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Linear Algebra Exam pack
Linear Algebra (University of South Africa)
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MAT1503
EXAM PACK
2017
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Welcome
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NOVEMBER 2016 SOLUTIONS
Question 1a
5𝑥 + 3𝑦 = 1 ⋯ ⋯ ⋯ ⋯ ⋯ (1)
𝑥 − 2𝑦 = 1 ⋯ ⋯ ⋯ ⋯ ⋯ (2)
From (2)
X = 8 +2y
Substituting for x in (1)
5(8 +2y)+3y =1
40+ 10y+3y= 1
13y = -39
Y= -3
Using equation x = 8 +2y
X = 8 +2(-3)
X=2
X = 2 and y = -3
Question 1b
1
𝐴 = [1
2
2 3
4 1]
1 9
Make zeros in column 1 excepts entry at row 1 (pivot entry)
Subtract R1from R2
1 2 3
[0 2 −2]
2 1 9
Subtract R1 x 2 for R3
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1 2
3
[0 2 −2]
0 −3 3
Make zeros in column 2 except pivot entry.
Subtract R2 from R1
1 0
5
[0 2 −2]
0 −3 3
Divide R2 by 2
1 0
5
[0 1 −1]
0 −3 3
Add R2 x 3 to R3
𝟏 𝟎 𝟓
𝒓𝒓𝒆𝒇(𝑨) = [𝟎 𝟏 −𝟏]
𝟎 𝟎 𝟎
Question 1ci
4 1
𝐴=[
3 1
−2 7
]
−1 5
1
5
3
−1
𝐵=[
]
−2 4
2 −3
1
5
4 1 −2 7 3 −1
𝐴𝐵 = [
]
][
3 1 −1 5 −2 4
2 −3
𝐴𝐵 = [
𝐴𝐵 = [
(4 ∗ 1) + (3 ∗ 1) + (−2 ∗ −2) + (7 ∗ 2) (4 ∗ 5) + (1 ∗ −1) + (−2 ∗ 4) + (7 ∗ −3)
]
(3 ∗ 1) + (1 ∗ 3) + (−1 ∗ −2) + (5 ∗ 2) (3 ∗ 5) + (1 ∗ −1) + (−1 ∗ 4) + (5 ∗ −3)
25 −10
]
18 −5
Question 1cii
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4 1
𝐴𝐶 = [
3 1
𝐴𝐶 = [
3 4
−2 7 2 1
][
]
−1 5 −2 3
1 −3
(4 ∗ 3) + (2 ∗ 1) + (−2 ∗ −2) + (7 ∗ 1) (4 ∗ 4) + (1 ∗ 1) + (−2 ∗ 3) + (7 ∗ −3)
]
(3 ∗ 3) + (1 ∗ 2) + (−1 ∗ −2) + (5 ∗ 1) (3 ∗ 4) + (1 ∗ 1) + (−1 ∗ 3) + (5 ∗ −3)
𝟐𝟓 −𝟏𝟎
𝑨𝑪 = [
]
𝟏𝟖 −𝟓
Question 1ciii
Observe that AB = AC
Question 1civ
D=B–C
3 4
1
5
3
−1
𝐷=[
] − [ 2 1]
−2 3
−2 4
2 −3
1 −3
1−3
5−4
3−2
−1 − 1
𝐷=[
−2 − (−2) 4 − 3 ]
2 − 1 −3 − (−3)
−2 1
𝐷 = [ 1 −2]
−0 1
1 0
4
𝐴𝐷 = [
3
−2 1
1 −2 7 1 −2
][
]
1 −1 5 −0 1
1 0
(4 ∗ −2) + (1 ∗ 1) + (−2 ∗ 0) + (7 ∗ 1) (4 ∗ 1) + (1 ∗ −2) + (−2 ∗ 1) + (7 ∗ 0)
𝐴𝐷 = [
]
(3 ∗ 2) + (1 ∗ 1) + (−1 ∗ 0) + (5 ∗ 1) (3 ∗ 1) + (1 ∗ −2) + (−1 ∗ 1) + (5 ∗ 0)
𝟎 𝟎
𝑨𝑪 = [
]
𝟎 𝟎
Question 1di
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A is a 2 x 2 matrix
B is a 2 x 3 matrix
Question 1dii
X should be a 2 x 3 matrix
Question 1diii
X = A-1B
𝐴=[
4 3
]
5 4
The determinant is not 0 so the inverse exist
4
[
5
1
4 −3
3 −1
]
[
] =
4
4 ∗ 4 − 3 ∗ 5 −5 4
𝐴−1 = [
4 −3
]
−5 4
X = A-1B
𝑋
4 −3 1
3 −5
=[
]
][
−5 4 −1 −2 5
𝑋
=[
(1 ∗ 4) + (−3 ∗ 1) (4 ∗ 3) + (−3 ∗ −2) (4 ∗ −5) + (−3 ∗ 5)
]
(−5 ∗ −1) + (4 ∗ 1) (−2 ∗ 4) + (−5 ∗ 3) (−5 ∗ −5) + (4 ∗ 5)
𝟏
𝟏𝟖 −𝟑𝟓
𝑿 =[
]
−𝟏 −𝟐𝟑 𝟒𝟓
Question 2a
From the equations we get the matrix
X1 X2 b
7
8
5
6
9
4
Find the determinant, Δ = (7*9)-(6*8) = 15
Replace the first column of the matrix with solution vector and find the determinant
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X1 X2
5
8
4
9
Δ1 = (5*9)-(4*8) = 13
Replace the second column with solution vector and find the determinant
X1 X2
7
5
6
4
Δ2 = (7*4)-(6*5) = -2
𝑿𝟏 =
𝑿𝟏 =
∆𝟏 𝟏𝟑
=
=𝒙
𝟏𝟓
∆
∆𝟐 −𝟐
=
=𝒚
𝟏𝟓
∆
Question 2b
2 0 0 −3
𝑑𝑒𝑡 [0 −1 0 0 ]
7 4 3 5
−6 2 2 4
Swap R1↔R3
7
[0
2
−6
4 3 5
−1 0 0 ]
0 0 −3
2 2 4
Cancel the leading coefficient in R3 by
R3← R3 -2/7R1
7
4
3
5
0
−1
0
0
]
[
0 −8/7 −6/7 −31/7
−6
2
2
4
Cancel the leading coefficient in R4
R4← R4 +6/7R1
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7
4
3
5
0
−1
0
0
[ 0 −8/7 −6/7 −31/7]
0 38/7 32/7
58/7
Swap rows R2↔R4
7
4
3
5
0 38/7 32/7 58/7
]
[
0 −8/7 −6/7 −31/7
0 −1
0
0
Cancel the leading coefficient in R3
R3← R3 +4/19R2
3
5
7
4
32/7 58/7
0
38/7
[
]
0
0 2/19 51/19
0 −1
0
0
Cancel the leading coefficient in R4
R4← R4 +7/38R2
3
5
7
4
32/7 58/7
0
38/7
[
]
2/19 51/19
0
0
0
0 16/19 29/19
Swap R3↔R4
3
5
7
4
32/7
58/7
[0 38/7
]
16/19 29/9
0
0
0
0 2/19 −51/19
Cancel the leading coefficient in R4
R4← R4 -1/8R3
3
5
7
4
32/7 58/7
[0 38/7
]
0
0 16/19 29/9
0
0
0 −23/8
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Determinant is the product of the diagonal
3
5
7
4
32/7
58/7
]
[0 38/7
0
0 16/19 29/9
0
0
0 −23/8
Δ = 7*38/7*16/19*-23/8= -92
Interchanging the 2 rows negate the determinant, therefore multiplying the result by (- 1)3
Δ = -92(- 1)3 = 92
Question 2c
Suppose that A is a square matrix of size n where the 2 rows s and t are equal.
