(7,1)
Contents
1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.A 1.B
Solutions to textbook problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Linear systems and Gaussian elimination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Vectors and vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Matrices and determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Eigenvectors and eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Unconstrained optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Constrained optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 Differential equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 Difference equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Systems of differential and difference equations . . . . . . . . . . . . . . . . . . . . . . . . . 95 Complex numbers and trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . 100 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11
Exam problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Final exam BI, January 2021 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Final exam BI, November 2019 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Final exam BI, November 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Final exam BI, November 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Final exam BI, December 2016. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Final exam BI, December 2015. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Final exam BI, December 2014. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Final exam BI, December 2013. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Final exam BI, December 2012. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Final exam BI, December 2011. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Final exam BI, December 2010. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
107 107 108 109 110 112 113 114 116 117 119 120
7
(8,1)
Contents
3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11
8
Solutions to exam problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to final exam BI, January 2021 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to final exam BI, November 2019 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to final exam BI, November 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to final exam BI, November 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to final exam BI, December 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to final exam BI, December 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to final exam BI, December 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to final exam BI, December 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to final exam BI, December 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to final exam BI, December 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to final exam BI, December 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
123 123 130 135 141 148 153 161 166 173 178 183
(9,1)
CHAPTER 1
Solutions to textbook problems
1.1 Linear systems and Gaussian elimination Solution 1.1 The coefficient matrix A and the augmented matrix ðAjbÞ are: 7 2 4 7 2 , ðAjbÞ ¼ a) A ¼ 4 4 7 4 4 1 0 0 1 0 1 1 1 0 1 1 1 1 b) A ¼ @ 1 1 1 2 A, ðAjbÞ ¼ @ 1 1 1 2 3 A 1 4 7 3 1 4 7 3 14 Solution 1.2 The linear systems are: a) 2x þ y ¼7 x þ 3y þ 4z ¼ 5 5x 4y þ 2z ¼ 13 b) x þ y þ z ¼ 7 x þ 2y þ 4z ¼ 12 x þ 3y þ 9z ¼ 19 c) x þ y z þ w ¼ 12 2x 3y þ 4z þ 7w ¼ 10 Solution 1.3 We first use substitution to solve the system. The first equation gives x ¼ 7 y z. We substitute this expression in the last two equations, and this gives the equations y þ 3z ¼ 5, 2y þ 8z ¼ 12 Next we solve the first of these new equations for y, and get y ¼ 5 3z. When we substitute this in the second equation, we get 2ð5 3zÞ þ 8z ¼ 12, or 2z ¼ 2. This means that z ¼ 1, that y ¼ 5 3ð1Þ ¼ 2, and that x ¼ 7 2 1 ¼ 4. The system has one solution ðx, y, zÞ ¼ ð4, 2, 1Þ.
9
(10,1)
1
Solutions to textbook problems
Then we solve the system by Gaussian elimination. We write down the augmented matrix and use the Gaussian process 1 1 0 0 0 1 1 1 1 7 1 1 1 7 1 1 1 7 @ 1 2 4 12 A ! @ 0 1 3 5 A ! @ 0 1 3 5 A 0 0 2 2 1 3 9 19 0 2 8 12 This is an echelon form, and back substitution gives 2z ¼ 2, or z ¼ 1 in the last equation, y þ 3z ¼ 5, or y ¼ 5 3ð1Þ ¼ 2 in the second equation and x þ y þ z ¼ 7, or x ¼ 7 2 1 ¼ 4 in the first equation. Hence we find one solution ðx, y, zÞ ¼ ð4, 2, 1Þ also with Gaussian elimination. Solution 1.4 We solve the system by Gaussian elimination, and mark the pivots: a) We use the Gaussian process 4 6 4 4 2 1 1 1
!
4 6 0 2
4 4 3 3
We find an echelon form, and we see from the pivot positions that there is one free variable, and infinitely many solutions. The free variable is z, and back substitution gives 2y ¼ 3 3z, or y ¼ 3=2 3z=2 in the second equation, and 4x ¼ 4 6y 4z ¼ 4 6ð3=2 3z=2Þ 4z ¼ 5 þ 5z, or x ¼ 5=4 5z=4 in the first equation. The solutions are ðx, y, zÞ ¼ ð5=4 5z=4, 3=2 3z=2, zÞ with z free. b) We use the Gaussian process 1 0 6 1 7 @ 3 1 4 A ! 6 2 1
1 6 1 7 @ 0 1=2 1=2 A 0 1 8 0
0 !
