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Solutions to MATH103 SEPTEMBER 2006 Examination
Note: all questions are similar to homework or examples done in class.
Section A
- z = 1 + 2i gives z − 2 i = 1 − 4 i so the required expressions is
1 (1 − 4 i)^2
− 15 − 8 i
−15 + 8i 289
and the real part is − 28915 while the imaginary part is 2898.
[1 mark for z, 3 marks for calculation.]
- |z| =
√ 12 + (−
3)^2 = 2 [1 mark]
Let θ = arg(z). Then tan θ = −
√ 3 1 =^ −
3, so θ = −π 3 or 23 π. Since z is in the 4rd quadrant, −π 3 is correct. Thus z = 2e−πi/^3.
1− i 3
y
0 1 x
−π/
[1 mark for diagram, 2 marks for arg of z]
By de Moivre’s theorem,
z^9 = 2^9 e^9 ×^
− 3 πi = 2^9 e−^3 πi^ = 512eπi^ = − 512.
The real part of z^9 is −512 and the imaginary part is 0. [2 marks]
- (3i − 2)^2 = 2^2 − 2 × 2 × (3i) + (3i)^2 = 4 − 12 i − 9 = − 5 − 12 i [1 mark]
Thus the square roots of − 5 − 12 i are ±(3i − 2). Using the quadratic formula,
z =
−8 + 6i ±
√ (8 − 6 i)^2 − 4(12 − 12 i) 2
= −4 + 3i ±
√ (4 − 3 i)^2 − (12 − 12 i)
= −4+3i±
16 − 24 i − 9 − 12 + 12i = −4+3i±
− 5 − 12 i = −4+3i±(3i−2) = −6+6i or − 2. [3 marks]
- p = 45 a + 15 b and m = (p + c)/2 = 104 a + 101 b + 105 c, so that 4
→ MA +
→ MB + 5
→ MC= 4(a − m) + b − m + 5(c − m) = (4a + b + 5c) − 10 m = (4a + b + 5c) − (4a + b + 5c) = 0. [4 marks]
- (i)
→ AB= (− 1 − (−2), 2 − 0 , 1 − 3) = (0, 2 , −2),
→ AC= (2, 4 , 0), [1 mark] → AB ×
→ AC= (8 − 0 , −4 + 0, 0 − 4) = (8, − 4 , −4). Checking perpendicularity: (8, − 4 , −4) · (0, 2 , −2) = 0 − 8 + 8 = 0 and (8, − 4 , −4) · (2, 4 , 0) = 16 − 16 + 0 = 0, as required. [3 marks]
(ii) The area of the triangle is 12 |
→ AB ×
→ AC | = (^12)
√ 82 + (−4)^2 + (−4)^2 = 2
- [1 mark] Let h be the length of the perpendicular from B to AC. Then the area of the triangle
is 12 h|
→ AC | = 12 h
22 + 4^2 + 0^2 = h
- Equating this to the area found in (i) we get
h = 2
√ √^6 5.^ [2 marks]
(iii) A normal to the plane ABC is given by
→ AB ×
→ AC= (8, − 4 , −4) as above. To reduce the numbers, it is better to take one qutaer of it: n = (2, − 1 , −1). An equation is then
(x − a) · n = 0, that is (x + 2, y, z − 3) · (2, − 1 , −1) = 0,
that is 2(x + 2) − y − (z − 3) = 0 giving 2x − y − z + 7 = 0. We can equally well use (x − b) · n = 0 or (x − c) · n = 0; these give the same answer. Alternatively, the plane will be 2x − y − z + k = 0 for some number k, since its normal is (2, − 1 , −1) and substituting the coordinates of A (or B or C) in this equation gives k = 7. [3 marks]
- The pairs given are of the form (x, y) so we have to solve the equations
p + q + r = 1 (1) p − q + r = 11 (2) p + 2q + 4r = 2 (3)
(1)–(2) gives 2q = −10 so q = −5. Then (3)–(1) gives q + 3r = 1 so 3r = 1 − q = 6 giving r = 2. Finally (1) gives p = 4. [2 marks for setting up the equations and 3 for solving them]
- (a) u = (2, − 8 , 4), v = (− 3 , 12 , −6) are linearly dependent (say, u = −^23 v), and 3 u + 2v can be taken for a non-trivial linear combination equalling the zero vector. The two vectors span the same as just either of them, that is only a line in R^3 , not the entire R^3. [3 marks]
(b) Putting the vectors, w, u, v as the rows of a matrix and using row reduction gives
−→
−→
using R 2 − 2 R 1 , R 3 +R 1 , and then R 3 + 85 R 2. The row echelon form contains no zero rows, hence the three vectors are linearly independent. Since they are linearly independent they do span R^3. [4 marks]
- For B we need only multiply the diagonal entries, since B is lower triangular. So det(B) = 2 × (−1) × 4 = −8. [1 mark] Expanding det(A) by the top row (which contains a zero and therefore gives a short calculation) gives det(A) = −4(2 − 15) − 1(− 6 − 6) = 52 + 12 = 64. [3 marks]
Using rules for determinants we now get det(AB−^2 ) = det(A)/(det(B))^2 = 64/(−8)^2 = 1. The matrix B + 2I is also lower triangular, and has diagonal entries 4, 1, 6. The deter- minant is the product of these, hence is 24. [2 marks]
