(2x03y0+ 6z0) + (2) + (12) + (18)|
22 + (3)2 + 62
but (x0, y0, z0) is a point on the plane so 2x03y0+ 6z0= 1 so we get
= |(1) + (2) + (12) + (18)|
22 + (3)2 + 62
Alternate solution: You can use the distance formula but do see the formula inthe study guide
2
-
8/10/2019 MAT1503 Exam Solutions
13/93
May/June 2011 Examination Paper
Question 8
8.1 Express the complex number 1 +i
3 in polar form
8.2 State De Moivres Theorem
8.3 Use De Moivres Theorem to find all cube roots of8
solution:
8.1 We need to find r and such that 1 +i3 = r (cos +i sin )
r =
12 + (
3)2 = 2
1 +i
3 = 2
1
2+i
3
2
and
cos = 1
2 sin =
3
2
= 3 + 2k
for integer k
Polar form :
For k = 0
1 +i
3 = 2
cos
3+i sin
3
8.2 Ifz= r (cos +i sin ) and n is a positive integer, then
zn = (r (cos +i sin ))n =rn (cos n+i sin n)
8.3 Let z3 = 8 and z = r(cos +i sin ) then
(r(cos +i sin ))3
= 8 (cos +i sin )
1
-
8/10/2019 MAT1503 Exam Solutions
14/93
Using De Moivres Theorem
r3 (cos3+i sin3)) = 8 (cos +i sin )
r= 81
3 = 2
and
cos3= cos , sin3= sin
3= + 2k
=
3 +
2k
3
for integer k .
zk= 2
cos
3 +
2k
3
+i sin
3+
2k
3
for integer k .
The distinct cube roots are for k = 0, 1, 2:
Ifk = 0 then
z0 = 2
cos
3
+i sin
3
= 2
1
2+ i
3
2
Ifk = 1
z1 = 2cos
3 +
2
3 +i sin
3 +
2
3 = 2 (1 +i0)Ifk = 2
z2 = 2
cos
3 +
4
3
+i sin
3 +
4
3
= 2
1
2 i
3
2
2
-
8/10/2019 MAT1503 Exam Solutions
15/93
Oct/Nov-2011- Examination Paper
Question 1
Consider the system of linear equations2x+y 3z = 22x 3y+z = 102x+y+z = 6
1.1 Is there a value of a such that (a, 1, 3) is a solution of the system? If yes,give the value of a. If no, explain why not.
1.2 Reduce the augmented matrix of the system to generalized row echelonform and find all solutions of the system.
solution:We begin by substituting x = a, y = 1 andz = 3 into the system for each linearequation we get
2a+ (1) + 3(3) = 2a = 5
2a 3(1) + 3 = 10a = 5
2a+ (1) + 3 =6a = 5
Yes, since there is a single value of a that satisfies all three equations, thus(5, 1, 3) is a solution of the system
1.2 The augmented matrix of the system is given by
2 1 3 | 22 3 1 | 10
2 1 1 | 6
Performing the elementary row operationsR1+R2 > R2
R1+R3 > R3 we get
2 1 3 | 20 4 4 | 80 2 2 | 4
1
2R2+R3 > R3
2 1 3 | 20 4 4 | 80 0 0 | 0
The augmented matrix is in generalized row echelon form.
1
-
8/10/2019 MAT1503 Exam Solutions
16/93
let z= t t R
From the equation4y+ 4z= 8we gety= 2 +tand from the equation2x+y 3z= 2we getx= 2 +t.Thus the solution is given by
(2 +t, 2 +t, t) where t R
2
-
8/10/2019 MAT1503 Exam Solutions
17/93
Oct/Nov-2011- Examination Paper
Question 2
Consider the nonhomogeneous system of linear equations
x 2y+ z = 4y z = 3
a2 a 2
z = a+ 1
2.1 Determine the value/(s) of a for which the system has
a) no solution
b) exactly one solution
c) infinitely many solutions
solution:
From the system, we get
(a 2) (a+ 1) z= (a+ 1)
a) Ifa= 2, we get 0z= 3. This equation has no solution, hence the system hasno solution
b) If a = 2 and a = 1, the equation has a unique solution, hence the sys-tem has a unique solution
c) Ifa= 1, then the equation 0z= 0, has infinitely many solutions, hence thesystem has infinitely many solutions
2.2 Is there any value of a for which the associated homogeneous system has nosolution ? Give a reason for your answer
solution:For a homogeneous system of three linear equations in three variable, exactlyone of the following is true
i) the system has only the trivial solution
ii) the system has infinitely many solutions,so there is no value of a for which the associated homogeneous system has nosolution
1
-
8/10/2019 MAT1503 Exam Solutions
18/93
2.3 Find all values of a for which the associated homogeneous system has
a) only trivial solution
b) non- trivial solution
solution:
a) if a= 2 and a=1, then the system has only a trivial solution
b) Ifa= 2 or a= 1 the system has non-trivial solution
2
-
8/10/2019 MAT1503 Exam Solutions
19/93
Oct/Nov-2011- Examination Paper
Question 3Suppose that
A =
2 1 30 1 2
3 4 1
Evaluate det(A) by expanding along the second column.
solution:det(A)
= 10 23 1+ (1) 2 33 1 4 2 30 2
= 1(0.(1) 3.2) 1(2.(1) 3.3) 4(2.(2) 3.0)
= 1(6) 1(2 9) 4(4)
= 6 + 11 16
= 10 + 11 = 1
3.2 Suppose that
B=2 0 31 1 4
3 2 1
C=
2 1 33 4 1
0 1 2
D=
4 2 60 2 4
6 8 2
F =
6 1 30 1 2
9 4 1
Then evaluate the following:
a) det(B)
b) det(C)
1
-
8/10/2019 MAT1503 Exam Solutions
20/93
c) det(D)
d) det(F)
solution:a) Since B= AT we get
det(B) = det(AT) = det(A) = 1
b) Comparing matrices C with A we have that R2 < > R3 in ma-trix C, thus we get
det(C) = det(A) = 1
c) Since D= 2A we get
det(D) = det(2A) = 23 det(A) = 8.1 = 8
d) Finding det(A) by expanding along the first column we get
det(A)
= 21 24 1
(0)1 34 1
+ 3 1 3
1 2
= 2
1 24 1+ 3
1 31 2
= 1Note: the determinant is the same no matter which row or column one uses
Now, find the determinant of Fby expanding along the first column we get that
det(F) = 6
1 24 1+ 9
1 31 2
= (3) 2 1 2
4 1+ 3 1 3
1 2
= 3det(A)
= 3.1 = 3
Question 4Suppose A is a 2 2 matrix and Ei i = 1, 2, 3 are elementary matrices such that
E3E2E1A = I2
2
-
8/10/2019 MAT1503 Exam Solutions
21/93
where
E1 =
1
2 0
0 1
E2 =
0 11 0
E3 =
1 03 1
4.1 Find E1i
i = 1, 2, 3.