Form the matrix B by swapping rows s and t.
As a consequence of our hypothesis A = B, then
Det(A) = ½ (det(A) + det(A))
= ½ (det(A) –det(B))
= ½ (det(A) - det(A))
:Theorem DRCS
:Hypothesis A = B
= ½ (0)
= 0 proven
Question 2d
A = PBP-1
Det(A) = det(PBP-1)
Det(A) = det(P).det(B).det(P-1)
:det(P).det(P-1) = 1
Det(A) = det(B) proven
Question 2e
𝐶=[
−1 2
]
3 4
K=2
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Det is a multiplier function of columns or rows of a matrix hence,
= det(kC1,kC2, ……………,kCn)
Det(kC)
= kdet(C1,kC2, ……………,kCn)
=kn det(C1,C2, ……………,Cn)
= kn det(C)
𝐶=[
−1 2
]
3 4
K=2
−1 2
det(𝑘𝐶) = det(2𝐶) = 𝑑𝑒𝑡 [2 ∗ [
] ]
3 4
−2 4
= 𝑑𝑒𝑡 [ [
] ]
6 8
= (-2*8)-(6*4)
=-40
kn det(C) = 22det(C)
: n =2 for 2x2 matrix
−1 2
= 4 𝑑𝑒𝑡 [ [
] ]
3 4
= 4*((-1*4)-(2*3))= -40
Therefore det(kC) = kn det(C)
Question 3ai
U=(2 , -1 , 1)
uv = u . v
v = (1 , 1, 2)
= (2 * 1) + (-1 * 1)+ (1 * 2) = 3
Question 3aii
u . v = ІuІІvІ Cos𝜃
Cos𝜃 =
u .v
ІuІІvІ
𝜃 = 𝐶𝑜𝑠 −1 [
3
√22 + −12 + 12 √12 + 12 + 22
]
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𝜃 = 𝐶𝑜𝑠 −1 [
𝜽 = 𝟓𝟐. 𝟐𝟎
3
√4√6
]
Question 3bi
𝑃𝑟𝑜𝑗𝑎 𝑢 =
𝑢. 𝑎
→
𝑎. 𝑎 𝑎
=
(3 , 7 , −7). (1 , 0 , 5)
→
(1 , 0 , 5). (1 , 0 , 5) 𝑎
=
(3 ∗ 1) + (1 ∗ 0) + (−7 ∗ 5)
→
(1 ∗ 1) + (0 ∗ 0) + (5 ∗ 5) 𝑎
=
−32
(1 , 0 , 5)
26
=
−32
→
26 𝑎
= (−𝟑𝟐/𝟐𝟔, 𝟎 , −𝟏𝟔𝟎/𝟐𝟔)
Question 3bii
𝑢 − 𝑃𝑟𝑜𝑗𝑎 𝑢 = (3 , 7 , −7) − (−
32
= (3 + 26) , (1 − 0) , (−7 +
𝟏𝟏𝟎
𝟐𝟐
=(
,𝟏 ,− )
𝟐𝟔
𝟐𝟔
32
, 0 , −160/26)
26
160
)
26
Question 3c
𝑑 = ‖→ − →‖
𝑢
𝑣
= √(7 − (−7) + (−5 − (−2) + (1 − (−1))
= √14 − 3 + 2
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= √𝟏𝟑
Question 3d
𝐴𝑟𝑒𝑎 = →∗ →
𝑢
𝑣
𝑖
𝑗
𝑘
= | 7 −5 1 |
−7 −2 −1
7
1
7 −5
−5 1
=|
|→
|→+ |
|→ −|
−7 −1 𝑗
−7 −2 𝑘
−2 −1 𝑖
= [(−5 ∗ −1) − (−2 ∗ 1)] → −[(7 ∗ −1) − (−7 ∗ 1)] → + [(7 ∗ −2) − (−7 ∗ −5)] →
= 7 → −0 → + 49 →
𝑖
𝑗
𝑘
𝑖
𝐴𝑟𝑒𝑎 = |→∗ →| = √7
𝐴𝑟𝑒𝑎 =
𝑢
𝑣
7√50 𝑢2
2
𝑗
+ 492
Question 3e
Let the equation of plane be
𝑎(𝑥 − 𝑥0 ) + 𝑏(𝑦 − 𝑦0 ) + 𝑐(𝑧 − 𝑧0 ) = 0
(𝑥0 ), (𝑦0 ), (𝑧0 ) is a point on the plane
Given (−4 , −1 , −1) (−2 , 0 , 1) and (−1 , −2 , −3)
→= ⟨(−4 + 2), (−1 − 0) , (−1 − 1)⟩ = ⟨−2, −1, −2⟩
𝑣1
→= ⟨(−4 + 1), (−1 + 2) , (−1 + 3)⟩ = ⟨−3,1,2⟩
𝑣2
To find the perpendicular vector, take the correspondence of 𝑣1 and 𝑣2
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𝑘
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𝑖
𝑗
𝑘
𝑣1 ∗ 𝑣2 = |−2 −1 −2 |
−3 1
2
−2 −1
−2 −2
−1 −2
=|
|→
|→+ |
|→ −|
𝑗
𝑖
−3 1 𝑘
−3 2
1
2
= [−2 − (−2)] → −[−4 − 6] → + [−2 − 3] →
𝑖
𝑗
= 0 → −10 → −5 →
𝑖
𝑗
𝑘
𝑘
= ⟨0, −10, −5⟩ = ⟨𝑎, 𝑏, 𝑐⟩
Using the plane equation
