May/June 2011 Examination Paper Question 1
Consider the following system of linear equations
x1 3x1 −x1
+ x2 + 2x2 − x2
− − +
x3 x3 2x3
= = =
2 3 −1
Write down the augmented matrix, reduce the augmented matrix to generalized row echelon form and then determine the solution of system
solution :
Augmented Matrix:
1 3 −1
1 2 −1
−1 −1 2
2 3 −1
Apply the elimination row operations −3R1 + R2 − − − − − − − − > R2 R1 + R3 − − − − − − − − > R3 we get 1 0 0
1 −1 0
−1 2 1
2 −3 1
The matrix is in generalized row echolen form so the solution is
x3 = 1, −x2 + 2x3 = −3, x1 + x2 − x3 = 2
1
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, x and b 2.3 Determine the inverse A−1 of matrix A
solution: 2.1 1 1 −1 2 −1 A= 3 −1 −1 2 2 b= 3 −1 2.2 Dimension of matrix A: 3×3 Dimension of matrix b: 3×1 Dimension of matrix x: 3×1
2.3
1 1 −1 1 0 3 2 −1 0 1 −1 −1 2 0 0
0 0 1
−3R1 + R2 − − − −− > R2 R1 + R3 − − − − − − − − > R3
2
1 0 0
1 0 0
1 −1 1 0 0 −1 2 −3 1 0 0 1 1 0 1 −R2 − − − − − − − − − − > R2 1 −1 1 0 0 1 −2 3 −1 0 0 1 1 0 1
2R3 + R2 − − − − − − − −− > R2 R3 + R1 − − − − − − − − − − − − > R1 1 1 0 2 0 1 0 1 0 5 −1 2 0 0 1 1 0 1 −R2 + R1 − − − − − − − − > R1 1 0 0 −3 1 −1 0 1 0 5 −1 2 0 1 0 0 1 1 Thus the inverse
A−1
−3 = 5 1
1 −1 −1 2 0 1
3
May/June 2011 Examination Paper Question 3 Consider the system Ax = b
Multiply both sides by the inverse of A
A−1 (Ax) = A−1 b Apply associative property for multiplication for matrices, we get A−1 A x = A−1 b But AA−1 = I we obtain
Ix = A−1 b
and so using the inverse calculated in Question 2 we get −3 1 −1 2 −2 x = A−1 b = 5 −1 2 3 = 5 1 0 1 −1 1
1
May/June 2011 Examination Paper Question 4 If
1 1 −1 0 −1 C = 3 −1 −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(CC −1 ) b) det(3C) c) det(CC T ) d) det(2C −1 )
solution: 4.1 det(C) 0 −1 0 − 1 3 −1 + (−1) 3 = 1 −1 2 −1 −1 −1 2
= 1(−1) − 1(5) + (−1)(−3) = −3
4.2 Consider the homogeneous system Cx = 0. Since the det(C) is nonzero, C has an inverse C −1 . Multiplying both sides of system by the inverse we get
C −1 Cx = C −1 0
x = C −1 0 = 0
1
that is x = 0 is the trivial solution (only solution)
4.3 a) det(CC −1 ) = det(C) det(C −1 )
Since det(C −1 ) =
1 we get det(C)
det(CC −1 ) = det(C)
1 det(C)
1 = −3 −3 =1
b) det(3C) = 33 det(C) = 27(−3) = −81 2
c) det(CC T ) = det(C) det(C T ) = (det(C)) = (−3)2 = 9 since det(C) = det(C T ) d) det(2C −1 ) = 23 det(C −1 ) = 23
8 1 = det(C) −3
2
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 in 5.1 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 the n × 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 elementary matrix.
b)
0 0 1
1 0 B = 0 3 0 0
B −1
1 = 0 0
0 0 C = 0 1 1 0 1
0 1 3
0
0 0 1
1 0 0
C −1
0 = 0 1
0 1 1 0 0 0
1 0 F = −2 1 0 0
F −1
1 = 2 0
2
0 0 1
0 0 1 0 0 1
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) while L2 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 vector v = 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 position vector of point B then the parametric equation for L1 : x = 4 + t,
y = 3 + 3t,
z = 0 − 2t
t∈R
L2 : direction vector v 1 = 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 of plane 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
v · v 1 = (4, −2, −1) · (1, 3, −2) = 4 − 6 + 2 = 0 therefore the lines are perpendicular.
6.3 The line L passes through the points P1 (2, 4, −1) and P2 (5, 0, 7). Determine the point of intersection of L 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 position 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 set z = 7 + 8t = 0, then solving to find t , t = −7/8 Substituting this valve in the remaining equations we get the x and y coordinate
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
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 − 3y − 2z − 4 = 0
solution: For the equation of plane x − 3y − 2z = 4 the normal to plane is n = (1, −3, −2) Two planes are parallel if their normals are scalar multiples of each other
Taking r 0 be the position vector of (1, 1, 1) and if r is the position vector of any point in the plane (x, y, z) the the vector r − r 0 is a vector that is orthogonal to the normal n.
We get n · (r − r 0 ) = 0
(1, −3, −2) · (x − 1, y − 1, z − 1) = 0
(x − 1) − 3(y − 1) − 2(z − 1) = 0
x − 3y − 2z + 4 = 0
7.2 Determine the volume of the parallelepiped determined by three vectors u = (1, 3, −1) , v = (1, 1, 2) and w = (3, −1, 2). Use the triple scalar product 1 u · (v × w) = 1 3
3 −1 1 2 = 20 −1 2
Volume of parallelepiped = |u · (v × w)| = 20 1
7.3 Calculate the distance between the plane 2x − 3y + 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 b be the vector b = AB = (x0 , y0 , z0 ) − (1, −4, 3) = (x0 − 1, y0 + 4, z0 − 3). The distance from A to the plane is equal to the absolute value of the scalar projection of b onto the normal vector n = (2, −3, 6) Distance = b |cos θ| b·n |n| |(x0 − 1, y0 + 4, z0 − 3) · (2, −3, 6)| p = 22 + (−3)2 + 62 |(2x0 − 3y0 + 6z0 ) + (−2) + (−12) + (−18)| p = 22 + (−3)2 + 62 =
but (x0 , y0 , z0 ) is a point on the plane so 2x0 − 3y0 + 6z0 = −1 so we get =
|(−1) + (−2) + (−12) + (−18)| p 22 + (−3)2 + 62
Alternate solution: You can use the distance formula but do see the formula in the study guide
2
May/June 2011 Examination Paper Question 8 √ 8.1 Express the complex number 1 + i 3 in polar form 8.2 State De Moivre’s Theorem 8.3 Use De Moivre’s Theorem to find all cube roots of −8 solution: √ 8.1 We need to find r and θ such that 1 + i 3 = r (cos θ + i sin θ) q √ r = 12 + ( 3)2 = 2 √ ! √ 3 1 1+i 3=2 +i 2 2 and 1 cos θ = 2 ∴θ=
π 3
√ 3 sin θ = 2
+ 2πk
for integer k Polar form : For k = 0 √ π π 1 + i 3 = 2 cos + i sin 3 3
8.2 If z = r (cos θ + i sin θ) and n is a positive integer, then
z n = (r (cos θ + i sin θ))n = rn (cos nθ + i sin nθ)
8.3 Let z 3 = −8 and z = r(cos θ + i sin θ) then 3
(r(cos θ + i sin θ)) = 8 (cos π + i sin π) 1