solution:
E11
=
2 00 1
E12
=
0 11 0
E13
=
1 03 1
4.2 Write A as a product of elementary matrices and determine A by multi-plying the elementary matrices.
E3E2E1A = I2
A
= E11
E12
E13
=
2 00 1
0 11 0
1 03 1
= 6 2
1 0
4.3 Write A1 as a product of elementary matrices and determine A1 bymultiplying the elementary matrices.
A1
= E3E2E1
=
1 03 1
0 11 0
1
2 0
0 1
3
-
8/10/2019 MAT1503 Exam Solutions
22/93
= 0 11
2
3
4.4 Use the inverse A1 to solve the system
Ax = b
where
x =
x1x2
b =
21
solution:x = A1b
=
0 11
2 3
21
=
12
thus x1= 1 and x2 = 2
4
-
8/10/2019 MAT1503 Exam Solutions
23/93
Oct/Nov-2011- Examination Paper
Question 5
5.1 Let L1 be a line through the points A(3,1, 2) and B(1,2,1) while L2is the line passing through points C(2, 3,3) and D(4,3,5)
a) determine the parametric equations of lines.
b) determine whether the lines are mutually perpendicular.
solution:
5.1 a)Direction vector:v1 = AB = (1,2,1) (3,1, 2) = (2,1,3)
If (x,y,z) is the position vector of any point on the line L1 then the para-metric equation of line is:
L1 :
x= 3 2t,
y= 1 t,
z = 2 3t
wheret R
Direction vector:v2 = CD= (4,3,5) (2, 3,3) = (6,6,2)
If (x,y,z) is the position vector of any point on the line L2 then the para-metric equation of line is:
L2 :
x= 2 + 6s,
1
-
8/10/2019 MAT1503 Exam Solutions
24/93
y= 3 6s,
z= 3 2s
wheres R
From 5.1 b)
5b)v1 = (2,1,3) andv2 = (6,6,2)
Find the dot product
v1 v2 = 12 + 6 + 6 = 0
and since the dot is zero it means that the lines are mutually perpendicular.
5.2 Does the point (8, 2, 1) lie on the plane x + 2y 3z 9 = 0
solution:
Substituting the point into the equation of plane
8 + 2(2) 3(1) 9 = 8 + 4 3 9 = 0
and since the equation is satisfied this implies that the point lies in the plane.
5.3 The line L passes through pointsP1(3,1, 2) and P2(1,2,1).Determine the point of intersection ofL and the xy-plane.
solution:Direction vector:
v= (1,2,1) (3,1, 2) = (2,1,3)
Parametric equation of line:
x= 1 2t, y= 2 t, z = 1 3t
wheretR
2
-
8/10/2019 MAT1503 Exam Solutions
25/93
Line will intersect the xy plane the z-component will be zero
z = 1 3t= 0
t = 1/3 substituting this for t we get
x= 1 2(1/3) = 5/3,
y = 2 (1/3) = 5/3,
z= 0
The lineL will intersect the xy-plane at point (5/3,5/3, 0)
3
-
8/10/2019 MAT1503 Exam Solutions
26/93
-
8/10/2019 MAT1503 Exam Solutions
27/93
6.2 Determine volume of the parallelepiped spanned by vectors
u= (1, 3, 1), v= (1, 1, 2) and w = (3, 1, 2) by evaluating
u(vw)
solution:
Triple scalar product
u(vw) =
1 3 11 1 23 1 2
= 20
Volume =|20|= 20
6.3 Calculate the distance between the plane 2x3y+ 6z= 1 and the point(1, 1, 0) and deduce the relationship between this point and the plane.
solution:
distance = |2(1)3(1) + 6(0)(1)|
22
+ (3)2
+ 62
= 0
Since the distance is zero this means that the point lies in the plane.
2
-
8/10/2019 MAT1503 Exam Solutions
28/93
Oct/Nov-2011- Examination Paper
Question 7
7.1 State De Moivres Theorem
solution:Ifz = r(cos +i sin ) and n is a positive integer, then
zn = [r(cos +i sin )]n =rn(cosn+i sinn)
7.2 a) Express sin 4 in terms of powers of sin and cos
b) Use De Moivres Theorem to find the cube roots of 1.
solution:7.2 a) Using De Moivres Theorem we get that
(cos +i sin )4
= cos 4+i sin4
Now, we get using the binomial theorem that
(cos +i sin )4
=4
0
cos4 +
4
1
cos3 i sin +
4
2
cos2 i2 sin2 +
4
3
cos i3 sin3 +
4
4
i4 sin4
Using that i2 = 1,4
k
=
4!
(4 k)!k !we get
(cos +i sin )4 = cos4 +i4cos3 sin 6cos2 sin2 i4cos sin3 + sin4
Finally, we get
sin4
=Im
cos4 +i4cos3 sin 6cos2 sin2 i4cos sin3 + sin4
= 4 cos3 sin 4cos sin3
7.2 b) letz3 = 1
1
-
8/10/2019 MAT1503 Exam Solutions
29/93
where z = r(cos +i sin ) Then
r3(cos +i sin )3 = 1(cos0 +i sin0)
we get that r = 1 and cos3= cos(0) and sin 3= sin (0)from which we get that
3= 0 + 2k
for k = 0, 1, 2
= 2k3
and so the cube roots are:
zk= cos 2k
3 +i sin 2k
3
k = 0, 1, 2
z0 = cos0 +i sin0 = 1
z1 = cos 2
3 +i sin 2
3 = 1
2+ i
3
2
z2 = cos 4
3 +i sin 4
3 = 1
2 i
3
2
2
-
8/10/2019 MAT1503 Exam Solutions
30/93
May/June 2012 Examination paper
Question 1
a) If
A=
1 02 3
and
B=
1 23 0
Check whether or notAB= BA.