𝑎(𝑥 − 𝑥0 ) + 𝑏(𝑦 − 𝑦0 ) + 𝑐(𝑧 − 𝑧0 ) and the point (−4 , −1 , −1)
Giving us
0(𝑥 + 4) + (−10)(𝑦 + 1) + (−5)(𝑧 + 1) = 0
(−10)(𝑦 + 1) + (−5)(𝑧 + 1) = 0
(10)(𝑦 + 1) + (5)(𝑧 + 1) = 0
The equation of the plane passing through the point is
(𝟏𝟎)(𝒚 + 𝟏) + (𝟓)(𝒛 + 𝟏) = 𝟎
Question 4a
𝑧 + (1 − 𝑖) = (3 − 2𝑖)
𝑧 = 3 + 2𝑖 − (1 − 𝑖)
𝒛 = 𝟐 + 𝟑𝒊
Question 4b
𝑧 = 1 + √3𝑖
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𝑟 = √𝑎2 + 𝑏 2
2
𝑟 = √12 + √3 = 2
𝑏
𝜃 = 𝑡𝑎𝑛−1 ( )
𝑎
𝜃 = 𝑡𝑎𝑛−1 (
𝜽=
𝝅
𝟑
√3
)
1
z in polar form is
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝒛 = 𝟐(𝒄𝒐𝒔
𝝅
𝝅
+ 𝒊𝒔𝒊𝒏 )
𝟑
𝟑
Question 4c
𝑖
1+𝑖
in the form 𝑎 + 𝑏𝑖
Multiply conjugate 1 − 𝑖
𝑖
1+𝑖
∗
1−𝑖
1−𝑖
=
𝑖−𝑖 2
1−𝑖+𝑖−𝑖 2
𝑖
1+𝑖 1 1
=
= + 𝑖
1+𝑖
2
2 2
=
𝑖−(−1)
1−(−1)
since 𝑖 2 = −1
Its real part = ½
Its imaginary part = ½
Question 4d
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De moivres theorem states that if n is any real number then to calculate powers of a complex
number
(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)𝑛 = cos(𝑛𝜃) + 𝑖𝑠𝑖𝑛(𝑛𝜃)
Question 4e
First convert the complex number to a polar form
𝑟 = √𝑎2 + 𝑏 2
𝑟 = √12 + 12 = √2
𝑏
𝜃 = 𝑡𝑎𝑛−1 ( )
𝑎
1
𝜃 = 𝑡𝑎𝑛−1 ( )
1
𝜋
𝜃=
4
z in polar form is
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
1 + 𝑖 = √2(𝑐𝑜𝑠
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛 )
4
4
De moivres theorem state that all nth roots of a complex number 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃) is given by
𝑛
√𝑟 [𝑐𝑜𝑠
𝜃 + 2𝜋𝑘
𝜃 + 2𝜋𝑘
+ 𝑖𝑠𝑖𝑛
]
𝑛
𝑛
We have
𝜋
𝑟 = √2, 𝜃 = 4 , 𝑛 = 2
Thus 𝑘 = 𝑜
𝜋
𝜋
+ 2𝜋(0)
+ 2𝜋(0)
4
√
= √2 [𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛 4
]
2
2
2
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4
= √2 [𝑐𝑜𝑠
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛 ]
8
8
√𝟐 𝟏 𝟒
√𝟐 𝟏
𝟒
= √𝟐√ + + √𝟐𝒊√−
+
𝟐
𝟐
𝟒
𝟒
for 𝑘 = 1
𝜋
𝜋
+ 2𝜋(1)
+ 2𝜋(1)
4
√
= √2 [𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛 4
]
2
2
2
4
= √2 [𝑐𝑜𝑠
9𝜋
9𝜋
+ 𝑖𝑠𝑖𝑛 ]
8
8
√𝟐 𝟏 𝟒
√𝟐 𝟏
𝟒
+
= − √𝟐√ + + √𝟐𝒊√−
𝟐
𝟐
𝟒
𝟒
The root are
√𝟐 𝟏 𝟒
√𝟐 𝟏
𝟒
+
√𝟏 + 𝒊 = √𝟐√ + + √𝟐𝒊√−
𝟒
𝟒
𝟐
𝟐
= 𝟏. 𝟎𝟗𝟖𝟔𝟖𝟒𝟏𝟏𝟑𝟒𝟔𝟕𝟖𝟏 + 𝟎. 𝟒𝟓𝟓𝟎𝟖𝟗𝟖𝟔𝟎𝟓𝟔𝟐𝟐𝟐𝟕𝒊
And
√𝟐 𝟏 𝟒
√𝟐 𝟏
𝟒
+
√𝟏 + 𝒊 = − √𝟐√ + + √𝟐𝒊√−
𝟐
𝟐
𝟒
𝟒
= −𝟏. 𝟎𝟗𝟖𝟔𝟖𝟒𝟏𝟏𝟑𝟒𝟔𝟕𝟖𝟏 + 𝟎. 𝟒𝟓𝟓𝟎𝟖𝟗𝟖𝟔𝟎𝟓𝟔𝟐𝟐𝟐𝟕𝒊
Converting to exponential form
𝒓 = √1.098684113467812 + 0.4550898605622272 = 1.1892071
0.455089860562227
𝜃 = 𝑡𝑎𝑛−1 ( 1.09868411346781 ) = 22𝑜 30′
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𝑧1 = √1 + 𝑖 = 1.1892071𝑖(22
0 30′)
𝑧2 = √1 + 𝑖 = 1.1892071𝑖(−157
0 30′)
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JUNE 2016 SOLUTIONS