Using De Moivre’s Theorem
r3 (cos 3θ + i sin 3θ)) = 8 (cos π + i sin π) 1
r = 83 = 2 and cos 3θ = cos π
, sin 3θ = sin π
3θ = π + 2πk ∴θ=
2πk π + 3 3
for integer k. π π 2πk 2πk zk = 2 cos + i sin + + 3 3 3 3 for integer k. The distinct cube roots are for k = 0, 1, 2: If k = 0 then π π z0 = 2 cos + i sin =2 3 3
√ ! 3 1 +i 2 2
If k = 1 π 2π 2π π + i sin = 2 (−1 + i0) + + z1 = 2 cos 3 3 3 3 If k = 2
π 4π 4π π + i sin =2 + + z2 = 2 cos 3 3 3 3
2
√ ! 1 3 −i 2 2
Oct/Nov-2011- Examination Paper Question 1 Consider the system of linear equations 2x + y − 3z = 2 2x − 3y + z = 10 −2x + 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 echelon form and find all solutions of the system. solution: We begin by substituting x = a, y = 1 and z = 3 into the system for each linear equation we get −2a + (1) + 3(3) = 2 ∴a=5 2a − 3(1) + 3 = 10 ∴a=5 −2a + (1) + 3 = −6 ∴a=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 | 2 2 −3 1 | 10 −2 1 1 | −6
Performing the elementary row operations −R1 + R2 − − − − − −− > R2 R1 + R3 − − − − − −− > R3 we get 2 1 −3 | 2 0 −4 4 | 8 0 2 −2 | −4 1 2 R2
2 0 0
+ R3 − − − −− > R3 1 −3 | 2 −4 4 | 8 0 0 | 0
The augmented matrix is in generalized row echelon form. 1
let z = t t ∈ R From the equation −4y + 4z = 8 we get y = −2 + t and from the equation 2x + y − 3z = 2 we get x = 2 + t. Thus the solution is given by (2 + t, −2 + t, t) where t ∈ R
2
Oct/Nov-2011- Examination Paper Question 2 Consider the nonhomogeneous system of linear equations x − 2y + z y−z a2 − a − 2 z
= = =
4 3 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) If a = 2, we get 0z = 3. This equation has no solution, hence the system has no solution b) If a 6= 2 and a 6= −1, the equation has a unique solution, hence the system has a unique solution c) If a = −1, then the equation 0z = 0, has infinitely many solutions, hence the system has infinitely many solutions
2.2 Is there any value of a for which the associated homogeneous system has no solution ? Give a reason for your answer
solution: For a homogeneous system of three linear equations in three variable, exactly one 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 no solution
1
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 6= 2 and a 6= −1, then the system has only a trivial solution b) If a = 2 or a = −1 the system has non-trivial solution
2
Oct/Nov-2011- Examination Paper Question 3 Suppose that
2 1 3 A = 0 −1 2 3 4 −1 Evaluate det(A) by expanding along the second column. solution: det(A) 0 2 + (−1) 2 = −1 3 3 −1
2 3 3 − 4 0 2 −1
= −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 2 B = 1 3
0 3 −1 4 2 −1
2 C = 3 0
1 3 4 −1 −1 2
4 D = 0 6
2 6 −2 4 8 −2
6 F = 0 9
1 3 −1 2 4 −1
Then evaluate the following: a) det(B) b) det(C) 1
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 matrix 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) −1 2 − (0) 1 = 2 4 −1 4 −1 2 1 +3 = 2 −1 4 −1 =1 Note: the determinant
1 3 3 + 3 −1 2 −1 3 2
is the same no matter which row or column one uses
Now, find the determinant of F by expanding along the first column we get that −1 2 + 9 1 3 det(F ) = 6 4 −1 −1 2 −1 2 + 3 1 3 = (3) 2 −1 2 4 −1 = 3 det(A) = 3.1 = 3 Question 4 Suppose A is a 2 × 2 matrix and Ei i = 1, 2, 3 are elementary matrices such that E3 E2 E1 A = I2
2
where 1
0 0 E2 = 1 1 E3 = −3
0 1 1 0 0 1
2 = 0 0 = 1 1 = 3
0 1 1 0 0 1
E1 =
2
4.1 Find Ei−1 i = 1, 2, 3.
solution: E1−1 E2−1 E3−1
4.2 Write A as a product of elementary matrices and determine A by multiplying the elementary matrices.
E3 E2 E1 A = I2 ∴ A = E1−1 E2−1 E3−1 2 0 0 1 1 = 0 1 1 0 3 6 2 = 1 0
0 1
4.3 Write A−1 as a product of elementary matrices and determine A−1 by multiplying the elementary matrices. A−1 = E3 E2 E1 1 0 0 = −3 1 1
1 1 2 0 0
0 1
3
=
0 1 2
1 −3
4.4 Use the inverse A−1 to solve the system Ax = b where x1 x= x2
b=
2 1
solution: −1 x= A b 0 1 2 = 1 −3 1 2 1 = −2 thus x1 = 1 and x2 = −2
4
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 L2 is 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: v 1 = 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 parametric equation of line is: L1 : x = 3 − 2t,
y = −1 − t,
z = 2 − 3t where t ∈ R
Direction vector: v 2 = 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 parametric equation of line is: L2 : x = −2 + 6s,
1
y = 3 − 6s,
z = −3 − 2s where s ∈ R From 5.1 b) 5b) v 1 = (−2, −1, −3) and v 2 = (6, −6, −2) Find the dot product v 1 · v 2 = −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 points P1 (3, −1, 2) and P2 (1, −2, −1). Determine the point of intersection of L 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 where tǫR 2
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 line L will intersect the xy-plane at point (5/3, −5/3, 0)
3
Oct/Nov-2011- Examination Paper Question 6 6.1 Determine an equation for the plane that passes through the origin and is parallel to the plane 7x + 4y − 2z + 3 = 0
solution: Since plane passes through the origin the point A(0, 0, 0) must lie in plane. Let B(x, y, z) be any point in the plane, then the vector b = AB = (x, y, z) − (0, 0, 0) = (x, y, z) is a vector in the plane and thus must be perpendicular to the normal of the plane. Since the plane is to be parallel to plane 7x + 4y − 2z + 3 = 0 we take the normal of the plane passing through the origin to have the same normal vector as the plane 7x + 4y − 2z + 3 = 0 which is n = (7, 4, −2) Thus we get b·n=0 (x, y, z) · (7, 4, −2) = 0 7x + 4y − 2z = 0.
1
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 · (v × w)
solution: Triple scalar product 1 u · (v × w) = 1 3
3 −1 1 2 = 20 −1 2
∴ Volume = |20| = 20
6.3 Calculate the distance between the plane 2x − 3y + 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)| p =0 22 + (−3)2 + 62
Since the distance is zero this means that the point lies in the plane.