solution:
AB=
1 02 3
1 23 0
=
1.1 + 0.3 1.2 + 3.02.1 + 3.3 2.2 + 3.0
=
1 211 4
BA =
3 63 0
Thus is follows that AB =BA
b) Reduce the following matrix to reduced row-echelon form
0 3 90 2 4
0 0 3
1/3R1 > R10 1 30 2 4
0 0 3
2R1+ R2 > R2
0 1 30 0 10
0 0 3
1/10R2 > R20 1 30 0 1
0 0 3
1/3R3 > R3
1
-
8/10/2019 MAT1503 Exam Solutions
31/93
0 1 30 0 10 0 1
R1+ R3 > R30 1 30 0 1
0 0 0
3R2+ R1 > R1
0 1 00 0 10 0 0
c) Solve the system
3x + 4y+ z = 12x + 3y = 0
4x + 3y z = 2
Augmented matrix:
3 4 1 12 3 0 04 3 1 2
R2 < > R12 3 0 03 4 1 1
4 3 1 2
1/2R1 > R11
3
2 0 0
3 4 1 14 3 1 2
3R1+ R2 > R2
4R1+ R3 > R31
3
2 0 0
0 12
1 10 3 1 2
2R2 > R21
3
2 0 0
0 1 2 20 3 1 2
2
-
8/10/2019 MAT1503 Exam Solutions
32/93
3R2+ R3 > R31
3
2 0 0
0 1 2 20 0 7 8
1/7R3 > R31
3
2 0 0
0 1 2 20 0 1 8
7
Thus
z = 8
7
y 2z= 2 y = 2
7
x +3
2y = 0 x =
3
7
d) Let
A=
2 11 0
Using row operations, find theA1 and verify it ie show thatAA1 =I=A1A
solution:
A I 2 1 1 01 0 0 1
1/2R1 > R1 1 1
2
1
2 0
1 0 0 1
R1+ R2 > R21 1
2
1
2 0
0 12
1
2 1
2R2 > R21 1
2
1
2 0
0 1 1 2
1/2R2+ R1 > R11 0 0 10 1 1 2
Thus we get
A1 =
0 11 2
3
-
8/10/2019 MAT1503 Exam Solutions
33/93
To verify the A1 is the inverse of matrixA we perform the product
AA1 =
2 11 0
0 11 2
=
1 00 1
A1A=
0 11 2
2 11 0
=
1 00 1
4
-
8/10/2019 MAT1503 Exam Solutions
34/93
May/June 2012 Examination paper
Question 2
a) Let
A=
4 1 56 2 10 2 1
and find its determinant
solution:
det(A) = 01 5
2 1 2
4 5
6 1+ 1
4 1
6 2
= 0 2(26) + 1(14) = 66
b) Show that the matrix
B=
cos sin 0 sin cos 0
0 0 1
is invertible for all values of
solution:
det(B) = 0
sin 0cos 0
0
cos 0 sin 0
+ 1
cos sin sin cos
= cos2 + sin2 = 1
since det(B)= 0
it follows that A is invertible ie. there exists a matrixA1 such that A1A =I=AA1
since the det(B) = 1 is a constant, independent of it follows that B is in-
vertible for all values of
c) Given
C=
2 1 31 2 45 3 6
show that det(C) = det(CT)
1
-
8/10/2019 MAT1503 Exam Solutions
35/93
solution:
det(C) = 2
2 43 6
(1) 1 45 6
+ 3 1 25 3
= 2(12 (12)) + 1(6 20) + 3(3 10)
= 2(24) 14 + 3(13) =5
CT =
2 1 51 2 33 4 6
det(CT) = 22 34 6
11 33 6
+ 51 23 4
= 2(12 (12)) 1(6 (9)) + 5(4 6)
= 2(24) 3 50 =5
Thus we get
det(C) =5 = det(CT)
d) Solve using Cramers rule
2x y= 43x + 2y= 13
solution:
x=
4 1
13 2
2 13 2
=
21
7 = 3
y=
2 43 13
2 13 2
=
14
7 = 2
2
-
8/10/2019 MAT1503 Exam Solutions
36/93
May/June 2012 Examination paper
Question 3
Consider the vectors
u= (1, 0, 4) v= (1, 1,12
)
a) projuv
solution:
projuv = v u|u|2u
= (1, 1, 1
2) (1, 0, 4)|(1, 0, 4)|2
=1 + 0 + 2
17 (1, 0, 4)
= 1
17(1, 0, 4)
b) Calculate the area of the parallelogram determined by u and v
solution:
u
v=
i j k
1 0 4
1 1 12
=i
0 41 1
2
j
1 41 1
2
+k
1 01 1
=4i j 92
+ k
Area of parallelogram
=
|u
v
|=
(
4)2 + (
9
2)2 + (1)2 =
149
2
c) Find an equation of the plane containing the point (1 , 1,1) and perpendic-ular to the line through the points (2, 0,1) and (1, 1, 0)
solution:v= (2, 0,1) (1, 1, 0) = (3,1,1)
Equation of plane:
Let (x , y , z) be any point on plane then
1
-
8/10/2019 MAT1503 Exam Solutions
37/93
(x 1, y 1, z+ 1) (3,1,1) = 03(x 1) (y 1) (z+ 1) = 0
3x y z = 3
d) Determine whether the points (5,6, 10) ,(3, 3, 8) are on the line
x= 2 +t, y= 3 3t, z= 4 + 2t, tR
solution:For the point (5,6, 10)
5 = 2 +t t = 3
6 = 3 3t t = 3
10 = 4 + 2t t = 3
Since there is a single value for t = 3 that satisfies all three parametric equationsof line, we conclude that the point (5,6, 10) lies on the line
For the point (3, 3, 8)
3 = 2 +t t = 1
3 = 3 3t t = 0
8 = 4 + 2t t = 2
Since there is a no single value for t that satisfies all three parametric equa-tions of line, we conclude that the point (3, 3, 8) does not lie on the line
2
-
8/10/2019 MAT1503 Exam Solutions
38/93
May/June 2012 Examination paper
Question 4a) Use De Moivres theorem to express cos 3 and sin 3 in terms of powers ofsin and cos
solution:
a) Using De Moivres theorem we get
(cos + i sin )3 = cos 3+ i sin3
Now,
(cos + i sin )3
= cos3 3cos sin2 + i3cos2 sin i sin3
Equating real and imaginary parts we get
cos3= cos3 3cos sin2
and
sin = 3 cos2 sin sin3
b) Determine the cube roots of i in the polar form
solution:
Let z3 = i and ifz= r(cos + i sin ) then
r3(cos3+ i sin3) = 1
cos
2 +i sin
2
then it follows that
r = (1)1
3
and
3=
2 + 2k for some integer k
=
6 +
2k
3
zk =
cos
6 +
2k
3
+i sin
6 +
2k
3
For distinct roots:
Ifk= 0
z0 =
cos
6
+i sin
6
1
-
8/10/2019 MAT1503 Exam Solutions
39/93
Ifk
= 1
z1 =
cos
5
6
+i sin
5
6
Ifk= 2
z2 =
cos
3
2
+i sin
3
2
c) If (a+ ib)3 = 8, prove that a2 + b2 = 4 (Hint: solve for b2 in terms ofa2 thensolve for a
solution:
If (a+ ib)3 = 8 we get
a3 +i3a2b 3ab2 ib3 = 8
Equating real and imaginary parts we get
a3 3ab2 = 8
3a2b b3 = 0
If b= 0 then
3a2b= b3
3a2 = b2
Substituting this into the first equation
a3 3a(3a2) = 8
8a3 = 8
a = 1
Ifa= 1 then b2 = 3(1)2 thus
a2 +b2 = (1)2 + 3 = 4
If b= 0 then a3 = 8
a = 2
a2 +b2 = 4 + 0 = 4
2
-
8/10/2019 MAT1503 Exam Solutions
40/93
October/November 2012
Question 1
a) Describe the properties of a matrix in reduced row echelon form
b) Solve the following system using Gauss-Jordan elimination
4x18x2 = 123x16x2 = 9
2x1+ 4x2 = 6
c) LetA =
0 11 1
.