6 @0 0
1 1 7 1 1 A 0 9
We find an echelon form, and since there is a pivot in the last column, there are no solutions. Solution 1.5 We solve the linear systems by Gaussian elimination, and give a geometric description of the solutions: a) The linear system 3x 4y ¼ 6 is already in echelon form, with augmented matrix ð 3 4 j 6 Þ. It has infinitely many solutions, since x is a basic variable and y is free, and we have that 3x ¼ 4y þ 6, which means that x ¼ 4y=3 þ 2. The set of solutions is V ¼ fð4y=3 þ 2, yÞ : y free g. This is a line in R2 .
10
(11,1)
1.1 Linear systems and Gaussian elimination
b) We use a Gaussian process to obtain an echelon form of the system, and get 1 0 1 0 1 0 1 1 1 4 1 1 1 4 1 1 1 4 @ 2 1 3 3 A ! @ 0 3 1 5 A ! @ 0 3 1 5 A 0 3 1 5 0 0 0 0 3 0 4 7 The system has infinitely many solutions, since x and y are basic variables and z is free. We have that 3y þ z ¼ 5, which gives y ¼ 5=3 þ z=3, and x þ y þ z ¼ 4, which gives x ¼ 4 ð5=3 þ z=3Þ z ¼ 7=3 4z=3. The set of solutions can be written V ¼ fð7=3 4z=3, 5=3 þ z=3, zÞ : z free g. This is a line in R3 . c) We use a Gaussian process to obtain an echelon form of the system, and get 6 3 12 9 2 1 4 3 6 3 12 9 ! ! 6 4 2 2 3 2 1 1 0 7 14 7 The system has infinitely many solutions, since x and y are basic variables and z is free. We have that 7y þ 14z ¼ 7, which gives y ¼ 1 þ 2z, and 6x þ 3y 12z ¼ 9, which gives 6x ¼ 9 3ð1 þ 2zÞ þ 12z ¼ 6 þ 6z, or x ¼ 1 þ z. The set of solutions is V ¼ fð1 þ z, 1 þ 2z; zÞ : z free g. This is a line in R3 . Solution 1.6 We solve the linear systems by Gaussian elimination and give a geometric description of the solutions for each value of h: a) The linear system hx hy ¼ 3 is already in echelon form, with augmented matrix ð h h j 3 Þ. If h ¼ 0, then there is a pivot in the last column, and there are no solutions. If h 6¼ 0, then there is a pivot in the x-column, and there are infinitely many solutions since x is a basic variable and y is free. In this case, we have that hx ¼ hy þ 3, which means that x ¼ y þ 3=h. The set of solutions is V ¼ fðy þ 3=h; yÞ : y free g. This is a line in R2 for each h 6¼ 0. b) We use a Gaussian process to obtain an echelon form of the system, and get h 1 2 h 1 2 ! 3 6 12 0 0 12 3h There is a pivot in the x-column, and if 12 3h 6¼ 0, then there is a pivot in the last column. This means that if h 6¼ 4, there are no solutions. If h ¼ 4, then there are infinitely many solutions, since x is a basic variable and y is free. We have that x þ 2y ¼ h with h ¼ 4, which gives x ¼ 4 2y. The set of solutions is V ¼ fð4 2y, yÞ : y free g. This is a line in R2 for h ¼ 4.