2
Oct/Nov-2011- Examination Paper Question 7 7.1 State De Moivre’s Theorem solution: If z = r(cos θ + i sin θ) and n is a positive integer, then z n = [r(cos θ + i sin θ)]n = rn (cos nθ + i sin nθ)
7.2 a) Express sin 4θ in terms of powers of sin θ and cos θ b) Use De Moivre’s Theorem to find the cube roots of 1. solution: 7.2 a) Using De Moivre’s Theorem we get that 4
(cos θ + i sin θ) = cos 4θ + i sin 4θ
Now, we get using the binomial theorem that (cos θ + i sin θ)4 =
4 0
cos4 θ +
4 1
cos3 θi sin θ +
Using that i2 = −1,
4 k
=
4 2
cos2 θi2 sin2 θ +
4 3
cos θi3 sin3 θ +
4 4
4 4 i sin θ
4! we get (4 − k)! k!
(cos θ + i sin θ)4 = cos4 θ + i4 cos3 θ sin θ − 6 cos2 θ sin2 θ − i4 cos θ sin3 θ + sin4 θ
Finally, we get sin 4θ = Im cos4 θ + i4 cos3 θ sin θ − 6 cos2 θ sin2 θ − i4 cos θ sin3 θ + sin4 θ = 4 cos3 θ sin θ − 4 cos θ sin3 θ
7.2 b) let z3 = 1 1
where z = r(cos θ + i sin θ) Then r3 (cos θ + i sin θ)3 = 1(cos 0 + i sin 0) we get that r = 1 and cos 3θ = cos (0) and sin 3θ = sin (0) from which we get that
3θ = 0 + 2πk for k = 0, 1, 2
∴θ=
2πk 3
and so the cube roots are: 2πk zk = cos 2πk 3 + i sin 3
k = 0, 1, 2 z0 = cos 0 + i sin 0 = 1 2π 1 z1 = cos 2π 3 + i sin 3 = − 2 + i
√ 3 2
4π 1 z2 = cos 4π 3 + i sin 3 = − 2 − i
√ 3 2
2
May/June 2012 Examination paper Question 1 a) If −1 0 A= 2 3 and
1 B= 3
2 0
Check whether or not AB = BA. solution: −1 0 1 2 −1.1 + 0.3 −1.2 + 3.0 −1 −2 AB = = = 2 3 3 0 2.1 + 3.3 2.2 + 3.0 11 4 BA =
3 6 −3 0
Thus is follows that AB 6= BA b) Reduce the following matrix to reduced row-echelon form 0 3 9 0 2 −4 0 0 3 1/3R1 − − − − > R1 0 1 3 0 2 −4 0 0 3 −2R1 + R2 − − − − > R2 0 1 3 0 0 −10 0 0 3 −1/10R2 − − − −− > R2 0 1 3 0 0 1 0 0 3
1/3R3 − − − − > R3
1
0 1 0 0 0 0
3 1 1
−R1 + R3 − − − − − − > R3 0 1 3 0 0 1 0 0 0 −3R2 + R1 − − − − > R1 0 1 0 0 0 1 0 0 0 c) Solve the system 3x + 4y + z 2x + 3y 4x + 3y − z
= = =
1 0 −2
Augmented matrix: 3 4 1 1 2 3 0 0 4 3 −1 −2
R2 < − − − > R1 0 2 3 0 3 4 1 1 4 3 −1 −2
1/2R1 − − − − > R1 0 0 1 32 3 4 1 1 4 3 −1 −2
−3R1 + R2 − − − −− > R2 −4R1 + R3 − − − − > R3 3 0 0 1 2 0 − 1 1 1 2 0 −3 −1 −2
−2R2 − −− > R2 0 0 1 23 0 1 −2 −2 0 −3 −1 −2
2
3R2 + R3 − − − −− > R3 0 0 1 23 0 1 −2 −2 0 0 −7 −8 −1/7R3 − − − − > R3 0 0 1 23 0 1 −2 −2 8 0 0 1 7
Thus 8 z= 7 y − 2z = −2 3 x+ y =0 2
2 7 3 ∴x=− 7 ∴y=
d) Let A=
2 1 −1 0
Using row operations, find the A−1 and verify it ie show that AA−1 = I = A−1 A solution: A I 2 1 −1 0
1 0 0 1
1/2R1 − − − − > R1 1 1 1 0 2 2 −1 0 0 1 R1 + R2 − − − − > R2 1 0 1 21 2 1 0 21 1 2 2R2 − − − − > R2 1 1 21 0 2 0 1 1 2 −1/2R2 + R1 − − − − > R1 0 −1 1 0 0 1 1 2 Thus we get A
−1
0 −1 = 1 2 3
To verify the A−1 is the inverse of matrix A we perform the product 2 1 0 −1 1 0 AA−1 = = −1 0 1 2 0 1 0 −1 2 1 1 0 A−1 A = = 1 2 −1 0 0 1
4
May/June 2012 Examination paper Question 2 a) Let
−1 5 2 1 2 1
4 A = 6 0 and find its determinant solution:
−1 5 4 5 4 −1 det(A) = 0 −2 +1 2 1 6 1 6 2
= 0 − 2(−26) + 1(14) = 66 b) Show that the matrix
cos θ B = − sin θ 0
sin θ cos θ 0
0 0 1
is invertible for all values of θ solution:
sin θ det(B) = 0 cos θ
cos θ 0 − 0 − sin θ 0
= cos2 θ + sin2 θ = 1
cos θ 0 + 1 − sin θ 0
sin θ cos θ
since det(B) 6= 0 it follows that A is invertible ie. there exists a matrix A−1 such that A−1 A = I = AA−1 since the det(B) = 1 is a constant, independent of θ it follows that B is invertible for all values of θ
c) Given
2 −1 3 C = 1 2 4 5 −3 6 show that det(C) = det(C T )
1
solution:
2 4 − (−1) 1 4 + 3 1 det(C) = 2 5 5 6 −3 6
2 −3
= 2(12 − (−12)) + 1(6 − 20) + 3(−3 − 10) = 2(24) − 14 + 3(−13) = −5
2 C T = −1 3 2 det(C ) = 2 4 T
1 5 2 −3 4 6
−1 2 −1 −3 −3 +5 −1 3 4 3 6 6
= 2(12 − (−12)) − 1(−6 − (−9)) + 5(−4 − 6) = 2(24) − 3 − 50 = −5 Thus we get det(C) = −5 = det(C T )
d) Solve using Cramer’s rule 2x − y = 4 3x + 2y = 13 solution: 4 −1 13 2 x = 2 −1 3 2 21 = =3 7 2 4 3 13 y= 2 −1 3 2 14 =2 = 7
2
May/June 2012 Examination paper Question 3 Consider the vectors 1 v = (−1, 1, ) 2
u = (1, 0, 4) a) proju v solution: v·u proju v = u |u|2 (−1, 1, 12 ) · (1, 0, 4) = |(1, 0, 4)|2 −1 + 0 + 2 = (1, 0, 4) 17 1 (1, 0, 4) = 17
b) Calculate the area of the parallelogram determined by u and v solution: i j u × v = 1 0 −1 1 0 = i 1
k 4 1 2
1 4 1 − j −1 2
1 0 4 1 + k −1 1 2
= −4i − j 29 + k
Area of parallelogram √ q 149 9 2 2 2 = |u × v| = (−4) + (− 2 ) + (1) = 2 c) Find an equation of the plane containing the point (1, 1, −1) and perpendicular 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
(x − 1, y − 1, z + 1) · (3, −1, −1) = 0 3(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, t ∈ R
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 equations of 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 equations of line, we conclude that the point (3, 3, 8) does not lie on the line