−α^2 + 5α + 6 = −(α + 1)(α − 6). Hence A is invertible if and only if α 6 = 6, − 1 , as required. [4 marks for evaluating the determinant, 1 for deducing when A is invertible]
(ii) We need the inverse of A 0 =
.^ The matrix of minors of^ A 0 is
. The matrix of cofactors of A 0 is
and its transpose is
. The determinant of A 0 is 6, using the formula found in (i). Dividing
all the terms of the last matrix by the determinant we get
A− 0 1 =
11 6 −
4 3
7 6
−^2323 −^13
1 2 −^1
1 2
[6 marks]
(iii) Using row operations on the augmented matrix of the equations we get
0 − 3 4 a 1 2 − 1 b 2 1 2 c
−→
1 2 − 1 b 0 − 3 4 a 2 1 2 c
−→
1 2 − 1 b 0 − 3 4 a 0 − 3 4 c − 2 b
1 2 − 1 b 0 − 3 4 a 0 0 0 c − 2 b − a
swapping the first two rows at the start, using R 3 − 2 R 1 after this and finishing with R 3 − R 2. The required condition is the vanishing of the last term in the final third row: −a − 2 b + c = 0. [4 marks]
- (i)
x + 4y − 2 z = 3 (4) 2 x + 7y + 3z = − 4 (5)
Taking (5) − 2 × (4) gives −y + 7z = −10, that is y = 7z + 10. Substituting in (4) gives x = −4(7z + 10) + 2z + 3 = − 26 z − 37. The parametric form of L is therefore (− 26 z − 37 , 7 z + 10, z). [It is equally valid to parametrize by x or y, but these give slightly more complicated expessions.] [4 marks]
(ii)
→ AB= (2, 0 , 1) so the general point of L′^ is (1, − 4 , 2) + λ(2, 0 , 1) = (1 + 2λ, − 4 , 2 + λ). [3 marks]
(iii) L′^ meets the plane x + y − z = −7 in the point whose parameter λ is obtained by substituting the general point as in (ii) into the equation of the plane. This gives 1 + 2λ − 4 − 2 − λ = − 7 , that is λ = −2. Thus the point of intersection is (1, − 4 , 2) − 2(2, 0 , 1) = (− 3 , − 4 , 0). [4 marks]
(iv) In order for L to meet L′^ we need there to exist values of z and λ which make the general points in (i) and (ii) equal. Thus we require that the three equations 1 + 2λ = − 26 z − 37 , −4 = 7z + 10, 2 + λ = z should have a common solution for z and λ. From the second equation z = −2, hence λ = −4 from the third. Putting these into the first equation, we get −7 = 15 which cannot be true. Therefore, the lines L and L′ do not meet. [4 marks]
- (i) Writing the vectors as the rows of a matrix (starting with v 2 ) and using row reduction gives
after applying R 2 + 2R 1 R 3 − 3 R 1 , R 4 − R 1 and then R 3 + 2R 1 , R 4 − R 2. Perform now R 3 /8 and R 4 /(−10), and R 4 = R 3 on the next step:
−→
.
We have ended up with a row of zeros; this means that the vectors are linearly dependent. [6 marks]
(ii) The three nonzero rows of the reduced matrix in (i) have the same span as the four given vectors, and they are linearly independent because this process always results in linearly independent vectors. To simplify and arrange more zeros (this is convenient for any further calculations), we can clear the last entries in the 1st and 2nd row using 3rd row (R 2 − 6 R 3 , R 1 − R 3 ) and get
−→
,
where the last move was R 1 − R 2. Therefore, w 1 = (1, − 1 , 0 , 0), w 2 = (0, 2 , 3 , 0), w 3 = (0, 0 , 3 , 1) are also suitable spanning vectors. We can extend these to a basis for R^4 by adding any row to the matrix with w 1 , w 2 , w 3 as rows in such a way as to give four independent rows. The simplest thing is to add (0, 0 , 0 , 1). [2 marks for the independent vectors spanning S, 3 for completing to a basis]
(iv) We need only check whether there are scalars a, b, c such that a(1, − 1 , 0 , 0) + b(0, 2 , 3 , 0) + c(0, 0 , 3 , 1) = (1, − 3 , 0 , 1). From the first components we have a = 1, and from the second components it follows that −1 + 2b = −3 so that b = −1. Comparing the third components we have −3 + 3c = 0, hence c = 1. After this, the forth components in both sides of the equation coincide. Thus the vector (1, − 3 , 0 , 1) does lie in S. It is (1, − 1 , 0 , 0) − (0, 2 , 3 , 0) + (0, 0 , 3 , 1). [4 marks]