11
(12,1)
1
Solutions to textbook problems
c) We use a Gaussian process to obtain an echelon form of the system, and get 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 @ 2 1 2 A ! @ 0 3 1 2A 1 3 4 A ! @ 0 3 0 6 2 h 1 0 0 0 hþ3 1 7 1 h There are pivots in the x- and y-columns, and a pivot in the last column if h þ 3 6¼ 0. This means that if h 6¼ 3, there are no solutions. If h ¼ 3, then there are infinitely many solutions, since x and y are basic variables and z is free. We have that 3y þ z ¼ 2, which gives 3y ¼ 2 z, or y ¼ z=3 2=3, and x þ y þ z ¼ 1, which gives x ¼ 1 ðz=3 2=3Þ z ¼ 5=3 4z=3. The set of solutions can be written V ¼ fð5=3 4z=3, z=3 2=3, zÞ : z free g. This is a line in R3 for h ¼ 3. Solution 1.7 We first use a Gaussian process that gives an echelon form of the matrix: 1 1 1 0 0 0 1 1 1 3 1 1 1 3 1 1 1 3 @1 2 4 7A ! @0 1 3 4A ! @0 1 3 4A 0 2 8 10 0 0 2 2 1 3 9 13 The pivot positions are marked in blue. Then we multiply the last row with 1=2 (or divide by 2) to get all pivots equal to one, and continue the Gaussian process to obtain a reduced echelon form: 1 1 1 0 0 0 1 1 1 3 1 1 0 2 1 0 0 1 @0 1 3 4A ! @0 1 0 1A ! @0 1 0 1A 0 0 1 1 0 0 1 1 0 0 1 1 This is a reduced echelon form, with all pivots equal to 1 and all entries over pivots equal to zero. Note that to get zeros over a pivot, we use the pivot in the same column, and we start with the last variable column and move towards the first. Solution pivots in 0 1 1 @ 2 1 1 4
1.8 We use a Gaussian process to find an echelon form, and mark the blue: 1 1 1 0 0 1 1 1 1 5 1 1 5 1 1 1 1 5 2 1 11 A ! @ 0 3 0 3 1 A ! @ 0 3 0 3 1 A 0 3 0 4 2 1 5 7 0 0 0 1 3
We see that there are infinitely many solutions, since x, y and w are basic and z is a free variable. We get w ¼ 3, 3y 3w ¼ 1, which gives 3y ¼ 1 þ 3ð3Þ ¼ 10, or y ¼ 10=3, and x þ y þ z þ w ¼ 5, which gives x ¼ 5 ð 10=3Þ z ð3Þ ¼ 16=3 z. The set of solutions is V ¼ fð16=3 z, 10=3, z, 3Þ : z free g.
12
(13,1)
1.1 Linear systems and Gaussian elimination
Solution 1.9 We consider the various possible pivot positions of the 4 4 linear system, where the first three equations are the same as in the previous problem. For each case that it is possible to obtain, we give an example of a non-degenerate equation ax þ by þ cz þ dw ¼ e that give the required number of solutions: a) There is one solution if we choose the equation z ¼ 1, since this would give the following echelon form when we switch the last two rows: 1 0 1 1 1 1 5 B 0 3 0 3 1 C C B @0 0 1 0 1 A 0 0 0 1 3 b) There are infinitely many solutions with one degree of freedom if we choose the equation 4x þ 4y þ 4z þ 5w ¼ 23 (the sum of the first three equations), since we would obtain the echelon form 1 0 1 1 1 1 5 B 0 3 0 3 1 C C B @0 0 0 1 3 A 0 0 0 0 0 c) There are no solutions if we choose the equation 4x þ 4y þ 4z þ 5w ¼ 0, since we would obtain the echelon form 1 0 1 1 1 1 5 B 0 3 0 3 1 C C B @0 0 0 1 3 A 0 0 0 0 23 d) Since the echelon form in the previous problem had three pivot positions, the augmented matrix of the 4 4 linear system has at least three pivot positions. It is therefore not possible to have infinitely many solutions with two degrees of freedom (this would mean that there are just two pivots).
13
(14,1)
1
Solutions to textbook problems
Solution 1.10 The matrices A and B are row equivalent if there is a Gaussian process that transforms A into B. We can check this by finding the reduced echelon form of A, and see if it is the same as the reduced echelon form of B. a) We see that A and B are row echelon form: 1 3 1 A¼ ! 0 5 0 1 1 1 B¼ ! 3 0 0
equivalent since they have the same reduced 3 1
!
1 3
!
1 0 0 1 1 1 0 1
!
1 0 0 1
b) We see that A and B are not row equivalent since they have different reduced echelon forms; 1 3 1 3 1 3 1 0 A¼ ! ! ! 4 5 0 7 0 1 0 1 1 2 1 2 B¼ ! 3 6 0 0 c) We see that A and B are not echelon forms; 1 3 2 1 A¼ ! 0 0 1 0 2 2 5 6 ! B¼ 0 4 0 0
row equivalent since they have different reduced
d) We see that A and B are not echelon forms; 1 3 2 1 A¼ ! 0 0 1 0 2 5 6 2 B¼ ! 0 0 4 0
row equivalent since they have different reduced
14
3 0 0 1 5 6 1 0
3 0 0 1 5 6 0 1
!
2 0
0 1
6 0
!
1 0 3 0 1 0
!
2 5 0 0 0 1
!