2
May/June 2012 Examination paper Question 4 a) Use De Moivre’s theorem to express cos 3θ and sin 3θ in terms of powers of sin θ and cos θ solution: a) Using De Moivre’s theorem we get (cos θ + i sin θ)3 = cos 3θ + i sin 3θ
Now, (cos θ + i sin θ)3 = cos3 θ − 3 cos θ sin2 θ + i3 cos2 sin θ − i sin3 θ Equating real and imaginary parts we get cos 3θ = cos3 θ − 3 cos θ sin2 θ and sin θ = 3 cos2 sin θ − sin3 θ b) Determine the cube roots of i in the polar form
solution: Let z 3 = i and if z = r(cos θ + i sin θ) then π π r3 (cos 3θ + i sin 3θ) = 1 cos + i sin then it follows that 2 2 1
r = (1) 3 and π + 2πk for some integer k 2 π 2πk ∴θ= + 6 3 π 2πk 2πk π + i sin + + zk = cos 6 3 6 3 3θ =
For distinct roots: If k = 0 π π + i sin z0 = cos 6 6 1
If k = 1 5π 5π + i sin z1 = cos 6 6 If k = 2 3π 3π + i sin z2 = cos 2 2 c) If (a + ib)3 = 8, prove that a2 + b2 = 4 (Hint: solve for b2 in terms of a2 then solve for a solution: If (a + ib)3 = 8 we get a3 + i3a2 b − 3ab2 − ib3 = 8 Equating real and imaginary parts we get a3 − 3ab2 = 8 3a2 b − b3 = 0 If b 6= 0 then 3a2 b = b3 3a2 = b2 Substituting this into the first equation a3 − 3a(3a2 ) = 8 −8a3 = 8 ∴ a = −1 If a = −1 then b2 = 3(−1)2 thus a2 + b2 = (−1)2 + 3 = 4
If b = 0 then a3 = 8 ∴a=2 a2 + b 2 = 4 + 0 = 4 2
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
4x1 − 8x2 3x1 − 6x2 −2x1 + 4x2
c) Let A =
= = =
12 9 −6
0 −1 . 1 −1
Find A3 .
2 d) Given B = 1 1
7 1 4 −1. 3 0
Use row operations, find B −1 , verify that BB −1 = I = B −1 B
solutions: a) • If a row does not consist entirely of zeros, then the first nonzero number in the row is a 1
• If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix
• in any two succesive rows that do not consist entirely of zeros, the leading 1 in the lower row occurs further to the right than the leading 1 in the higher row
• each column that contains a leading 1 has zeros everywhere else
1
b) Augmented matrix 12 4 −8 3 −6 9 −6 −2 4 1/4R1 − −− > R1 1 −2 3 3 −6 9 −2 4 −6 −3R1 + R2 − − − − > R2 2R1 + R3 − − − −− > R3 1 −2 0 0 0 0
3 0 0
Let x2 = t, t ∈ R then we get
x1 − 2x2 = 3
x1 = 2x2 + 3 = 2t + 3
solution set = {(2t + 3; t) : t ∈ R}
c) The product −1 1 A2 = AA = −1 0 A3 = A2 A
=
−1 1 0 −1 0 1
2 d) 1 1
7 1 1 4 −1 0 3 0 0
−1 1 0 = −1 0 1 0 0 1 0 0 1
2
R1 < − − − > R3
1 3 1 4 2 7
0 0 0 −1 0 1 1 1 0
1 0 0
0 0 0 −1 0 1 1 1 0
1 −1 −2
−R1 + R2 − − − − > R2 −2R1 + R3 − − − − > R3
1 3 0 1 0 1
−R2 + R3 − − − − > R3
1 3 0 1 0 0
1 0 0 0 −1 0 1 −1 2 1 −1 −1
0.5R3 − − − −− > R3
1 3 0 1 0 0
0 0 −1 0 1 12
0 1 − 21
1 −1 − 21
R3 + R2 − − − − > R2
1 3 0 1 0 0
0 0 0 12 1 1 2
0 1 2 − 21
1 − 23 − 21
−3R2 + R1 − − − − > R1 1 0 0 − 23 − 23 11 2 1 0 1 0 1 − 23 2 12 1 − 2 − 21 0 0 1 2
B −1
3 −2 = 1 2 1 2
− 32
1 2 − 12
11 2 − 23 − 21
3
BB −1
2 = 1 1
3 −2 B −1 B = 12 1 2
3 −2 7 1 4 −1 21 1 3 0 2 − 23
1 2 − 21
11 2 2 − 23 1 − 12 1
− 32
1 2 − 21
7 4 3
11 2 − 23 − 21
1 0 = 0 1 0 0
1 1 0 −1 = 0 1 0 0 0
4
0 1 1
0 1 1
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
4x1 − 8x2 3x1 − 6x2 −2x1 + 4x2
c) Let A =
= = =
12 9 −6
0 −1 . 1 −1
Find A3 .
2 d) Given B = 1 1
7 1 4 −1. 3 0
Use row operations, find B −1 , verify that BB −1 = I = B −1 B
solutions: a) • If a row does not consist entirely of zeros, then the first nonzero number in the row is a 1
• If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix
• in any two succesive rows that do not consist entirely of zeros, the leading 1 in the lower row occurs further to the right than the leading 1 in the higher row
• each column that contains a leading 1 has zeros everywhere else
1
b) Augmented matrix 12 4 −8 3 −6 9 −6 −2 4 1/4R1 − −− > R1 1 −2 3 3 −6 9 −2 4 −6 −3R1 + R2 − − − − > R2 2R1 + R3 − − − −− > R3 1 −2 0 0 0 0
3 0 0
Let x2 = t, t ∈ R then we get
x1 − 2x2 = 3
x1 = 2x2 + 3 = 2t + 3
solution set = {(2t + 3; t) : t ∈ R}
c) The product −1 1 A2 = AA = −1 0 A3 = A2 A
=
−1 1 0 −1 0 1
2 d) 1 1
7 1 1 4 −1 0 3 0 0
−1 1 0 = −1 0 1 0 0 1 0 0 1
2
R1 < − − − > R3
1 3 1 4 2 7
0 0 0 −1 0 1 1 1 0
1 0 0
0 0 0 −1 0 1 1 1 0
1 −1 −2
−R1 + R2 − − − − > R2 −2R1 + R3 − − − − > R3
1 3 0 1 0 1
−R2 + R3 − − − − > R3
1 3 0 1 0 0
1 0 0 0 −1 0 1 −1 2 1 −1 −1
0.5R3 − − − −− > R3
1 3 0 1 0 0
0 0 −1 0 1 12
0 1 − 21
1 −1 − 21
R3 + R2 − − − − > R2
1 3 0 1 0 0
0 0 0 12 1 1 2
0 1 2 − 21
1 − 23 − 21
−3R2 + R1 − − − − > R1 1 0 0 − 23 − 23 11 2 1 0 1 0 1 − 23 2 12 1 − 2 − 21 0 0 1 2
B −1
3 −2 = 1 2 1 2
− 32
1 2 − 12
11 2 − 23 − 21
3
BB −1
2 = 1 1
3 −2 B −1 B = 12 1 2
3 −2 7 1 4 −1 21 1 3 0 2 − 23
1 2 − 21
11 2 2 − 23 1 − 12 1
− 32
1 2 − 21
7 4 3
11 2 − 23 − 21
1 0 = 0 1 0 0
1 1 0 −1 = 0 1 0 0 0
4
0 1 1
0 1 1
October/November 2012 Question 2 a) Given
3 2 −1 3 A = 1 6 2 −4 0 find Adj(A)
b) Let k = −2, 2 B = 3 1
−1 3 2 1 4 5
find det(B) and det(kB) and compare them.