FindA3.
d) Given B =
2 7 11 4 1
1 3 0
.
Use row operations, find B1 , verify thatBB1 =I=B1B
solutions:
a)
If a row does not consist entirely of zeros, then the first nonzero numberin the row is a 1
If there are any rows that consist entirely of zeros, then they are groupedtogether at the bottom of the matrix
in any two succesive rows that do not consist entirely of zeros, the leading1 in the lower row occurs further to the right than the leading 1 in thehigher row
each column that contains a leading 1 has zeros everywhere else
1
-
8/10/2019 MAT1503 Exam Solutions
41/93
b) Augmented matrix
4 8 123 6 9
2 4 6
1/4R1 > R1 1 2 33 6 9
2 4 6
3R1+ R2 > R2
2R1+ R3 > R3
1 2 30 0 0
0 0 0
Let x2 = t, t Rthen we get
x12x2= 3
x1 = 2x2+ 3 = 2t + 3
solution set ={(2t + 3; t) :t R}
c) The product
A2 =AA =
1 11 0
A3 =A2A
=
1 11 0
0 11 1
=
1 00 1
d)
2 7 11 4 1
1 3 0
1 0 00 1 00 0 1
2
-
8/10/2019 MAT1503 Exam Solutions
42/93
R1 < > R3
1 3 01 4 1
2 7 1
0 0 10 1 01 0 0
R1+ R2 > R22R1+ R3 > R3
1 3 00 1 10 1 1
0 0 10 1 11 0 2
R2+ R3 > R3
1 3 00 1 1
0 0 2
0 0 10 1 11 1 1
0.5R3 > R3
1 3 00 1 10 0 1
0 0 1
0 1 11
2 1
2 1
2
R3+ R2 > R2
1 3 00 1 0
0 0 1
0 0 11
2
1
2 3
21
2 1
2 1
2
3R2+ R1 > R1
1 0 00 1 0
0 0 1
3
2 3
2
11
21
2
1
2 3
21
2 1
2 1
2
B1 =
3
2 3
2
11
21
2
1
2 3
21
2 1
2 1
2
3
-
8/10/2019 MAT1503 Exam Solutions
43/93
BB
1
=
2 7 1
1 4 11 3 0
32
32
11
21
2
1
2
3
21
2 1
2 1
2
=
1 0 0
0 1 10 0 1
B1B =
3
2 3
2
11
21
2
1
2 3
21
2 1
2 1
2
2 7 11 4 1
1 3 0
=
1 0 00 1 1
0 0 1
4
-
8/10/2019 MAT1503 Exam Solutions
44/93
October/November 2012
Question 1
a) Describe the properties of a matrix in reduced row echelon form
b) Solve the following system using Gauss-Jordan elimination
4x18x2 = 123x16x2 = 9
2x1+ 4x2 = 6
c) LetA =
0 11 1
.
FindA3.
d) Given B =
2 7 11 4 1
1 3 0
.
Use row operations, find B1 , verify thatBB1 =I=B1B
solutions:
a)
If a row does not consist entirely of zeros, then the first nonzero numberin the row is a 1
If there are any rows that consist entirely of zeros, then they are groupedtogether at the bottom of the matrix
in any two succesive rows that do not consist entirely of zeros, the leading1 in the lower row occurs further to the right than the leading 1 in thehigher row
each column that contains a leading 1 has zeros everywhere else
1
-
8/10/2019 MAT1503 Exam Solutions
45/93
b) Augmented matrix
4 8 123 6 9
2 4 6
1/4R1 > R1 1 2 33 6 9
2 4 6
3R1+ R2 > R2
2R1+ R3 > R3
1 2 30 0 0
0 0 0
Let x2 = t, t Rthen we get
x12x2= 3
x1 = 2x2+ 3 = 2t + 3
solution set ={(2t + 3; t) :t R}
c) The product
A2 =AA =
1 11 0
A3 =A2A
=
1 11 0
0 11 1
=
1 00 1
d)
2 7 11 4 1
1 3 0
1 0 00 1 00 0 1
2
-
8/10/2019 MAT1503 Exam Solutions
46/93
R1 < > R3
1 3 01 4 1
2 7 1
0 0 10 1 01 0 0
R1+ R2 > R22R1+ R3 > R3
1 3 00 1 10 1 1
0 0 10 1 11 0 2
R2+ R3 > R3
1 3 00 1 1
0 0 2
0 0 10 1 11 1 1
0.5R3 > R3
1 3 00 1 10 0 1
0 0 1
0 1 11
2 1
2 1
2
R3+ R2 > R2
1 3 00 1 0
0 0 1
0 0 11
2
1
2 3
21
2 1
2 1
2
3R2+ R1 > R1
1 0 00 1 0
0 0 1
3
2 3
2
11
21
2
1
2 3
21
2 1
2 1
2
B1 =
3
2 3
2
11
21
2
1
2 3
21
2 1
2 1
2
3
-
8/10/2019 MAT1503 Exam Solutions
47/93
BB
1
=
2 7 1
1 4 11 3 0
32
32
11
21
2
1
2
3
21
2 1
2 1
2
=
1 0 0
0 1 10 0 1
B1B =
3
2 3
2
11
21
2
1
2 3
21
2 1
2 1
2
2 7 11 4 1
1 3 0
=
1 0 00 1 1
0 0 1
4
-
8/10/2019 MAT1503 Exam Solutions
48/93
October/November 2012
Question 2
a) Given
A=
3 2 11 6 32 4 0
find Adj(A)
b) Let k =
2,
B=
2 1 33 2 11 4 5
find det(B) and det(kB) and compare them.