1 5=2 0 0 0 1
(15,1)
1.1 Linear systems and Gaussian elimination
Solution 1.11 We use Gaussian elimination, and obtain the following echelon form of the augmented matrix: 1 0 1 0 1 0 3 3 1 1 1 1 1 1 1 1 1 3 @2 3 a bA ! @0 1 a 2 b 6 A ! @ 0 1 a 2 b 6 A 4 B 0 a 1 0 0 0 A 1 a 1 7 We have used the symbols A ¼ ða 2Þða 1Þ and B ¼ 4 ða 1Þðb 6Þ to write the echelon form in a more compact way. If A 6¼ 0, then there are pivots in all variable columns, and there is one solution. If A ¼ 0 and B 6¼ 0 there is a pivot in the last column, and there are no solutions. If A ¼ B ¼ 0 then there are infinitely many solutions, and one degree of freedom. We have that A ¼ 0 for a ¼ 1 and a ¼ 2, and A ¼ B ¼ 0 for a ¼ 2 and b ¼ 10. Therefore we have one solution if a 6¼ 1 and a 6¼ 2, infinitely many solutions (with one degree of freedom) if a ¼ 2 and b ¼ 10, and no solutions in all other cases. Solution 1.12 We use Gaussian elimination to solve the system, and find the following echelon form of the augmented matrix: 1 7 7 3 1 1 7 7 3 1 ! 1 6 6 4 2 0 1 1 1 1 We see that there are infinitely many solutions, with z and w as free variables. We solve for the basic variables x and y using back substitution: The last equation is y þ z þ w ¼ 1, w h i c h g i v e s y ¼ z þ w 1. T h e f i r s t eq u a t i o n i s x þ 7y 7z þ 3w ¼ 1, a n d t h i s g i v e s x ¼ 1 7ðz þ w 1Þ þ 7z 3w ¼ 8 10w. We can therefore express the basic variables in terms of the free as x ¼ 8 10w y ¼ 1 þ z þ w Solution 1.13 The system is homogeneous, therefore there are solutions, and the number of degrees of freedom is given by n rkðAÞ ¼ 4 rkðAÞ, where A is the 3 4 coefficient matrix. Since there are three rows, we have that rkðAÞ 3, and therefore there is at least one free variable. This means that there are infinitely many solutions, and therefore non-trivial solutions.
15
(16,1)
1
Solutions to textbook problems
Solution 1.14 We have the following Gaussian process for all values of p and q: 0 1 0 1 1 1 1 1 1 1 A ¼ @ 1 p p2 A ! @ 0 p 1 p2 1 A 0 q 1 q2 1 1 q q2 If p ¼ 1, then we obtain the echelon form 0 1 1 1 1 @ 0 q 1 q2 1 A 0 0 0 It has one pivot position if q ¼ 1, and two pivot positions otherwise. If p 6¼ 1, we continue the Gaussian process by dividing the second row by p 1, or multiplying it by 1=ðp 1Þ 6¼ 0. This gives an echelon form 0 1 0 1 1 1 1 1 1 1 @0 1 p þ 1 A ! @0 1 p þ 1A 0 q 1 q2 1 0 0 P where we use the symbol P ¼ q2 1 ðp þ 1Þðq 1Þ to write the echelon form in a more compact way. Since P ¼ ðq 1Þðq þ 1Þ ðp þ 1Þðq 1Þ ¼ ðq 1Þðq pÞ, we have two pivot positions if q ¼ 1 or q ¼ p, and three pivot positions otherwise. This means that 8 < 1, p ¼ q ¼ 1 rkðAÞ ¼ 2, p ¼ 1, q 6¼ 1 or q ¼ 1, p 6¼ 1 or p ¼ q 6¼ 1 : 3, otherwise Solution 1.15 We use Gaussian elimination to compute the rank of the matrix: a) We find an 0 2 @4 6
echelon form using the Gaussian 1 0 5 3 4 8 2 A @ 7 4 3 9 ! 0 9 5 2 4 0 0 2 @ 0 ! 0
process
1 5 3 4 8 3 2 5 7 A 6 4 10 20 1 5 3 4 8 3 2 5 7 A 0 0 0 6
There are three pivot positions, therefore the rank is 3.