c) Let
2 1 C = 3 4 0 0
0 0 2
and 1 D = 7 5
−1 3 1 2 0 1
det(CD) = det(C) det(D)
d) Given the following system of equations 5x1 + x2 − x3 9x1 + x2 − x3 x1 − x2 + 5x3
= = =
4 1 2
Apply Cramer’s Rule to find x1
1
solution:
a) Co-factors:
1+1
c11 = (−1)
1+2
c12 = (−1)
1+3
c13 = (−1)
2+1
c21 = (−1)
2+2
c22 = (−1)
6 −4
3 = 12 0
1 2
3 =6 0
1 2
6 = −16 −4
2 −4 3 2
−1 =4 0 −1 =2 0
3 c23 = (−1)2+3 2
2 = 16 −4
2 6
−1 = 12 3
3 1
−1 = −10 3
3 1
2 = 16 6
3+1
c31 = (−1)
3+2
c32 = (−1)
3+3
c33 = (−1)
Let C be the co-factor matrix 12 6 −16 C =4 2 16 12 −10 16
2
T 12 6 −16 12 4 12 2 16 = 6 2 −10 Adj (A) = C T = 4 12 −10 16 −16 16 16
2 b) det(B) = 2 4
3 2 3 1 1 = 56 +3 − (−1) 1 4 1 5 5
−4 2 −6 −2B = −6 −4 −2 −2 −8 −10 −6 −4 −6 −2 −4 −2 = −448 −6 −2 det (−2B) = −4 −2 −8 −2 −10 −8 −10 det(−2B) = −448 = (−2)3 56 = (−2)3 det(B) 4 c) det(C) = 2 0 1 det(D) = 1 0
3 0 − 1 2 0
3 4 0 = 10 +0 2 0 0
7 2 7 1 2 = −17 − (−1) +3 1 5 1 5 0
9 −1 8 CD = 31 1 17 10 0 2
1 det(CD) = 9 0
31 17 31 1 17 = −170 − (−1) +8 2 10 2 10 0
det(CD) = −170 = 10(−17) = det(C) det(D) 5 d) 9 1
1 −1 x1 4 1 −1 x2 = 1 −1 5 x3 2
5 1 −1 A = 9 1 −1 1 −1 5 det A = −16
3
Ax1
4 1 = 1 1 2 −1
det Ax1 = 12 x1 =
det(Ax1 ) = det(A)
12 −16
4
−1 −1 5
October/November 2012 Question 3
a) Consider the vectors u = (3, 0, 4) and v = (1, 1, 0)
i) Determine the orthogonal projection proju v 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 perpendicular to 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 parameteric equation are x = 2 − t, y = 2 − 3t, z = 4 + t solution: a) i) proju v =
u·v
2 u |u| 3 (3, 0, 4) · (1, 1, 0) (3, 0, 4) = (3, 0, 4) = 32 + 42 25
ii) First we find the cross product of u and v u×v i = 3 1
j k 0 4 1 0
= −4i + 4j + 3k Area of parallelogram = |u × v| = 4i + 4j + 3k =
√ √ 42 + 42 + 32 = 41units2
iii) Perimeter of parallelogram 1
= 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 line is 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 we get −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 equations it means that the point lies on the line
2
October/November 2012 Question 4
a) Use De Moivre’s Theorem to express sin 4θ in terms of sin θ and cos θ b) Determine the cube roots of −8 c) Let a, b be real numbers such that (a − ib)2 = 4 i) Prove the b = 0 ii) Show that a4 + a2 − 20 = 0 solution: a) Using De Moivre’s theorem we get 4
(cos θ + i sin θ) = cos 4θ + i sin 4θ Now, we get using the binomial theorem that
(cos θ + i sin θ) =
4 0
cos4 θ +
4
4 1
cos3 θi sin θ +
4 2
cos2 θi2 sin2 θ +
Simplifying and use that i2 = −1 we get (cos θ + i sin θ)
4 3
cos θi3 sin3 θ +
4
= cos4 θ + i4 cos3 θ sin θ − 6 cos2 θ sin2 θ − i4 cos θ sin3 θ + sin4 θ Finally, we have
sin 4θ = Im cos4 θ + i4 cos3 θ sin θ − 6 cos2 θ sin2 θ − i4 cos θ sin3 θ + sin4 θ = 4 cos3 θ sin θ − 4 cos θ sin3 θ b)Let z 3 = −8 and if z = r(cos θ + i sin θ) then (r(cos θ + i sin θ))3 = 8 (cos π + i sin π)
1
4 4
4 4 i sin θ
Using De Moivre’s Theorem
r3 (cos 3θ + i sin 3θ) = 8 (cos π + i sin π) 1
r = 8 3 = 2 and cos 3θ = cos π, sin 3θ = sin π
3θ = π + 2πk ∴θ=
π 2πk + 3 3
where k is an integer. π 2πk π 2πk zk = 2 cos + + + i sin 3 3 3 3 where k is an integer The distinct cube roots are: for k = 0 π π + i sin =2 z0 = 2 cos 3 3
√ ! 1 3 +i 2 2
for k = 1
π 2π 2π π + i sin = 2 (−1 + i0) + + z1 = 2 cos 3 3 3 3 for k = 2
π π 4π 4π z2 = 2 cos + i sin =2 + + 3 3 3 3
√ ! 1 3 −i 2 2
2
c) i) if (a − ib) = 4 then a2 − b2 − i2ab = 4.
