MAT1503 Exam Solutions

8/10/2019 MAT1503 Exam Solutions

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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

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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

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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

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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

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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

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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

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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

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C1 =

0 0 10 1 01 0 0

F =

1 0 02 1 00 0 1

F1 =

1 0 02 1 00 0 1

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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

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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)

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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

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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

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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 )

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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

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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.

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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

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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

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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

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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)

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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

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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

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= 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

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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,

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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

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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)

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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.

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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(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

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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

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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

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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

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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


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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


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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


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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

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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


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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

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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


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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

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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

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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

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Ax1 =

4 1 11 1

1

2 1 5

detAx1 = 12

x1 =det(Ax1)

det(A) = 12

16

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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

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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

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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)

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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

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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

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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

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= 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

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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

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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


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

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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

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= 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

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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


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

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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.

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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

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y=2 1

3

22 13 7

=4 3

14 3 =

7

11

6


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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

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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

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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

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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

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det(Ax) = 9 + 4 = 5

x = 5

11

Ay =

3 92 1

det(Ay) = 3 18 = 21

y =21

11

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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

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4 = 2

3t

6 = 4 +tSolving the system of equations we get t= 2

the point lies on the line.

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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

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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

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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

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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

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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

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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

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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


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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)

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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

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2.d)

det

1

40 4

= 0

( 1)( 4) = 0

= 1 or = 4

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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

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=

2(

2)

3(5) + 1(3) =

8

x3=det(A3)

det(A) =

8

2= 4

2


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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

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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

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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

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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


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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

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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

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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

MAT1503 Exam Solutions

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