Question 1ai
2𝑥 − 𝑦 = 5 ⋯ ⋯ ⋯ ⋯ ⋯ (1)
𝑥 + 2𝑦 = 0 ⋯ ⋯ ⋯ ⋯ ⋯ (2)
From (2)
X = - 2y
Substituting for x in (1)
5(-2y) – y =5
-4y – y = 5
Y= -1
Using equation X = - 2y
X = -2(-1)
X=1
X = 1 and y = -1
Question 1aii
𝑥 + 𝑦 = 1 ⋯ ⋯ ⋯ ⋯ ⋯ (1)
2𝑥 + 2𝑦 = 3 ⋯ ⋯ ⋯ ⋯ ⋯ (2)
From (1)
X=1–y
Substituting for x in (2)
2(1- y) +2 y =3
2 - 2y+2 y =3
2 ≠ 3 therefore the equation has no solution
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Question 1b
1 2
𝐴 = [3 8
2 7
1 4
7 20 ]
9 23
Make zeros in column 1 except entry at row 1, column 1( pivot entry)
Subtract row 1 multiplied by 3 from row 2
R2 = R2 – 3R1
1 2 1 4
= [0 2 4 8 ]
2 7 9 23
Subtract row 1 multiplied by 2 from row 3
R3 = R3 – 2R1
1 2 1 4
= [0 2 4 8 ]
0 3 7 15
Make zeros in column 2 except entry at row 2, column 2( pivot entry)
Subtract row 2 from row 1
R1 = R1 – R2
1 0 −3 −4
= [0 2 4
8 ]
0 3 7 15
Divide row 2 by 2
R2 = R2/2
1 0 −3 −4
= [0 1 2
4 ]
0 3 7 15
Subtract row 2 multiplied by 3 from row 3
R3 = R3 – 3R2
1 0 −3 −4
= [0 1 2
4 ]
0 0 1
3
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Make zeros in column 3 except entry at row 3, column 3( pivot entry)
Add row 3 multiplied by 3 to rows 1
R1 = R1 + 3R3
1 0 0 5
= [0 1 2 4 ]
0 0 1 3
Subtract row 3 multiplied by 2 from row 2
R2 = R2 – 2R3
1 0
𝑟𝑟𝑒𝑓(𝐴) = [0 1
0 0
0 5
0 −2 ]
1 3
Question 1ci
𝐵2 = (𝑎 + 𝑑)𝐵 − (𝑎𝑑 − 𝑏𝑐)𝐼
𝐼= [
1 0
𝑎
],𝐵 = [
0 1
𝑐
𝑏
]
𝑑
𝐵2 = (𝑎 + 𝑑)𝐵 − (𝑎𝑑 − 𝑏𝑐)𝐼
𝐵2 = [
2
𝑎
𝑐
𝑏 𝑎
][
𝑑 𝑐
= [𝑎 + 𝑏𝑐
𝑐𝑎 + 𝑑𝑐
𝑎∗𝑎+𝑏∗𝑐
𝑏
]=[
𝑐∗𝑎+𝑑∗𝑐
𝑑
𝑏𝑎 + 𝑏𝑑 ]
𝑐𝑏 + 𝑑 2
(𝑎 + 𝑑)𝐵 = (𝑎 + 𝑑) [𝑎
𝑐
𝑏∗𝑎+𝑏∗𝑑
]
𝑐∗𝑏+𝑑∗𝑑
2
𝑏
] = [𝑎 + 𝑎𝑑
𝑑
𝑐𝑎 + 𝑑𝑐
𝑏𝑎 + 𝑏𝑑]
𝑎𝑑 + 𝑑2
(𝑎𝑑 − 𝑏𝑐)𝐼 = (𝑎𝑑 − 𝑏𝑐) [1 0] = [𝑎𝑑 − 𝑏𝑐
0 1
0
𝐵2 = (𝑎 + 𝑑)𝐵 − (𝑎𝑑 − 𝑏𝑐)𝐼
2
[𝑎 + 𝑏𝑐
𝑐𝑎 + 𝑑𝑐
𝑏𝑎 + 𝑏𝑑] = [𝑎2 + 𝑎𝑑
𝑐𝑎 + 𝑑𝑐
𝑐𝑏 + 𝑑2
0
]
𝑎𝑑 − 𝑏𝑐
𝑏𝑎 + 𝑏𝑑] − [𝑎𝑑 − 𝑏𝑐
0
𝑎𝑑 + 𝑑2
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0
]
𝑎𝑑 − 𝑏𝑐
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2
(𝑎 + 𝑑)𝐵 − (𝑎𝑑 − 𝑏𝑐)𝐼 = [𝑎 + 𝑎𝑑
𝑐𝑎 + 𝑑𝑐
(𝑎 + 𝑑)𝐵 − (𝑎𝑑 − 𝑏𝑐)𝐼 = [
𝑏𝑎 + 𝑏𝑑 ] − [𝑎𝑑 − 𝑏𝑐
0
𝑎𝑑 + 𝑑 2
0
]
𝑎𝑑 − 𝑏𝑐
𝑎2 + 𝑎𝑑 − (𝑎𝑑 − 𝑏𝑐)
𝑏𝑎 + 𝑏𝑑 − 0
]
𝑐𝑎 + 𝑑𝑐 − 0
𝑎𝑑 + 𝑑2 − (𝑎𝑑 − 𝑏𝑐)
2
(𝑎 + 𝑑)𝐵 − (𝑎𝑑 − 𝑏𝑐)𝐼 = [𝑎 + 𝑏𝑐
𝑐𝑎 + 𝑑𝑐
𝑏𝑎 + 𝑏𝑑 ]
𝑐𝑏 + 𝑑 2
∴ 𝑩𝟐 = (𝒂 + 𝒅)𝑩 − (𝒂𝒅 − 𝒃𝒄)𝑰 shown
Question 1cii
𝐶 2 = (𝑎 + 𝑑)𝐶 − (𝑎𝑑 − 𝑏𝑐)𝐼
𝐶 3 = (𝑎 + 𝑑)𝐶 2 − (𝑎𝑑 − 𝑏𝑐)𝐶
𝐶 4 = (𝑎 + 𝑑)𝐶 3 − (𝑎𝑑 − 𝑏𝑐)𝐶 2
𝐶 2 = (𝑎 + 𝑑)𝐶 − (𝑎𝑑 − 𝑏𝑐)𝐼
𝐶= [
1
2 1
],1= [
0
1 2
2
𝐶 2 = (𝑎 + 𝑑) [
1
0
]
1
1 0
1
]
] − (𝑎𝑑 − 𝑏𝑐) [
0 1
2
=[
2(𝑎 + 𝑑)
𝑎+𝑑
𝑎𝑑 − 𝑏𝑐
]−[
𝑎+𝑑
2(𝑎 + 𝑑)
0
=[
2𝑎 + 2𝑑 − (𝑎𝑑 − 𝑏𝑐)
𝑎+𝑑−0
]
𝑎+𝑑−0
2𝑎 + 2𝑑 − (𝑎𝑑 − 𝑏𝑐)
=[
0
]
𝑎𝑑 − 𝑏𝑐
𝟐𝒂 + 𝟐𝒅 − 𝒂𝒅 + 𝒃𝒄)
𝒂+𝒅
]
𝒂+𝒅