c) Let
C=
2 1 0
3 4 00 0 2
and
D=
1 1 37 1 25 0 1
det(CD) = det(C)det(D)
d) Given the following system of equations
5x1+ x2 x3 = 49x1+ x2 x3 = 1x1 x2+ 5x3 = 2
Apply Cramers Rule to find x1
1
-
8/10/2019 MAT1503 Exam Solutions
49/93
solution:
a) Co-factors:
c11 = (1)1+1
6 34 0
= 12
c12 = (1)1+2
1 32 0
= 6
c13 = (1)1+3
1 62 4
= 16
c21 = (1)2+1
2 14 0
= 4
c22 = (1)2+2
3 12 0
= 2
c23 = (1)2+3
3 22 4 = 16
c31 = (1)3+1
2 16 3
= 12
c32 = (1)3+2
3 11 3
= 10
c33 = (1)3+3
3 21 6
= 16
Let Cbe the co-factor matrix
C=
12 6 164 2 16
12 10 16
2
-
8/10/2019 MAT1503 Exam Solutions
50/93
Adj (A) =CT
=
12 6 16
4 2 1612 10 16
T
=
12 4 12
6 2
1016 16 16
b) det(B) = 2
2 14 5
(1)3 11 5
+ 33 21 4
= 56
2B=
4 2 66 4 22 8 10
det(2B) = 4
4 28 10
2
6 22 10
6
6 42 8
= 448
det(2B) = 448 = (2)356 = (2)3 det(B)
c) det(C) = 2
4 00 2
13 00 2
+ 03 40 0
= 10
det(D) = 1
1 20 1
(1)7 25 1
+ 37 15 0
= 17
CD =
9 1 8
31 1 1710 0 2
det(CD) = 9
1 170 2
(1)31 1710 2
+ 831 110 0
= 170
det(CD) = 170 = 10(17) = det(C)det(D)
d)
5 1 19 1 11 1 5
x1x2x3
=
412
A=
5 1 19 1 11 1 5
detA= 16
3
-
8/10/2019 MAT1503 Exam Solutions
51/93
Ax1 =
4 1 11 1
1
2 1 5
detAx1 = 12
x1 =det(Ax1)
det(A) = 12
16
4
-
8/10/2019 MAT1503 Exam Solutions
52/93
October/November 2012
Question 3
a) Consider the vectors u = (3, 0, 4) and v = (1, 1, 0)
i) Determine the orthogonal projection projuv
ii) Calculate the area of the parallelogram bounded by these vectors.
iii) Determine the perimeter of the parallelogram bounded by these vectors.
b) Find an equation of the plane containing the point (0 , 1, 1) and perpendicularto the line passing through the points (2, 1, 0) and (1,1, 0)
c) Determine whether the point (2,10, 8) lies on the line whose parametericequation arex = 2 t, y= 2 3t, z = 4 +t
solution:
a) i) projuv= u v|u|2u
= (3, 0, 4) (1, 1, 0)32 + 42
(3, 0, 4) = 325
(3, 0, 4)
ii) First we find the cross product ofu and v
u v
=
i j k
3 0 41 1 0
= 4i+ 4j+ 3k
Area of parallelogram = |u v|
=4i+ 4j+ 3k
=
42 + 42 + 32 =
41units2
iii) Perimeter of parallelogram
1
-
8/10/2019 MAT1503 Exam Solutions
53/93
= 2 |u| + 2 |v|= 2
32 + 42 + 2
12 + 12 = 2.5 + 2
2
b) Direction vector of line
v= (1,1, 0) (2, 1, 0) = (1,2, 0)
Let (x , y , z) be the position vector of any point on the plane and since the lineis perpendicular to plane we can use the direction vector v as the normal vector
(x, y 1, z 1) (1,2, 0) = 0
and so we get
x 2(y 1) = 0
x 2y= 2
c) Substituting the point (2,10, 8) into the parametric equations of line weget
2 = 2 t
t = 4
10 = 2 3t
t = 4
8 = 4 +t
t = 4
since there is a single value for t that satisfies all the parametric equationsit means that the point lies on the line
2
-
8/10/2019 MAT1503 Exam Solutions
54/93
October/November 2012
Question 4
a) Use De Moivres Theorem to express sin4 in terms of sin and cos
b) Determine the cube roots of8
c) Let a, bbe real numbers such that (a ib)2 = 4
i) Prove the b= 0
ii) Show that a4 +a2 20 = 0
solution:
a) Using De Moivres theorem we get
(cos +i sin )4
= cos 4+i sin4
Now, we get using the binomial theorem that
(cos +i sin )4
= 4
0 cos4 +
4
1 cos3 i sin +
4
2 cos2 i2 sin2 +
4
3 cos i3 sin3 +
4
4i4 sin4
Simplifying and use that i2 = 1 we get
(cos +i sin )4
= cos4 +i4cos3 sin 6cos2 sin2 i4cos sin3 + sin4
Finally, we have
sin4
=Im
cos4 +i4cos3 sin 6cos2 sin2 i4cos sin3 + sin4 = 4 cos3 sin 4cos sin3
b)Let z3 = 8 and ifz = r(cos +i sin ) then
(r(cos +i sin ))3 = 8 (cos+i sin)
1
-
8/10/2019 MAT1503 Exam Solutions
55/93
Using De Moivres Theorem
r3(cos3+i sin3) = 8 (cos+i sin)
r = 81
3 = 2 and cos3= cos, sin3= sin
3= + 2k
=
3 +
2k
3
where k is an integer.
zk = 2
cos
3+
2k
3
+i sin
3 +
2k
3
where k is an integer
The distinct cube roots are:
for k = 0
z0 = 2
cos
3
+i sin
3
= 21
2+ i
3
2
for k = 1
z1 = 2
cos
3 +
2
3
+i sin
3 +
2
3
= 2 (1 +i0)
for k = 2
z2 = 2
cos
3 +
4
3
+i sin
3 +
4
3
= 2
1
2 i
3
2
c) i) if (a ib)2 = 4 then
a2 b2 i2ab= 4.
Two complex numbers are equal if there real parts and imaginary parts arethe same. So equating we get
2
-
8/10/2019 MAT1503 Exam Solutions
56/93
a2 b2 = 4 and 2ab= 0
From the second equation we get a = 0 or b = 0
Ifa = 0 then from first equation we getb2 = 4
which means that b must be complex, but since b is real, we cannot have thata= 0. Thus a = 0.
Thus we get that b= 0 must be zero.
ii) since b = 0 we get that a2 = 4 thus we get that
a4 +a2 20 = 42 + 4 20 = 0
3
-
8/10/2019 MAT1503 Exam Solutions
57/93
May/June 2013- Examination paper
Question 1
1.1 Describe the elementary row operations on a matrix.solution:
Multiply a row through by a nonzero constant
Add a multiple of one row to another
Interchange any two rows
1.2 Verify that
x= 19t 35
y= 25 13t
z= t
is a solution of
2x+ 3y+z = 5
5x+ 7y 4z= 0
solution:
Substituting for x, y and z in terms oft, we get
2(19t 35) + 3(25 13t) +t= 38t 70 + 75 39t+t= 5
and
5(19t 35) + 7(25 13t) 4t= 95t 175 + 175 91t 4t= 4t 4t
1
-
8/10/2019 MAT1503 Exam Solutions
58/93
= 0
Thus we conclude that x = 19t 35, y = 25 13t, z = t, t R is a solu-tion of the system
1.3 a) Compute
3 2 15 1 0
5
3 0 21 1 2
b) Find A in terms ofB if 2A B= 5(A+ 2B)
solution:
a)
3 2 15 1 0
5
3 0 21 1 2
=
3 2 15 1 0
15 0 105 5 10
=
12 2 11
0 6 10
1.3 b)2A B= 5(A+ 2B)
2A B= 5A+ 10B
2A 5A= B + 10B
3A= 11B
A = 11
3B
1.4 a) Given
A=
3 10 2
Compute A2 A 6I2
solution:A2
=
3 10 2
3 10 2
=
3.3 + 1.0 3. 1 + 1. 20.3 + 2.0 0. 1 + 2. 2
=
9 10 4
2
-
8/10/2019 MAT1503 Exam Solutions
59/93
A2
A
6I2
=
9 10 4
3 10 2
6
1 00 1
=
9 3 6 1 + 1 + 0
0 4 + 2 6
=
0 00 0
b) Given
B=
6 94 6
Compute B2
and say what you observe about B2
in relation to B.