Two complex numbers are equal if there real parts and imaginary parts are the same. So equating we get
2
a2 − b2 = 4 and 2ab = 0 From the second equation we get a = 0 or b = 0
If a = 0 then from first equation we get −b2 = 4 which means that b must be complex, but since b is real, we cannot have that a = 0. Thus a 6= 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
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 of t, 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
=0 Thus we conclude that x = 19t − 35, y = 25 − 13t, z = t, tion of the system
1.3 a) Compute 3 2 1 3 0 −2 −5 5 1 0 1 −1 2 b) Find A in terms of B if 2A − B = 5(A + 2B) solution: 3 2 a) 5 1 3 2 = 5 1 −12 = 0
1 3 0 −2 −5 0 1 −1 2 1 15 0 −10 − 0 5 −5 10 2 11 6 −10
1.3 b) 2A − B = 5(A + 2B) 2A − B = 5A + 10B 2A − 5A = B + 10B −3A = 11B ∴A=−
11 B 3
1.4 a) Given A=
3 0
Compute A2 − A − 6I2 solution: A2 3 −1 3 −1 = 0 −2 0 −2 3.3 + −1.0 3. − 1 + −1. − 2 = 0.3 + −2.0 0. − 1 + −2. − 2 9 −1 = 0 4 2
−1 −2
t ∈ R is a solu-
A2 − A − 6I2 9 −1 3 −1 1 0 = − −6 0 4 0 −2 0 1 9 − 3 − 6 −1 + 1 + 0 0 0 = = 0 4+2−6 0 0 b) Given B=
6 −4
9 −6
Compute B 2 and say what you observe about B 2 in relation to B. solution: 6 9 6 B2 = −4 −6 −4 0 0 = 0 0
9 −6
Relation of B 2 to matrix B is: B 2 = 0B where 0 is a scalar B + B2 = B 0 1.5 If A = 1
1 −1 1 and B = 1 1 0
Show that A and B are inverses of each other. solution: 0 AB = 1
1 −1 1 1 0 = 1 1 0 0 1
−1 1 0 1 1 0 BA = = 1 0 1 1 0 1
3
May/June 2013- Examination paper b) Given B=
6 −4
9 −6
Compute B 2 and say what you observe about B 2 in relation to B. solution: 6 9 6 2 B = −4 −6 −4 0 0 = 0 0
9 −6
Relation of B 2 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 6= 0 then b12 = −
(b11 ) b21
2
and so the matrix has the form 2 (−b22 ) B = −b22 − b21 b21 b22
For the matrix B 2 are entries are zero despite the fact that all entries of B are non zero In this example, we that product BB = 0 but the matrix B 6= 0
1
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 of t, 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
=0 Thus we conclude that x = 19t − 35, y = 25 − 13t, z = t, tion of the system
1.3 a) Compute 3 2 1 3 0 −2 −5 5 1 0 1 −1 2 b) Find A in terms of B if 2A − B = 5(A + 2B) solution: 3 2 a) 5 1 3 2 = 5 1 −12 = 0
1 3 0 −2 −5 0 1 −1 2 1 15 0 −10 − 0 5 −5 10 2 11 6 −10
1.3 b) 2A − B = 5(A + 2B) 2A − B = 5A + 10B 2A − 5A = B + 10B −3A = 11B ∴A=−
11 B 3
1.4 a) Given A=
3 0
Compute A2 − A − 6I2 solution: A2 3 −1 3 −1 = 0 −2 0 −2 3.3 + −1.0 3. − 1 + −1. − 2 = 0.3 + −2.0 0. − 1 + −2. − 2 9 −1 = 0 4 2
−1 −2
t ∈ R is a solu-
A2 − A − 6I2 9 −1 3 −1 1 0 = − −6 0 4 0 −2 0 1 9 − 3 − 6 −1 + 1 + 0 0 0 = = 0 4+2−6 0 0 b) Given B=
6 −4
9 −6
Compute B 2 and say what you observe about B 2 in relation to B. solution: 6 9 6 B2 = −4 −6 −4 0 0 = 0 0
9 −6
Relation of B 2 to matrix B is: B 2 = 0B where 0 is a scalar B + B2 = B det(B) = det(B 2 ) = 0 In this example we that product BB = 0 but the matrix B 6= 0 0 1 −1 1 1.5 If A = and B = 1 1 1 0 Show that A and B are inverses of each other. solution: 0 AB = 1 BA =
1 −1 1 1 0 = 1 1 0 0 1
−1 1 0 1 1 0 = 1 0 1 1 0 1
Question 2 2.1 Find det(A) if
3
a) cos θ A= sin θ
− sin θ cos θ
solution: det(A) = cos2 θ + sin2 θ = 1
2.2 b) a+1 a A= a a−1 solution: det(A) = (a + 1)(a − 1) − a2 = a2 − 1 − a2 = −1 2.2 Using the cofactor expansion find the 3 0 5 1 B= 2 6 −6 3 solution:
1 2 0 5 2 det(B) = 3 6 0 −1 − 0 2 0 3 1 0 −6 1 0 −1 6 −1 −2 + 0 6 = 3 1 3 1 0 3 0 = 3 [(1).(1) − (2).(3)]
det(B) where 0 0 2 0 0 −1 1 0
5 1 0 −1 + 0 2 6 −6 3 0 0 1
= 3(−5) = −15 2.3 Let C=
4 3
1 2
and show that
det(C −1 ) =
1 det(C)
solution: det(C) = 8 − 3 = 5 Since det(C) 6= 0, the matrix C has an inverse.
4
5 1 2 0 −1 − 0 2 6 0 −6 3 1 0
C −1 =
1 5
2 −1 −3 4
Now, we find the determinant, we get 5 4 3 1 1 1 2 −1 − = = = det(C ) = 5 5 5 5 25 5 det(C)
2.4 Let 3 2 D= 1 −1 find det(2D) and compare it to det(D) solution: det(D) = −3 − 2 = −5
2D =
6 2
4 −2
det(2D) = (6)(−2) − (4)(2) = −12 − 8 = −20 = 22 (−5) = 22 det(D)
2.5 Solve the system using Cramer’s rule 2x + y = 1 3x + 7y = −2
solution 2 1 x 1 = 3 7 y −2 1 1 −2 7 x = 2 1 3 7 9 7+2 = = 14 − 3 11
5
2 1 3 −2 y = 2 1 3 7 −7 −4 − 3 = = 14 − 3 11
6
May/June 2013- Examination paper Question 3 Consider the vectors √ u = (1, 0, 3) √ v = (1, 3, 0)
3.1 Find proju v solution: proju v v·u u k u k2 √ √ √ (1, 3, 0) · (1, 0, 3) √ (1, 0, 3) = 2 k (1, 0, 3) k √ √ √ 1.1 + 3.0 + 0. 3 √ (1, 0, 3) = 2 ( 1 + 0 + 3) √ 1 = (1, 0, 3) 4 =