𝟐𝒂 + 𝟐𝒅 − 𝒂𝒅 + 𝒃𝒄)
𝐶 3 = (𝑎 + 𝑑)𝐶 2 − (𝑎𝑑 − 𝑏𝑐)𝐶
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𝐶= [
2 1
],
1 2
𝐶2 = [
2 1 2 1
5 4
]=[
][
]
1 2 1 2
4 5
𝐶 3 = (𝑎 + 𝑑)𝐶 2 − (𝑎𝑑 − 𝑏𝑐)𝐶
5
𝐶 3 = (𝑎 + 𝑑) [
4
=[
2 1
4
]
] − (𝑎𝑑 − 𝑏𝑐) [
1 2
5
5𝑎 + 5𝑑) 4𝑎 + 4𝑑
2𝑎𝑑 − 2𝑏𝑐
]−[
4𝑎 + 4𝑑 5𝑎 + 5𝑑)
𝑎𝑑 − 𝑏𝑐
𝑎𝑑 − 𝑏𝑐
]
2𝑎𝑑 − 2𝑏𝑐
𝟓𝒂 + 𝟓𝒅 − 𝟐𝒂𝒅 + 𝟐𝒃𝒄 𝟒𝒂 + 𝟒𝒅 − 𝒂𝒅 + 𝒃𝒄
=[
]
𝟒𝒂 + 𝟒𝒅 − 𝒂𝒅 + 𝒃𝒄 𝟓𝒂 + 𝟓𝒅 − 𝟐𝒂𝒅 + 𝟐𝒃𝒄
𝐶 4 = (𝑎 + 𝑑)𝐶 3 − (𝑎𝑑 − 𝑏𝑐)𝐶 2
𝐶 2= [
𝐶 3= [
5 4
]
4 5
14 13
5 4 2 1
]
]=[
][
13 14
4 5 1 2
14 13
5
𝐶 4 = (𝑎 + 𝑑) [
] − (𝑎𝑑 − 𝑏𝑐) [
13 14
4
=[
4
]
5
14𝑎 + 14𝑑) 13𝑎 + 13𝑑
5𝑎𝑑 − 5𝑏𝑐
]−[
13𝑎 + 13𝑑 14𝑎 + 14𝑑)
4𝑎𝑑 − 4𝑏𝑐
4𝑎𝑑 − 4𝑏𝑐
]
5𝑎𝑑 − 5𝑏𝑐
𝟏𝟒𝒂 + 𝟏𝟒𝒅 − 𝟓𝒂𝒅 + 𝟓𝒃𝒄 𝟏𝟑𝒂 + 𝟏𝟑𝒅 − 𝟒𝒂𝒅 + 𝟒𝒃𝒄
=[
]
𝟏𝟑𝒂 + 𝟏𝟑𝒅 − 𝟒𝒂𝒅 + 𝟒𝒃𝒄 𝟏𝟒𝒂 + 𝟏𝟒𝒅 − 𝟓𝒂𝒅 + 𝟓𝒃𝒄
Question 2ai
𝑎
det(𝐴1 ) = 𝑑𝑒𝑡 [ 1
𝑐
det(𝐴1 ) = 𝑎1 𝑑 − 𝑐𝑏1
𝑏1
]
𝑑
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Question 2aii
𝑎
det(𝐴2 ) = 𝑑𝑒𝑡 [ 2
𝑐
𝑏2
]
𝑑
det(𝐴2 ) = 𝑎2 𝑑 − 𝑐𝑏2
Question 2aiii
𝑎 + 𝑎2
det(𝐴3 ) = 𝑑𝑒𝑡 [ 1
𝑐
𝑏1 + 𝑏2
]
𝑑
det(𝐴3 ) = 𝑑(𝑎1 + 𝑎2 ) − 𝑐(𝑏1 + 𝑏2 )
Question 2aiv
det(𝐴3 ) = det(𝐴1 ) + det(𝐴2 )
Question 2b
𝑎−𝑥
det(𝐷) = 𝑑𝑒𝑡 [ 1
0
−𝑥
det(𝐷) = (𝑎 − 𝑥) |
1
𝑏
−𝑥
1
𝑐
0]
−𝑥
1
1 0
0
|+𝑐|
|−𝑏|
0
0 −𝑥
−𝑥
−𝑥
|
1
det(𝐷) = (𝑎 − 𝑥)(−𝑥)(−𝑥) − (1)(0)(𝑎 − 𝑥) − 𝑏(1)(𝑥) − 0 + 𝑐(1) − 𝑐(0)(−𝑥)
det(𝐷) = (𝑎 − 𝑥)𝑥 2 + 𝑏𝑥
𝐝𝐞𝐭(𝑫) = 𝒙𝟑 + 𝒂𝒙 + 𝒃𝒙 + 𝒄
Question 2c
4
det(𝐸) = 𝑑𝑒𝑡 [5
2
1
0 2 1
0 4 2]
0 3 4
0 2 3
Divide R1 by R4
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1 0
= 5 0
2 0
[1 0
1
2
4
3
2
1
4
2
4
3]
Multiply R1 by 5 and subtract R1 from R2and restore it
1 0
= 0 0
2 0
[1 0
1
2
3
2
3
2
1
4
3
4
4
3]
Multiply R1 by 2 and subtract R3 from R2 and restore it
1 0
= 0 0
0 0
[1 0
1
4
3
4
7
2
2 3]
1
2
3
2
2
Subtract R1 from R4
1
4
3
4
7
2
3 11
0 0
[
2 4]
1
1 0
2
3
0
0
=
2
0 0 2
Restore R1 to original view;
=
4 0
0 0
0 0
[
0 0
2
3
2
2
3
2
1
3
4
7
2
11
4]
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Divide R3 by 2
=
4 0
0 0
0 0
[
0 0
2
3
2
1
3
2
1
3
4
7
4
11
4]
Multiply R3 by 3/2
4 0
2
3 1
0 0
3
2 4
3 21
=
0 0
2 8
3 11
0 0
[
2 4]
Subtract R3 from R4 and restore it
1
3
0 0
4
=
7
0 0
4
1
0 0 0
[
8]
4 0
2
3
2
1
Restore R3 to original view
1
3
0
4
7
0
2
1
0 0 0
[
8]
4
0 2
3
0
2
0 2
Multiply the main diagonal elements
∆= 4 ∗ 0 ∗ 2 ∗
1
=0
8
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𝐷𝑒𝑡(𝐸) = 0
Question 2d
1 1
𝐹 = [1 9
1 𝐶
1
𝐶]
3
① 1
⟹(1 9
1 𝑐
11 0
𝑐 |0 1
30 0
0
0) * (-1)
1
⤸
R2 – R1=>R2
1 0
① 1
1
( 0 8 𝑐 − 1|−1 1
0 0
1 𝑐
3
R3 – R1=>R3
0
0) * (-1)
1
1
1
1
1 0
(0
⑧
𝑐 − 1|−1 1
0 𝑐−1
2 −1 0
1/8R2 =>R2
⤸
0
0) * (-1/8)
1
⤸
1
1
1
1
0