solution:
B2 =
6 94 6
6 94 6
=
0 00 0
Relation ofB2 to matrix B is:
B2 = 0B where 0 is a scalar
B+B2 =B
1.5 IfA=
0 11 1
and B=
1 11 0
Show that A and B are inverses of each other.
solution:
AB=
0 11 1
1 11 0
=
1 00 1
BA =
1 11 0
0 11 1
=
1 00 1
3
-
8/10/2019 MAT1503 Exam Solutions
60/93
May/June 2013- Examination paper
b) Given
B=
6 94 6
Compute B2 and say what you observe about B2 in relation to B.
solution:
B2 =
6 94 6
6 94 6
=
0 00 0
Relation ofB2 to matrix B is:
In this Question it is not clear as to what the examiner is getting at nevertheless
For matrix B the entries have the relation:
b11 = b22
If b21 = 0 then b12 = (b11)
2
b21
and so the matrix has the form
B=
b22 (b22)
2
b21b21 b22
For the matrix B2 are entries are zero despite the fact that all entries of Bare non zero
In this example, we that product BB = 0butthe matrix B = 0
1
-
8/10/2019 MAT1503 Exam Solutions
61/93
May/June 2013- Examination paper
Question 1
1.1 Describe the elementary row operations on a matrix.solution:
Multiply a row through by a nonzero constant
Add a multiple of one row to another
Interchange any two rows
1.2 Verify that
x= 19t 35
y= 25 13t
z= t
is a solution of
2x+ 3y+z = 5
5x+ 7y 4z= 0
solution:
Substituting for x, y and z in terms oft, we get
2(19t 35) + 3(25 13t) +t= 38t 70 + 75 39t+t= 5
and
5(19t 35) + 7(25 13t) 4t= 95t 175 + 175 91t 4t= 4t 4t
1
-
8/10/2019 MAT1503 Exam Solutions
62/93
= 0
Thus we conclude that x = 19t 35, y = 25 13t, z = t, t R is a solu-tion of the system
1.3 a) Compute
3 2 15 1 0
5
3 0 21 1 2
b) Find A in terms ofB if 2A B= 5(A+ 2B)
solution:
a)
3 2 15 1 0
5
3 0 21 1 2
=
3 2 15 1 0
15 0 105 5 10
=
12 2 11
0 6 10
1.3 b)2A B= 5(A+ 2B)
2A B= 5A+ 10B
2A 5A= B + 10B
3A= 11B
A = 11
3B
1.4 a) Given
A=
3 10 2
Compute A2 A 6I2
solution:A2
=
3 10 2
3 10 2
=
3.3 + 1.0 3. 1 + 1. 20.3 + 2.0 0. 1 + 2. 2
=
9 10 4
2
-
8/10/2019 MAT1503 Exam Solutions
63/93
A
2
A 6I2
=
9 10 4
3 10 2
6
1 00 1
=
9 3 6 1 + 1 + 0
0 4 + 2 6
=
0 00 0
b) Given
B=
6 94 6
Compute B2
and say what you observe about B2
in relation to B.
solution:
B2 =
6 94 6
6 94 6
=
0 00 0
Relation ofB2 to matrix B is:
B2 = 0B where 0 is a scalar
B+B2 =B
det(B) = det(B2) = 0
In this example we that product BB = 0butthe matrix B = 0
1.5 IfA=
0 11 1
and B =
1 11 0
Show that A and B are inverses of each other.
solution:AB=
0 11 1
1 11 0
=
1 00 1
BA =
1 11 0
0 11 1
=
1 00 1
Question 2
2.1 Find det(A) if
3
-
8/10/2019 MAT1503 Exam Solutions
64/93
a)
A=
cos sin sin cos
solution:det(A) = cos2 + sin2 = 1
2.2 b)
A=
a+ 1 a
a a 1
solution:
det(A) = (a+ 1)(a 1) a2 =a2 1 a2 =1
2.2 Using the cofactor expansion find the det(B) where
B =
3 0 0 05 1 2 02 6 0 16 3 1 0
solution:
det(B) = 3
1 2 06 0 1
3 1 0
0
5 2 02 0 1
6 1 0
+ 0
5 1 02 6 1
6 3 0
0
5 1 22 6 0
6 3 1
= 3
1
0 11 0 2
6 13 0+ 0
6 03 1
= 3 [(1).(1) (2).(3)]
= 3(5) = 15
2.3 Let
C=
4 13 2
and show that
det(C1) = 1
det(C)
solution:det(C) = 8 3 = 5
Since det(C)= 0, the matrix Chas an inverse.