3.2 Calculate the area of the parallelogram determined by u and v. solution: Find the cross product: i j √k u × v = 1 √0 3 1 3 0 √ √ 0 1 1 3 3 −j + k = i √ 1 1 0 3 0 √ √ = i(−3) − j(− 3) + k 3
0 √ 3
√ √ Area of parallelogram = i(−3) − j(− 3) + k 3
=
√ √ 9 + 3 + 3 = 15
3.3 Find an equation of the plane containing u and v solution: Normal vector: √ √ n = u × v = i(−3) − j(− 3) + k 3 Equation of plane: 1
Let (x, y, z) be a point on the plane then (x − 1, y −
√ √ √ 3, z) · (−3, 3, 3) = 0
−3(x − 1) + −3x +
√ √ √ 3(y − 3) + 3(z) = 0
√ √ 3y + 3z = 0
3.4 Determine the parametric equations of the plane in 3.3 solution: From 3.3, we get √ √ 3 3 x= y+ z 3 3 If we let y = s and z = t then the parametric equation of plane is: √ √ 3 3 s+ t x= 3 3 y=s z=t where t, s ∈ R
2
May/June 2013- Examination paper Question 4 4.1 Use De Moivre’s theorem to express cos 2θ in terms of powers of sin θ and cos θ solution: cos 2θ = Re[(cos 2θ + i sin 2θ)] = Re[(cos θ + i sin θ)2 ] = Re[cos2 θ + 2i cos θ sin θ − sin2 θ] = cos2 θ − sin2 θ 4.2 Determine the cube root of −1 in the form a + ib , a, b ∈ R solution: Let z 3 = −1 If z = r(cos θ + i sin θ) then z 3 = r3 (cos 3θ + i sin 3θ) = 1(cos π + i sin π)
It follows that r = 11/3 = 1 and 3θ = π + 2πk for some integer k
∴ zk = cos
π + 2πk 3
For distinct roots: If k = 0 z0 = cos
π 3
+ i sin
π 3
√ 1 3 = +i 2 2
1
+ i sin
π + 2πk 3
If k = 1 z1 = cos
π + 2π 3
+ i sin
π + 2π 3
= −1 + 0i
π + 4π 3
+ i sin
π + 4π 3
=
If k = 2 z2 = cos
√ 1 3 −i 2 2
4.3 Use De Moivre’s theorem to determine (−1 + i)134 in the form x + iy solution: Polar form | − 1 + i| =
p √ (−1)2 + (1)2 = 2
−1 + i =
√
2
1 −1 √ + i√ 2 2
It follows that 1 1 cos θ = − √ and sin θ = √ 2 2 3π + 2πk for some integer k ∴θ= 4 The polar form is: √ 3π 3π −1 + i = 2 cos + i sin 4 4 Using De Moivre’s theorem we get that (−1 + i)134 = √ 134 3π 3π cos 134 = ( 2) + i sin 134 4 4 π π 67 =2 cos 201 + i sin 201 2 2 π π 67 cos(200 + 1) + i sin(200 + 1) =2 2 2 π π 67 cos(2π(50) + ) + i sin(2π(50) + ) =2 2 2 π π = 267 cos + i sin 2 2 = 0 + i267
2
Question 2 a+1 a a) i) A = a a−1 det(A) = (a + 1)(a − 1) − a2 = a2 − 1 − a2 = −1 ii) det(B 2 C −1 AB −1 C T ) 1 1 det(A) det(C) det(C) det(B) 1 1 = 22 (−1) (3) = −2 3 2 1 a b b) A = −a 1 c −b −c 1 = (det(B))2
det(A) −a 1 −a c 1 c − a = 1 −b 1 + b −b −c −c 1 = 1(1 + c2 ) − a(−a + bc) + b(ac + b)
= 1 + c2 + b2 + a2 c)i) (det B)n = det(B n ) = det(0) = 0 ∴ det(B) = 0 it follows that matrix B is invertible. ii) det(I) = det(C 2 ) = (det(C))2 ∴ det(C) = ±1 3 4 d) let A = 2 −1 det(A) = −3 − 8 = −11 9 4 Ax = −1 −1
1
det(Ax ) = −9 + 4 = −5 5 ∴x= 11 3 9 Ay = 2 −1 det(Ay ) = −3 − 18 = −21 ∴y=
21 11
2
Question 3 a) i) proj u v u·v u |u|2 (1, 0, 2) · (1, 1, 0) (1, 0, 2) = |(1, 0, 2)|2 1 = (1, 0, 2) 5
=
ii) u × v i j k = 1 0 2 1 1 0 0 2 1 2 1 0 −j +k = i 1 0 1 0 1 1 = −2i + 2j + k
q √ 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 position vector 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 = 2 − 3t, z = 4 + t then there is a t such that 0=2−t 1
−4 = 2 − 3t 6=4+t Solving the system of equations we get t = 2 ∴ the point lies on the line.
2
Question 4 a) (cos θ + i sin θ)3 = cos 3θ + i sin 3θ cos 3θ = Re(cos θ + i sin θ)3 = Re =
3 0
3
0
cos3 θ +
cos3 θ −
3 2
3 1
cos2 θ sin θi −
3 2
cos θ sin2 θ −
cos θ sin2 θ
= cos3 θ − 3 cos θ sin2 θ b) Let z 3 = 8, where z = r(cos θ + i sin θ) Then z 3 = r 3 (cos 3θ + i sin 3θ) = 8(cos 0 + i sin 0) r = 81/3 ∴θ=
cos 3θ = cos 0 and
sin 3θ = sin 0
2πk for integer k 3
The distinct roots is given by 2πk 2πk zk = 2 cos for k = 0, 1, 2 + i sin 3 3
z0 = 2 (cos 0 + i sin 0) 2π 2π + i sin z1 = 2 cos 3 3 4π 4π + i sin z2 = 2 cos 3 3 c) (a − ib)2 = 4i a2 − 2iab − b2 = 4i a2 − b2 + i(−2ab − 4) = 0
1
3 3
i sin3 θ
since two complex numbers are equal iff the real parts and imaginary part are equal, it follows i) a2 − b2 = 0 ii) −2ab − 4 = 0 ∴ ab = −2
2
May-June 2014 Examination paper Question 1 1 2 0 a)(i) 0 1 1 (ii) x = −2y ∴ x = −2 b) 1 1 −1 4 2 1 3 0 0 1 −5 8
−2R1 + R2 → R2 1 1 −1 4 0 −1 5 −8 0 1 −5 8 −R2 → R2 1 1 −1 4 0 1 −5 8 0 1 −5 8
−R2 + R3 → R3 1 1 −1 4 0 1 −5 8 0 0 0 0
1
Question 1. c)
1 −2 2 1 2 −4 1 i) AB = 2 1 1 −1 −1 3 = 0 1 0 1 −1 −2 5 0 1 2 −4 1 −2 2 1 0 BA = −1 −1 3 2 1 1 = 0 1 −1 −2 5 1 0 1 0 0
0 0 1 0 = I3 0 1 0 0 = I3 1
∴ B = A−1 ii) AX = Y A−1 (AX) = A−1 Y IX = A−1 Y X = A−1 1 2 −4 3 11 X = −1 −1 3 0 = −9 ∴ x1 = 11, x2 = −9, x3 = −13 −1 −2 5 −2 −13
1
Question 1 d) Since C is an inverse of B, we have CB = I.