0
(0
①
(𝑐 − 1)/8|−1/8 1/8 0) * (-C+1))
−1
0
1
0 𝑐−1
2
⤸
R3 – (c -1)R2 =>R3
1
(0
0
1
1
1
0
0
1
(𝑐 − 1)/8
1/8
0) * (-8/(−𝐶 2 + 2𝐶 + 15))
| −1/8
0 (−𝐶 2 + 2𝐶 + 15)/8 (𝑐 − 9)/8 (−𝑐 + 1)/8 1
⤸
−𝐶 2 +2𝐶+15
=>R3
8
3/
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1 1
1
1
0
0
0
1
(𝑐
−
1)/8
−1/8
1/8
0
)
(
|
2
2
2
(−𝑐 + 9)/(𝐶 − 2𝐶 − 15) (𝑐 − 1)/(𝐶 − 2𝐶 − 15) −8/(𝐶 − 2𝐶 − 15)
0 0
①
𝐶−1
R3=>R2
8
R2-
−𝐶 2 + 2𝐶 + 15 ≠ 0
1
(0
0
1
0
0
1 1
1 0 |(−𝑐 + 3)/(𝐶 2 − 2𝐶 − 15) −16/(8𝐶 2 − 16𝐶 − 120) (𝑐 − 1)/(𝐶 2 − 2𝐶 − 15)) ⤣*-1
0 ① (−𝑐 + 9)/(𝐶 2 − 2𝐶 − 15) (𝑐 − 1)/(𝐶 2 − 2𝐶 − 15)
−8/(𝐶 2 − 2𝐶 − 15)
R1 – R3 =>R1
2
2
2
8/(𝐶 2 − 2𝐶 − 15)
1 1 0 𝐶 − 𝐶 − 24/(𝐶 − 2𝐶 − 15) −𝐶 + 1/(𝐶 − 2𝐶 − 15)
(0 ① 0| (−𝑐 + 3)/(𝐶 2 − 2𝐶 − 15)
−16/(8𝐶 2 − 16𝐶 − 120) (𝑐 − 1)/(𝐶 2 − 2𝐶 − 15))
2
0 0 1 (−𝑐 + 9)/(𝐶 − 2𝐶 − 15)
(𝑐 − 1)/(𝐶 2 − 2𝐶 − 15)
−8/(𝐶 2 − 2𝐶 − 15)
R1 – R2 =>R1
2
2
2
2
1 0 0 𝐶 − 27/(𝐶 − 2𝐶 − 15) −𝐶 + 3/(𝐶 − 2𝐶 − 15) −𝐶 + 9/(𝐶 − 2𝐶 − 15)
(0 ① 0|(−𝑐 + 3)/(𝐶 2 − 2𝐶 − 15) −16/(8𝐶 2 − 16𝐶 − 120) (𝑐 − 1)/(𝐶 2 − 2𝐶 − 15) )
0 0 1 (−𝑐 + 9)/(𝐶 2 − 2𝐶 − 15) (𝑐 − 1)/(𝐶 2 − 2𝐶 − 15)
−8/(𝐶 2 − 2𝐶 − 15)
∴𝐹
−1
𝐶 2 − 27/(𝐶 2 − 2𝐶 − 15) −𝐶 + 3/(𝐶 2 − 2𝐶 − 15) −𝐶 + 9/(𝐶 2 − 2𝐶 − 15)
((−𝑐 + 3)/(𝐶 2 − 2𝐶 − 15) −16/(8𝐶 2 − 16𝐶 − 120) (𝑐 − 1)/(𝐶 2 − 2𝐶 − 15) )
(−𝑐 + 9)/(𝐶 2 − 2𝐶 − 15) (𝑐 − 1)/(𝐶 2 − 2𝐶 − 15)
−8/(𝐶 2 − 2𝐶 − 15)
It doesn’t have an inverse if
𝐶 2 − 2𝐶 − 15 = 0
(𝑐 + 3)(𝑐 − 5) = 0
∴ 𝒄 = −𝟑 𝒐𝒓 𝟓
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Question 3a
2U=(-4 , 0 , 8)
3v = (9 , -3, 18)
3𝑣 − 2𝑢 = [(9 − (−2)) , (−3 − 0), (18 − 8)] = (13 , −3 , 10)
Question 3b
‖𝑢 + 𝑣 + 𝑤‖
𝑢 + 𝑣 + 𝑤 = [(−2 + 3 + 2), (0 − 1 − 5), (4 + 6 − 5)] = (3, −6,5)
‖𝑢 + 𝑣 + 𝑤‖ = √32 + −62 + 52 = √70 = 8.37
Question 3c
−3𝑢 = (6,0, −12)
𝑣 + 5𝑤 = (3, −1,6) + (10, −25, −25) = (13, −26,19)
Finding the dot product of 2 vectors
(−3𝑢). (𝑣 + 5𝑤) = (6,0, −12). (13, −26,19)
= (6.13) + (0. (−26)) + ((−12). (19)
= - 150 (distance)
Question 3d
𝑃𝑟𝑜𝑗𝑤 𝑣 =
→.→
𝑤 𝑣
‖→‖
𝑤
2→
𝑤
→.→= (3, −1,6). (2, −5, −5) = 6 + 5 − 30 = −19
𝑤 𝑣
‖→‖ = √22 + −52 + −52 = √54
𝑤
2
‖→‖ = 54
𝑤
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𝑃𝑟𝑜𝑗𝑤 𝑣 =
−19
38 95 95
.(2, −5, −5)= (−
,
, )
54
54 54 54
Question 3e
𝐴𝑟𝑒𝑎 = ‖𝑢 ∗ 𝑣‖
𝑖
𝑎
𝑢∗𝑣 = | 𝑥
𝑏𝑥
𝑗
𝑎𝑦
𝑏𝑦
𝑘
𝑖
𝑗 𝑘
𝑎𝑧 | = |−2 0 4|
𝑏𝑧
3 −1 6
= 𝑖(0 ∗ 6 − (4 ∗ −1)) − 𝑗(6 ∗ −2 − 4 ∗ 3) + 𝑘(−2 ∗ −1 − 0 ∗ 3)
= 4𝑖 + 24𝑗 + 2𝑘
‖𝒖 ∗ 𝒗‖ = √𝟒𝟐 + 𝟐𝟒𝟐 + 𝟐𝟐 = 𝟐𝟒. 𝟒𝟏
Question 3f
→=→∗→
ℎ
𝑤
𝑣
Is the orthogonal vector to → and → so the equation of the plane orthogonal to the tip of → is
𝑣
𝑤
→. (→ − →) where → is a point on the plane.