16
(17,1)
1.1 Linear systems and Gaussian elimination
b) We find an echelon form of A using the Gaussian process 0 1 0 1 2 10 6 8 2 10 6 8 @ 1 5 4 11 A ! @ 0 0 1 7A ! 3 15 7 2 0 0 2 14
0
2 @0 0
1 10 6 8 0 1 7A 0 0 0
There are two pivot positions, therefore the rank is 2. c) We find an echelon form of A using the Gaussian process 1 0 1 0 1 2 5 0 1 1 2 5 0 1 B B 2 1 2 4 5C 5 8 4 3C C C ! B0 B @ 0 3 6 7 5 A @ 3 9 9 7 2 A 0 4 8 11 10 3 10 7 11 7 1 0 1 2 5 0 1 B0 1 2 4 5C C ! B @0 0 0 5 10 A 0 0 0 5 10 1 0 1 2 5 0 1 B0 1 2 4 5C C ! B @0 0 0 5 10 A 0 0 0 0 0 There are three pivot positions, therefore the rank is 3. Solution 1.16 We use a Gaussian process to find an echelon form: a) For any 0 1 1 @1 t t 1
t, we have the Gaussian process: 1 0 1 t 1 1 t 1A ! @0 t 1 1 t A 1 0 1 t 1 t2
0 !
1 1 1 t @0 t 1 1 t A 0 0 2 t t2
We notice that t 1 ¼ 0 for t ¼ 1 and 2 t t2 ¼ 0 for t ¼ 1 and t ¼ 2. Therefore, we have an echelon form with three pivots if t 6¼ 1; 2. If t ¼ 2, we have an echelon form with two pivots, and if t ¼ 1, we have an echelon form with one pivot. It follows that 0 1 8 1 1 t < 1, t ¼ 1 rk @ 1 t 1 A ¼ 2, t ¼ 2 : t 1 1 3, t 6¼ 1, 2
17
(18,1)
1
Solutions to textbook problems
b) For any t, we have the 0 1 3 1 @2 4 0 t 1 5
Gaussian process: 1 0 4 1 A @ 6 ! 0 3 0 0 1 @ ! 0 0 0 1 @ ! 0 0
3 2 1 3t
1 2 5þt
3 1 1 3t
1 1 5þt
3 1 1 1 0 4 2t
1 4 2 A 3 4t 1 4 1 A 3 4t 1
4 1 A 4 t
We notice that 4 2t ¼ 0 for t ¼ 2 and 4 t ¼ 0 for t ¼ 4, so for all values of t, we have a an echelon form with three pivots (the two marked in blue, and one of the two expressions in the third row). It follows that for all values of t, 0 1 1 3 1 4 rk @ 2 4 0 6A ¼ 3 t 1 5 3 c) For any t, we have the Gaussian process: 0 1 0 1 2 3 2 1 @ 2 1 A @ 1 3 ! 0 1 t tþ1 9 0 0 1 @ ! 0 0 0 1 @ 0 ! 0
1 2 3 2 5 5 7 A t 2 t 2 7 1 2 3 2 1 1 7=5 A t 2 t 2 7 1 2 3 2 1 1 7=5 A 0 0 7ð7 tÞ=5
We notice that 7ð7 tÞ=5 ¼ 0 for t ¼ 7, so for t 6¼ 7, we have an echelon form with three pivots, and for t ¼ 7, we have an echelon form with two pivots. It follows that 0 1 1 3 1 4 2, t ¼ 7 @ A rk 2 4 0 6 ¼ 3, t 6¼ 7 t 1 5 3
18
(19,1)
1.2 Vectors and vector spaces
Solution 1.17 No, it is not true, and here is counter-example: Consider the 2 3 matrices A and B, given by 1 0 1 1 0 0 , B¼ A¼ 0 1 1 0 1 0 We see that A and B have the same pivot positions ð1, 1Þ and ð2, 2Þ marked in blue, but A and B are not row equivalent: If we apply any elementary row operations to B, the third column would remain a zero column. Hence it is not possible to get from B to A using elementary row operations.
1.2 Vectors and vector spaces Solution 2.1 The vectors are shown in the figure below. In coordinates, we have that u þ v ¼ ð5, 2Þ, u w ¼ ð3, 1Þ, and 2w ¼ ð4, 4Þ. y 4 3 2 w –4
–3
–2
–1
1 –1
u+v
u 1
u–w
2
v
3
4
x
–2 –3 –4
–2w
Solution 2.2 When u ¼ ð1, 3Þ, v ¼ ð4, 1Þ and w ¼ ð 2, 2Þ, we get: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi a) kuk ¼ 12 þ 32 ¼ 10 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi b) kvk ¼ 42 þ ð 1Þ2 ¼ 17 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi c) kwk ¼ ð 2Þ2 þ 22 ¼ 8 d) u v ¼ 1 4 þ 3 ð 1Þ ¼ 1 e) u w ¼ 1 ð 2Þ þ 3 2 ¼ 4 f) v w ¼ 4 ð 2Þ þ ð 1Þ 2 ¼ 10
19