Multiplying both sides on the right by D gives (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
Question 1 e) Let T be a m × n matrix T = −T ⇐⇒ T + (T ) = −T + (T ) ⇐⇒ T + T = 0 ( 0 is the zero m × n matrix) ⇐⇒ 2T = 0 1 ⇐⇒ T = 0 2 ⇐⇒ T = 0
1
Question 2 a) Finding the determinant using co-factor method by expanding along the second row det(F ) 8 1 2 = 3 0 9 1 2 −1 8 1 8 2 1 2 +0 = −3 1 −1 − 9 1 2 2 −1 8 8 2 1 2 − 3 + 0 = 3 − 1 1 −1 2 −1
= 3 det(E)
1 2
Since matrix F results when the second row of E is multiplied by scalar 3, then det(F ) = 3 det(E)
1
2.b) i)det(G) = 4 − 6 = −2
det(H) = 4 + 1 = 5
1 2 GH = 3 4
4 3 2 −1 = 10 5 1 2
det(GH) = 20 − 30 = −10
∴ det(GH) = −10 = (−2)(5) = det(G) det(H) 2 −1 1 2 −1 0 = ii) HG = 1 2 3 4 7 10 det(HG) = −10 ∴ det(GH) = −10 = det(HG)
1
Question 2 c) If J, K are n × n matrices then JJ −1 = I and KK −1 = I.
Then det(J) det(J −1 ) = 1
and det(K) det(K −1 ) = 1
Thus det(J −1 K −1 JK)
= det(J −1 ) det(K −1 ) det(J) det(K)
=
1 1 det(J) det(K) det(J) det(K)
=1
1
2.d) λ − 1 −4 =0 det 0 λ−4 (λ − 1)(λ − 4) = 0 ∴ λ = 1 or λ = 4
1
Question 2 e) Let −2 3 −1 2 −1 A= 1 −2 −1 1 1 1 −1 2 −1 2 − 1 − 3 det(A) = −2 −2 −1 −2 1 −1 1 = −2(1) − 3(−1) − 1(3) = −2 1 3 −1 2 −1 A1 = 4 −3 −1 1 4 4 −1 2 −1 2 − 3 − 1 det(A1 ) = 1 −3 −1 −1 1 −3 1 = 1(1) − 3(1) − 1(2) = −4
x1 =
det(A1 ) −4 = =2 det(A) −2
−2 1 −1 4 −1 A2 = 1 −2 −3 1 1 1 −1 4 −1 4 −1 − 1 det(A2 ) = −2 −2 −3 −2 1 −3 1 = −2(1) − 1(−1) − 1(5) = −6
x2 =
det(A2 ) −6 = =3 det(A) −2
−2 3 1 2 4 A3 = 1 −2 −1 −3 1 1 2 2 4 4 +1 − 3 det(A3 ) = −2 −2 −1 −2 −3 −1 −3 1
= −2(−2) − 3(5) + 1(3) = −8 x3 =
det(A3 ) −8 = =4 det(A) −2
2
Question 3 a) i) u · a = 4.2 + (−1)(−1) + 3.2 = 8 + 1 + 6 = 15 ii) i j k u × a = 2 −1 3 = i + 8j + 2k 4 −1 2 iii) p √ kuk = 22 + (−1)2 + 32 = 14 kak =
p
42 + (−1)2 + 22 =
√
21
iv) Vector projection of u in the direction of a 15 u·a (4, −1, 2) = 2 a = 21 kak Vector projection of u perpendicular to a u·a 15 (4, −1, 2) =u− 2 a = (2, −1, 3) − 21 kak
v) The vector u × a is perpendicular to both u and a u × a = i + 8j + 2k
1
Question 3 b (i) Find a point in plane 2x − 3y + 6z = 1 Let z = 0 and y = 1 then 2x − 3y = 1 ∴x=2 (2, 1, 0) is a point on the plane 2x − 3y + 6z = 1 Form a vector from this point A(2, 1, 0) to point B(1, −4, −3) Let v = (1, −4, −3) − (2, 1, 0) = (−1, −5, −3) The normal vector of plane 2x − 3y + 6z = 1 is : n = (2, −3, 6) The distance between the plane and point is: Distance = ||v| cos θ| v · n = |v| |v| |n| |v · n| = |n| |(−1, −5, −3) · (2, −3, 6)| √ = 22 + 32 + 62 |−5| 5 =√ = 7 49 Alternate solution Use the formula
1
Question 3 ii) Let u = P1 P2 = (−1, −1, 1) − (−2, 1, 3) = (1, −2, −2)
v = P1 P3 = (3, 0, 2) − (−2, 1, 3) = (5, −1, −1)
The normal vector of plane: i j k n = u × v = 1 −2 −2 = −9j + 9k 5 −1 −1 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
Question 4 a) cos 3θ = Re (cos 3θ + i sin 3θ) = Re (cos θ + i sin θ)3 = Re
3 0
cos3 θ +
3 1
cos2 θ(i) sin θ +
3 2
cos θ(i)2 sin2 θ +
3 3
= Re cos3 θ + 3 cos2 θ(i) sin θ + 3 cos θ(i)2 sin2 θ + (i)3 sin3 θ = Re cos3 θ + 3 cos2 θ(i) sin θ − 3 cos θ sin2 θ − i sin3 θ = cos3 θ − 3 cos θ sin2 θ
1
(i)3 sin3 θ
Question 4 b) Let z 4 = −16 Then this equation in Polar form is given by : if z = r(cos θ + i sin θ) we get
r 4 (cos θ + i sin θ)4 = 16(cos π + i sin π) Using De Moivre’s Theorem we get
r 4 (cos 4θ + i sin 4θ) = 16(cos π + i sin π) r 4 = 16 and cos 4θ = cos(π); sin 4θ = sin(π) ∴ r = 161/4 = 2 4θ = π + 2πk for k ∈ Z ∴θ=
π kπ + 4 2
Thus the roots are given by: π kπ π kπ zk = 2 cos + i sin + + 4 2 4 2 For distinct roots k = 0, 1, 2, 3 z0 z1 z2 z3
π π 1 1 + i sin = 2 √ + i√ = 2 cos 4 4 2 2 π π π π 1 1 + i sin = 2 −√ + i√ + + = 2 cos 4 2 4 2 2 2 π π 1 1 = 2 cos + π + i sin +π = 2 −√ − i√ 4 4 2 2 π 3π 1 1 π 3π + i sin = 2 √ − i√ + + = 2 cos 4 2 4 2 2 2
1
Question 4 c) √ z1 = 1 + i 3 Represent in polar form: |z1 | =
√
12 + 3 = 2 √ ! 1 3 +i 2 2
√ |z1 | (1 + i 3) = 2 z1 = |z1 |
√ 3 1 cos θ = and sin θ = 2 2 ∴ tan θ =
θ=
√ sin θ = 3 cos θ
π 3
π π Polar form: z1 = 2 cos + i sin 3 3 √
z2 =
3+i
Represent in polar form: |z2 | =
√
12 + 3 = 2
|z2 | √ ( 3 + i) = 2 z2 = |z2 | cos θ =
√
3 1 +i 2 2
!
√
3 1 and sin θ = 2 2
1
∴ tan θ =
θ=
1 sin θ =√ cos θ 3
π 6
π π Polar form: z2 = 2 cos + i sin 6 6 z1 z2 π π π π .2 cos + i sin = 2 cos + i sin 3 π π 3 π 6 π 6 2 = 2 cos + sin + + 3 6 3 6 π π = 4 cos + i sin 2 2 z1 z2 π π 2 cos + i sin 3 3 = π π 2 cos + i sin π 6 π 6 π π = cos + i sin − − 3 6 6 π π 3 = cos + i sin 6 6
2