ℎ
𝑥
𝑢
𝑥
Question 4ai
𝑧1 = 1 + √3i
𝑟 2 = 𝑎2 + 𝑏 2
2
𝑟 = √12 + √3 = 2
𝜃 = 𝑡𝑎𝑛−1 (
𝜋
√3
)=
3
1
The polar form is
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𝑧1 = 2(𝑐𝑜𝑠
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛 )
3
3
Question 4aii
𝑧2 = √3 +i
𝑟 2 = 𝑎2 + 𝑏 2
2
𝑟 = √√3 + 12 = 2
𝜃 = 𝑡𝑎𝑛−1 (
1
√3
)=
The polar form is
𝝅
𝜋
6
𝝅
𝒛𝟐 = 𝟐(𝒄𝒐𝒔 𝟔 + 𝒊𝒔𝒊𝒏 𝟔 ) shown
Question 4aiii
𝜋
𝜋
𝜋
𝜋
𝑧1 𝑧2 = 2(𝑐𝑜𝑠 3 + 𝑖𝑠𝑖𝑛 ). 2(𝑐𝑜𝑠 + 𝑖𝑠𝑖𝑛 )
3
6
6
𝜋 𝜋
𝜋 𝜋
= 4(cos ( + ) + 𝑖𝑠𝑖𝑛 ( + ))
3 6
3 6
𝜋
𝜋
= 4(cos ( ) + 𝑖𝑠𝑖𝑛 ( ))
2
2
𝜋
2
cos ( ) = 0
𝜋
2
𝑠𝑖𝑛 ( ) = 1
and
𝒛𝟏 𝒛𝟐 = 𝟒(𝟎 + 𝒊) = 𝟒𝒊
Question 4aiv
𝜋
𝜋
𝑧1 2(𝑐𝑜𝑠 3 + 𝑖𝑠𝑖𝑛 3 )
=
𝑧2 2(𝑐𝑜𝑠 𝜋 + 𝑖𝑠𝑖𝑛 𝜋)
6
6
𝜋 𝜋
𝜋 𝜋
= cos ( − ) + 𝑖𝑠𝑖𝑛 ( − )
3 6
3 6
𝜋
𝜋
= cos ( ) + 𝑖𝑠𝑖𝑛 ( )
6
6
𝜋
6
cos ( ) =
√3
2
𝜋
6
and 𝑠𝑖𝑛 ( ) =
1
2
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𝒛𝟏
𝒛𝟐
=
𝟏
√𝟑
𝟐
- 𝒊
shown
𝟐
Question 4b
𝑧 = −16
𝑟 = √−162 + 02 = 16
𝜃 = 𝑡𝑎𝑛−1 (
0
)= 𝜋
16
∴ −16 = 16(𝑐𝑜𝑠𝜋 + 𝑖𝑠𝑖𝑛𝜋)
De Moivres theorem states that
𝑛
√𝑟 [𝑐𝑜𝑠
𝜃 + 2𝜋𝑘
𝜃 + 2𝜋𝑘
+ 𝑖𝑠𝑖𝑛
]
𝑛
𝑛
4
= √16 [𝑐𝑜𝑠
𝜋
𝜋
+ 𝑖𝑠𝑖𝑛 ]
2
2
√2
√2
= 2[ + 𝑖 ]
2
2
= √𝟐 + √𝟐𝒊
For k = 1
1
4
𝑧 4 = √16 [𝑐𝑜𝑠
= 2 [𝑐𝑜𝑠
𝑐𝑜𝑠
𝑐𝑜𝑠
3𝜋
4
3𝜋
4
=
𝜋 + 2𝜋
𝜋 + 2𝜋
+ 𝑖𝑠𝑖𝑛
]
4
4
3𝜋
3𝜋
+ 𝑖𝑠𝑖𝑛 ]
4
4
1
√2
=−
and
1
√2
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=−
𝟐
√𝟐
+
𝟐
√𝟐
𝒊
For k = 3
1
4
𝑧 4 = √16 [𝑐𝑜𝑠
= 2 [𝑐𝑜𝑠
𝑐𝑜𝑠
𝑐𝑜𝑠
7𝜋
4
7𝜋
4
7𝜋
7𝜋
+ 𝑖𝑠𝑖𝑛 ]
4
4
=−
=
𝜋 + 6𝜋
𝜋 + 6𝜋
+ 𝑖𝑠𝑖𝑛
]
4
4
2
√2
𝟐
√𝟐
and
2
2
= 2( −
𝑖)
√2 √2
= √𝟐 − √𝟐𝒊
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NOVEMBER 2015 SOLUTIONS
Question 1a
−2
3
2
5
𝑥1 [ ] + 𝑥2 [ ] + 𝑥3 [ ] = [ ]
2
−4
5
6
2𝑥1 + 3𝑥2 − 2𝑥3 = 5 … … … … … … … … . (1)
5𝑥1 − 4𝑥2 + 2𝑥3 = 6 … … … … … … … … . (2)
The augmented matrix is
2 3 −2 5
(
| )
5 −4 2 6
R2 -5/2R1 =>R2
(
2
3
−2 5
| 13)
23
7 −
5 − 2
2
From the matrix above let 𝑥 = 𝑥1 𝑦 = 𝑥2 and 𝑧 = 𝑥3
2𝑥 + 3𝑦 − 2𝑧 = 5……………………………(4)
−
23
𝑦
2
+ 7𝑧 = −
13
……….………………….…(3)
2
From equation 3
𝑦=
14
13
𝑧+
23
23
Stubstitute in equation 4
13
14
2𝑥 = 5 − 3 ( 𝑧 + ) + 2𝑧
23
23
𝑥=
38 2
+ 𝑧
23 23
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∴ 𝑥 = 𝑥1 =
𝑦 = 𝑥2 =
38 2
+ 𝑧
23 23
14
13
𝑧+
23
23
𝑧 = 𝑥3 = 𝑧
If 𝑥1 = 2 𝑥2 = 3 and 𝑥3 = 4, substituting in above solutions
Taking 𝑧 = 𝑥3 = 𝑧 then 𝑧 = 𝑥3 = 4
38
2
Therefore using 𝑥 = 𝑥1 = 23 + 23 𝑧
𝑥 = 𝑥1 =
38 2
+ (4)
23 23
𝑥 = 𝑥1 = 2
Taking
𝑦 = 𝑥2 =
14
13
𝑧+
23
23
Therefore
𝑦 = 𝑥2 =
14
13
(4) +
23
23
𝑦 = 𝑥2 = 3
∴ 𝑥1 = 2 , 𝑥2 = 3 𝑎𝑛𝑑 𝑥3 = 4 is a solution to the system of linear equations.