4
-
8/10/2019 MAT1503 Exam Solutions
65/93
C1 =1
5
2 13 4
Now, we find the determinant, we get
det(C1) =
2
5
4
5
3
5
1
5
=
5
25=
1
5 =
1
det(C)
2.4 Let
D= 3 2
1 1
find det(2D) and compare it to det(D)
solution:det(D) =3 2 =5
2D=
6 42 2
det(2D) = (6)(2) (4)(2) =12 8 = 20 = 22(5) = 22 det(D)
2.5 Solve the system using Cramers rule
2x+y = 1
3x+ 7y= 2
solution
2 13 7
x
y
=
12
x=
1 12 72 13 7
= 7 + 2
14 3 =
9
11
5
-
8/10/2019 MAT1503 Exam Solutions
66/93
y=2 1
3
22 13 7
=4 3
14 3 =
7
11
6
-
8/10/2019 MAT1503 Exam Solutions
67/93
-
8/10/2019 MAT1503 Exam Solutions
68/93
Let (x,y,z) be a point on the plane then
(x 1, y 3, z) (3,3,3) = 0
3(x 1) + 3(y 3) + 3(z) = 0
3x+ 3y+ 3z = 0
3.4 Determine the parametric equations of the plane in 3.3
solution:
From 3.3, we get
x=
3
3 y+
3
3 z
If we let y = s and z = t then the parametric equation of plane is:
x=
3
3 s+
3
3 t
y = s
z = t
where t, sR
2
-
8/10/2019 MAT1503 Exam Solutions
69/93
May/June 2013- Examination paper
Question 44.1 Use De Moivres theorem to express cos 2 in terms of powers of sin andcos
solution:
cos2
=Re[(cos2+i sin2)]
=Re[(cos +i sin )2]
=Re[cos2 + 2i cos sin sin2 ]
= cos2 sin2
4.2 Determine the cube root of1 in the form a +ib , a, b R
solution:Let z3 =1
Ifz = r(cos +i sin ) then
z3 =r3(cos 3+i sin3) = 1(cos+i sin )
It follows that
r = 11/3 = 1 and
3= + 2k
for some integer k
zk = cos
+ 2k
3
+i sin
+ 2k
3
For distinct roots:
Ifk= 0
z0 = cos
3
+i sin
3
=
1
2+i
3
2
1
-
8/10/2019 MAT1503 Exam Solutions
70/93
Ifk= 1
z1 = cos
+ 2
3
+i sin
+ 2
3
=1 + 0i
Ifk= 2
z2 = cos
+ 4
3
+i sin
+ 4
3
=
1
2 i
3
2
4.3 Use De Moivres theorem to determine (1 +i)134 in the form x +iy
solution:Polar form
| 1 +i| =
(1)2 + (1)2 = 2
1 +i = 21
2+i
12
It follows that
cos = 12
and sin = 1
2
= 3
4
+ 2k for some integer k
The polar form is:
1 +i= 2
cos3
4 +i sin
3
4
Using De Moivres theorem we get that
(1 +i)134 =
= (
2)134
cos1343
4 +i sin134
3
4
= 267
cos2012
+i sin2012
= 267
cos(200 + 1)
2 +i sin(200 + 1)
2
= 267
cos(2(50) +
2) +i sin(2(50) +
2)
= 267
cos
2+i sin
2
= 0 +i267
2
-
8/10/2019 MAT1503 Exam Solutions
71/93
Question 2
a) i) A=
a+ 1 a
a a 1
det(A) = (a+ 1)(a 1) a2 =a2 1 a2 = 1
ii) det(B2C1AB1CT)
= (det(B))2 1
det(C)det(A)
1
det(B)det(C)
= 221
3(1)
1
2(3) = 2
b) A=
1 a ba 1 cb c 1
det(A)
= 1
1 cc 1 a
a cb 1 +b
a 1b c
= 1(1 +c2) a(a+bc) +b(ac+b)
= 1 +c2 +b2 +a2
c)i) (det B)n = det(Bn) = det(0) = 0
det(B) = 0
it follows that matrix B is invertible.
ii) det(I) = det(C2) = (det(C))2
det(C) = 1
d) let A=
3 42 1
det(A) = 3 8 = 11
Ax =
9 41 1
1
-
8/10/2019 MAT1503 Exam Solutions
72/93
det(Ax) = 9 + 4 = 5
x = 5
11
Ay =
3 92 1
det(Ay) = 3 18 = 21
y =21
11
2
-
8/10/2019 MAT1503 Exam Solutions
73/93
Question 3
a) i) proj u v
= u v|u|2u
=(1, 0, 2) (1, 1, 0)
|(1, 0, 2)|2 (1, 0, 2)
=1
5(1, 0, 2)
ii) u v
=
i j k
1 0 21 1 0
= i
0 21 0
j1 21 0
+k1 01 1
= 2i+ 2j+ k
Area of parallelogram = |u v| =
(2)2 + 22 + 1 = 9 = 3
iii) perimeter of parallelogram = 2(|u| + |v|) = 2(5 + 2)b) Normal vector
n= (2, 1, 1) (1, 1, 0) = (1, 0, 1)
Equation of plane containing point (1, 1, 1) is: let (x , y , z ) be the positionvector of any point on plane then
(1, 0, 1) (x 1, y 1, z 1) = 0
(x 1) + (z 1) = 0
i.e., x+z= 2
c) If point (0,4, 6) lies on the line x = 2 t, y= 23t, z= 4 + tthen thereis a tsuch that
0 = 2 t
1
-
8/10/2019 MAT1503 Exam Solutions
74/93
4 = 2
3t
6 = 4 +tSolving the system of equations we get t= 2
the point lies on the line.
2
-
8/10/2019 MAT1503 Exam Solutions
75/93
Question 4
a) (cos +i sin )3 = cos 3+i sin3
cos3
= Re(cos + i sin )3
= Re
3
0
cos3 +
3
1
cos2 sin i
3
2
cos sin2
3
3
i sin3
=3
0
cos3
3
2
cos sin2
= cos3
3cos sin2
b) Let z3 = 8, where z=r(cos +i sin ) Then
z3 = r3(cos3+i sin3) = 8(cos 0 +i sin 0)
r= 81/3 cos3= cos 0 and sin 3= sin 0
=2k
3 for integer k
The distinct roots is given by
zk = 2
cos
2k
3 +i sin
2k
3
for k= 0, 1, 2
z0= 2 (cos 0 +i sin 0)
z1= 2
cos
2
3 +i sin
2
3
z2= 2
cos43
+i sin43
c)(a ib)2 = 4i
a2 2iab b2 = 4i
a2 b2 +i(2ab 4) = 0
1
-
8/10/2019 MAT1503 Exam Solutions
76/93
since two complex numbers are equal iff the real parts and imaginary partare equal, it follows
i) a2 b2 = 0
ii) 2ab 4 = 0 ab = 2
2
-
8/10/2019 MAT1503 Exam Solutions
77/93
May-June 2014 Examination paper
Question 1
a)(i)
1 2 00 1 1
(ii) x= 2y x = 2
b)
1 1 1 42 1 3 00 1 5 8
2R1+ R2 R21 1 1 40 1 5 8
0 1 5 8
R2 R2
1 1 1 40 1 5 80 1 5 8
R2+ R3 R31 1 1 40 1 5 8
0 0 0 0
1
-
8/10/2019 MAT1503 Exam Solutions
78/93
Question
1.c)
i) AB =
1 2 22 1 11 0 1
1 2 41 1 31 2 5
=
1 0 00 1 00 0 1
= I3
BA =
1 2 41 1 31 2 5
1 2 22 1 11 0 1
=
1 0 00 1 00 0 1
= I3
B =A1
ii) AX=Y
A1(AX) =A1Y
IX=A1Y
X=A1
X=
1 2 41 1 3
1
2 5
30
2
=
119
13
x1= 11, x2= 9, x3= 13
1
-
8/10/2019 MAT1503 Exam Solutions
79/93
Question
1 d)Since C is an inverse ofB,
we haveCB= I.
Multiplying both sides on the right by Dgives
(CB)D= ID= D.
