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# The University of Sydney MATH1902 Linear Algebra (Advanced)

Semester 1

## Longer Solutions to Selected Exercises for Week 2

2013

11.

8 d

By Pythagoras d = 82 + 62 = 10 . If is the angle to the horizontal then cos = 6/10 , yielding an angle 53 . Thus the resultant force is 10 newtons in a direction 53 to the horizontal, towards the right. 12.

A Observe that

C 1 AC . 2

1 1 1 DE = DB + BE = AB + BC = AB + BC 2 2 2

This tells us that the line segment joining D to E is parallel to and half the length of the line segment joining A to C . 13. The associative law for addition of vectors says that, for any vectors u, v and w, u+ v+w = u+v +w .

To verify this, we suppose that the vectors have been lined up so that the point P is at the tail of u, the point Q is both at the tip of u and at the tail of v, the point R is 1

both at the tip of v and at the tail of w, and the point S is at the tip of w. Then u + v + w = P Q + QR + RS = P Q + QS = PS = P R + RS = P Q + QR + RS = u + v + w . To explain the Triangle Inequality, consider the following diagram: B v

The vectors v and w have been placed tip-to-tail so that they label two directed edges of a triangle ABC , so that v = AB , w = BC . Then v + w = AC . The shortest distance between two points is a straight line, so that travelling from A to C via B is at least as far as travelling directly from A to C . Thus |v + w| = AC AB + BC = |v| + |w| ,

which veries the triangle inequality. This becomes equality precisely when B falls on the direct path joining A to C , so that the triangle becomes degenerate.

14.

Later in the course we learn a general method for solving systems of equations, called Gaussian elimination. For this problem, we can nd a quick ad hoc solution, by eliminating z , by taking multiples of the rst two equations 6 x + 9 y + 12 z = 12 10 x + 10 y + 12 z = 6 and subtracting one from the other to get 4x + y = 6. By subtracting twice the original third equation from the rst, we get 4x + y = 2. Solving these two equations in x and y quickly yields x = 1 and y = 2. From any of the original equations, we nd z = 3. Thus the intersection point is (1, 2, 3).

15. To nd the reected line is a simple three part process: (i) we rst translate the line ax + by = c vertically k units to get the line ax + b(y + k ) = c, which becomes ax + by = c bk ; (ii) we then reect this line in y = x, yielding bx + ay = c bk ; (iii) we nish the job by translating this last line vertically k units, to get the line bx + a(y k ) = c bk , that is bx + ay = c + k (a b) . 2

17.

B b

E CD = b a , 18. DE = a ,

D EF = b , B FA = a b .

60

45 C

30 A

We have AB = 20 and BC = 10. By the Cosine Rule, 202 + 102 2(10)(20) cos 105 25 . AC = By the Sine Rule, sin(30 ) = from which it follows that 10 sin 105 , AC |

so that 7 . Hence the nal distance and direction of the aircraft from the starting point are approximately 25 km and 7 north of east respectively. 19. 1 1 AD = AB + BD = AB + BC = AB + BA + AC 2 2 1 = AB + AB + AC 2 20. = 1 1 1 AB + AC . AB + AC = 2 2 2

30 23 ,

Let P QRS be a parallelogram and T the midpoint of the diagonal P R. Then 1 1 QT = QP + P T = QP + P R = QP + (P Q + QR) 2 2 1 1 1 1 = QP + (QP + QR) = (QP + QR) = (RS + QR) = QS , 2 2 2 2 which proves that T is the midpoint of the diagonal QS , so that the diagonals bisect each other. 3

21.

Let P , Q, R, S be the respective midpoints of the edges AB , BC , CD , DA of the quadrilateral ABCD . Then, by two applications of Exercise 12, rstly to the triangle ABC , and then secondly to the triangle ADC , 1 P Q = AC = SR , 2 which is sucient to prove that P QRS is a parallelogram. B Q P C A S D

22.

Consider a triangle ABC such that X is the midpoint of BC , Y the midpoint of AC and Z the midpoint of AB . The medians are represented by the vectors AX , BY and CZ . But, using Exercise 19, the vector sum is 1 1 1 AX + BY + CZ = (AB + AC ) + (BA + BC ) + (CA + CB ) 2 2 2 1 = (AB + BA + BC + CB + AC + CA) 2 1 = ( 0 + 0 + 0) = 0 , 2 which proves that the medians can be shifted parallel to themselves to form another triangle.

23. To nd the rotated line is a simple three part process: (i) we rst translate the line ax + by = c horizontally x0 units and vertically y0 units, to get the line a(x + x0 ) + b(y + y0 ) = c , which becomes (ii) we then rotate this line ninety degrees anticlockwise about the origin, yielding bx + ay = c ax0 by0 ; (iii) we nish the job by translating this last line horizontally x0 units and vertically y0 units, to get the line b(x x0 ) + a(y y0 ) = c ax0 by0 , that is bx + ay = c + a(y0 x0 ) b(x0 + y0 ) . 4 ax + by = c ax0 by0 ;

24.

A quick solution uses similar triangles: Let r = |ST | and s = |T R| so that |P Q| = |SR| = r + s. But P QU is similar to T RU , so corresponding ratios of lengths of sides are equal: |P U | r+s |P U | = = , |UR| |UR| s so that the ratio in which U divides the diagonal is r + s : s. An alternative solution using vectors does not require any geometric insight: suppose that and are scalars such that P U = P R and QU = QT . Then (P S + SR) = P R = P U = P Q + QU = P Q + QT r = P Q + (QP + P S + ST ) = P Q + (QP + P S + SR) , r+s r )P Q . But P Q and P S are not so that, rearranging, ( )P S = (1 + r+ s parallel so the scalar coecients must be zero: r = (1 + ) = 0, r+s r+s , so that the ratio in which U divides from which it follows quickly that = = r + 2s the diagonal is r + s : s.

25. The mindreader always produces the number 0 by the instructions. To see this let x be any positive integer, so that X becomes the integer part of 2x + 1 and Y becomes 2 the xth number along the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6 We want to prove that X = Y , so that when we take Y away from X we get zero. Observe that 1 + 2 + . . . + (Y 1) + 1 x 1 + 2 + . . . + Y giving Y (Y + 1) (Y 1)Y +1 x 2 2 whence Thus Y 2 Y + 2 2x Y 2 + Y . 1 Y 2 so Y whence 1 2 Y < 2x
2

7 2x < + 4

1 Y + 2

7 1 < 2x < Y + , 4 2

1 < Y +1. 2 Thus if we throw away everything to the right of the decimal point and the decimal point, in the decimal expansion of 2x + 1 , we must be left with the integer Y , that 2 is, X = Y , voila! 2x + 5
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