But we also have by the associative property, that
(CB)D= C(BD) = CI= C
since D is an inverse and so we get that C= D
1
-
8/10/2019 MAT1503 Exam Solutions
80/93
Question 1
e)Let T be a m nmatrix
T = T
T+ (T) = T+ (T)
T+ T= 0 ( 0 is the zero m nmatrix)
2T = 0
T =
1
2
0
T = 0
1
-
8/10/2019 MAT1503 Exam Solutions
81/93
Question 2
a) Finding the determinant using co-factor method by expanding along thesecond row
det(F)
=
8 1 23 0 91 2 1
= 3
1 22 1
+ 0
8 21 1
9
8 11 2
= 3
1 22 1 + 0
8 21 1 3
8 11 2
= 3 det(E)
Since matrix F results when the second row of E is multiplied by scalar3, then det(F) = 3 det(E)
1
-
8/10/2019 MAT1503 Exam Solutions
82/93
2.b)
i)det(G) = 4 6 = 2
det(H) = 4 + 1 = 5
GH=
1 23 4
2 11 2
=
4 310 5
det(GH) = 20
30 =
10
det(GH) = 10 = (2)(5) = det(G)det(H)
ii) HG =
2 11 2
1 23 4
=
1 07 10
det(HG) = 10
det(GH) =
10 = det(HG)
1
-
8/10/2019 MAT1503 Exam Solutions
83/93
Question 2
c) IfJ,Kare n n matrices then JJ1 =I and KK1 =I .
Then det(J) det(J1) = 1
and det(K)det(K1) = 1
Thus det(J1
K1
JK)
= det(J1)det(K1)det(J)det(K)
= 1
det(J)
1
det(K)det(J)det(K)
= 1
1
-
8/10/2019 MAT1503 Exam Solutions
84/93
2.d)
det
1
40 4
= 0
( 1)( 4) = 0
= 1 or = 4
1
-
8/10/2019 MAT1503 Exam Solutions
85/93
Question 2
e)Let
A=
2 3 11 2 12 1 1
det(A) = 2
2 11 1
3
1 12 1
1
1 22 1
= 2(1) 3(1) 1(3) = 2
A1=
1 3 14 2 13 1 1
det(A1) = 1
2 11 1
3
4 13 1
1
4 23 1
= 1(1) 3(1) 1(2) = 4
x1=det(A1)
det(A) =
4
2= 2
A2=
2 1 11 4 12 3 1
det(A2) = 2
4 13 1
1
1 12 1
1
1 42 3
= 2(1) 1(1) 1(5) = 6
x2=det(A2)
det(A) =6
2= 3
A3=
2 3 11 2 42 1 3
det(A3) = 2
2 41 3
3
1 42 3
+ 1
1 22 1
1
-
8/10/2019 MAT1503 Exam Solutions
86/93
=
2(
2)
3(5) + 1(3) =
8
x3=det(A3)
det(A) =
8
2= 4
2
-
8/10/2019 MAT1503 Exam Solutions
87/93
Question 3
a) i)u a= 4.2 + (1)(1) + 3.2 = 8 + 1 + 6 = 15
ii)
u a=
i j k
2 1 34 1 2
= i + 8j+ 2k
iii)
u = 22 + (1)2 + 32 = 14
a =
42 + (1)2 + 22 = 21
iv)Vector projection ofu in the direction ofa
= u a
a
2
a=15
21 (4,1, 2)
Vector projection ofu perpendicular to a
= u u aa2 a= (2,1, 3) 15
21 (4,1, 2)
v) The vector u ais perpendicular to both u and a
u a= i + 8j+ 2k
1
-
8/10/2019 MAT1503 Exam Solutions
88/93
Question 3
b (i)
Find a point in plane 2x3y+ 6z= 1
Let z= 0 and y = 1 then
2x3y = 1
x = 2
(2, 1, 0) is a point on the plane 2x3y+ 6z= 1Form a vector from this point A(2, 1, 0) to point B(1,4,3)
Letv= (1,4,3)(2, 1, 0) = (1,5,3)
The normal vector of plane 2x3y+ 6z= 1 is :
n= (2,3, 6)The distance between the plane and point is:
Distance
=||v| cos |
=
|v| vn|v| |n|
=|v
n
||n|=
|(1,5,3)(2,3, 6)|22 + 32 + 62
=|5|
49=
5
7
Alternate solution Use the formula
1
-
8/10/2019 MAT1503 Exam Solutions
89/93
Question 3
ii) Let u= P1P2= (1, 1, 1) (2, 1, 3) = (1, 2, 2)
v=P1P3= (3, 0, 2) (2, 1, 3) = (5, 1, 1)
The normal vector of plane:
n= u v=
i j k
1 2 2
5 1 1
= 9j+ 9k
Equation of plane containing all points is given by:
Let (x , y , z ) be any point in the plane then
(0, 9, 9) (x+ 2, y 1, z 3) = 0
9(y 1) + 9(z 3) = 0
y z+ 2 = 0
1
-
8/10/2019 MAT1503 Exam Solutions
90/93
Question 4
a)cos3
= Re (cos3+ i sin3)
= Re (cos + i sin )3
= Re
3
0
cos3 +
3
1
cos2 (i)sin +
3
2
cos (i)2 sin2 +
3
3
(i)3 sin3
= Re
cos3
+ 3 cos2
(i)sin + 3 cos (i)2
sin2
+ (i)3
sin3
= Re
cos3 + 3 cos2 (i)sin 3cos sin2 i sin3
= cos3 3cos sin2
1
-
8/10/2019 MAT1503 Exam Solutions
91/93
Question 4
b)Let z4 = 16
Then this equation in Polar form is given by :
ifz= r(cos +i sin ) we get
r4(cos +i sin )4 = 16(cos + i sin )
Using De Moivres Theorem we get
r4(cos 4+ i sin4) = 16(cos + i sin )
r4 = 16 and cos 4= cos(); sin 4= sin()
r = 161/4 = 2
4= + 2k for k Z
=
4+
k
2
Thus the roots are given by:
zk = 2
cos
4+
k
2
+i sin
4+
k
2
For distinct roots k= 0, 1, 2, 3
z0= 2
cos
4
+i sin
4
= 2
1
2+i
12
z1= 2
cos4
+ 2
+i sin
4
+ 2
= 2
1
2+i 1
2
z2= 2
cos
4+
+i sin
4
+
= 2
1
2 i 1
2
z3= 2
cos
4+
3
2
+i sin
4+
3
2
= 2
1
2 i 1
2
1
-
8/10/2019 MAT1503 Exam Solutions
92/93
Question 4
c)z1 = 1 + i
3
Represent in polar form:
|z1| =
12 + 3 = 2
z1 =|z1|
|z1
|
(1 + i
3) = 2
1
2+i
3
2
cos =1
2and sin =
3
2
tan = sin
cos =
3
=
3
Polar form: z1 = 2
cos
3+ i sin
3
z2=
3 + i
Represent in polar form:
|z2| =
12 + 3 = 2
z2 =|z2||z2|(
3 + i) = 2
3
2 +i
1
2
cos =
3
2 and sin =
1
2
1
-
8/10/2019 MAT1503 Exam Solutions
93/93
tan = sin
cos =
1
3
=
6
Polar form: z2 = 2
cos
6+ i sin
6
z1z2
= 2cos 3
+i sin
3 .2cos
6+ i sin
6
= 22
cos
3+
6
+ sin
3+
6
= 4
cos
2+i sin
2
z1
z2
=2
cos
3+ i sin
3
2cos
6+ i sin
6= cos
3
6
+ i sin
3
6
= cos
6
+ i sin
6