Question 1b
𝑥 + 2𝑦 = [1,3, −2]…………………………(1)
𝑥 + 𝑦 = [2,0,1]………………………………(2)
Equation 1 – equation 2
𝑦 = [(1 − 2), (3 − 0), (−2 − 1)]
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𝑦 = [−1,3, −3]
From equation 2
𝑥 + 𝑦 = [2,0,1]
𝑥 = [2,0,1] − 𝑦
𝑥 = [2,0,1] − [−1,3, −3]
𝑥 = [3, −3,4]
∴ 𝒙 = [𝟑, −𝟑, 𝟒]
𝒚 = [−𝟏, 𝟑, −𝟑]
Question 1ci
𝐴=[
3 1
]
5 2
Determinant is not zero hence the inerse exist
𝐴1
𝐴2
𝐵1
𝐵2
3
5
1
2
1
0
0
1
𝐵1 𝐵2 𝑖𝑠 𝑡ℎ𝑒 𝑖𝑑𝑒𝑛𝑡𝑖𝑡𝑦 𝑚𝑎𝑡𝑟𝑖𝑥 𝑓𝑜𝑟 2𝑥2
Make pivot in column 1 by dividing R1 by 3
𝐴1
𝐴2
𝐵1
𝐵2
1
5
1/3 1/3 0
2
0
1
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Multiply R1 x 5
𝐴1
𝐴2
𝐵1
𝐵2
5
5
5/3 5/3 0
2
0
1
Subtract R1 fro R2 and restore it
𝐴1
𝐴2
𝐵1
1
0
1/3 1/3
1/3 -5/3
𝐵2
0
1
Make the pivot in second column by R2 by 1/3
𝐴1
𝐴2
𝐵1
1
0
1/3 1/3
1/3 -5/3
𝐵2
0
1
Subtract R2 from R1 and restore it
𝐴1
𝐴2
𝐵1
𝐵2
1
0
0
1
2
-5
-1
3
The inverse matrix is now on the right of 𝐵1 𝐵2
𝐴−1 = [
2 −1
]
−5 3
Question 1cii
AL = B
L = B/A
𝐿 = 𝐵𝐴−1
1 2 2 −1
=[
]
][
3 4 −5 3
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1 ∗ 2 + 2 ∗ 5 1 ∗ −1 + 2 ∗ 3
=[
]
3 ∗ 2 + 4 ∗ −5 3 ∗ −1 + 4 ∗ 3
−8 5
=[
]
−14 9
Question 1ciii
MA= B
M=B/A
𝑀 = 𝐵𝐴
1 2 2 −1
=[
]
][
3 4 −5 3
1 ∗ 2 + 2 ∗ 5 1 ∗ −1 + 2 ∗ 3
=[
]
3 ∗ 2 + 4 ∗ −5 3 ∗ −1 + 4 ∗ 3
−8 5
=[
]
−14 9
Question 1d
Let
𝑐=[
𝑎
𝑐
𝑏
]
𝑑
𝐾𝑐 = [
𝐾𝑎
𝐾𝑐
𝐾𝑏
]
𝐾𝑑
𝐾𝑐 = [
𝐾𝑎
𝐾𝑐
𝐾𝑏
]=0
𝐾𝑑
If
𝐾𝑐 = 0
For the above equation to be true, if K=0 then kC = 0. If c = 0 the Kc = 0 also.
Therefore for Kc =0 either K = 0 or c =0
Question 2ai
Let
𝑎
𝐷=[
𝑐
𝑏
]
𝑑
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Then
𝐷 −1 =
1
𝑑
[
𝑎𝑑 − 𝑏𝑐 −𝑐
𝑑
− 𝑏𝑐
= [𝑎𝑑−𝑐
𝑎𝑑 − 𝑏𝑐
−𝑏
]
𝑎
−𝑏
𝑎𝑑 − 𝑏𝑐 ]
𝑎
𝑎𝑑 − 𝑏𝑐
𝑎
−1
𝐷 −1 =
[𝑎𝑑 − 𝑏𝑐
𝑐
𝑎
−𝑏
−𝑐
𝑑
)(
)−(
)(
)
(
𝑎𝑑 − 𝑏𝑐 𝑎𝑑 − 𝑏𝑐 𝑎𝑑 − 𝑏𝑐
𝑎𝑑 − 𝑏𝑐 𝑎𝑑 − 𝑏𝑐
1
𝑑
− 𝑏𝑐
=
[𝑎𝑑−𝑐
𝑎𝑑
𝑏𝑐
(
)−(
)
(𝑎𝑑 − 𝑏𝑐)2
(𝑎𝑑 − 𝑏𝑐)2 𝑎𝑑 − 𝑏𝑐
1
𝑎
1
=
[𝑎𝑑 − 𝑏𝑐
𝑐
𝑎𝑑 − 𝑏𝑐
(
)
(𝑎𝑑 − 𝑏𝑐)2 𝑎𝑑 − 𝑏𝑐
𝑎
(𝑎𝑑 − 𝑏𝑐)2 𝑎𝑑 − 𝑏𝑐
=
[
𝑐
𝑎𝑑 − 𝑏𝑐
𝑎𝑑 − 𝑏𝑐
𝑎
= 𝑎𝑑 − 𝑏𝑐 [𝑎𝑑 − 𝑏𝑐
𝑐
𝑎𝑑 − 𝑏𝑐
𝒂 𝒃
=[
] shown
𝒄 𝒅
𝑏
𝑎𝑑 − 𝑏𝑐]
𝑑
𝑎𝑑 − 𝑏𝑐
−𝑏
𝑎𝑑 − 𝑏𝑐]
𝑎
𝑎𝑑 − 𝑏𝑐
𝑏
𝑎𝑑 − 𝑏𝑐]
𝑑
𝑎𝑑 − 𝑏𝑐
𝑏
𝑎𝑑 − 𝑏𝑐]
𝑑
𝑎𝑑 − 𝑏𝑐
𝑏
𝑎𝑑 − 𝑏𝑐]
𝑑
𝑎𝑑 − 𝑏𝑐
Question 2aii
If D and E are invertible matrices, then according to the law below DE will also be invertible.
𝐷𝐸 −1 = 𝐸 −1 𝐷 −1
Question 2b
The matrix for the system becomes
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2
1 −3 0
(4
5
1 |8)
−2 −1 4 2
Using the Cramers rule a matix like
𝑎1
(𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑎1
𝑎
𝐷=( 2
𝑎3
𝑑1
𝐷𝑥 = (𝑑2
𝑑3
𝑐1 𝑑1
𝑐2 |𝑑2 ) is divided into 3x3 matrices
𝑐3 𝑑3
𝑏1
𝑏2
𝑏3
𝑏1
𝑏2
𝑏3
𝑎1
𝐷𝑦 = (𝑎2
𝑎3
𝑎1
𝐷𝑧 = (𝑎2
𝑎3
∴𝑥=
𝑑1
𝑑2
𝑑3
𝑑𝑒𝑡𝐷𝑥
𝑑𝑒𝑡𝐷
𝑏1
𝑏2
𝑏3
𝑐1
𝑐2 )
𝑐3
𝑐1
𝑐2 )
𝑐3
𝑐1
𝑐2 )
𝑐3
𝑑1
𝑑2 )
𝑑3
,𝑦 =
𝑑𝑒𝑡𝐷𝑦
𝑑𝑒𝑡𝐷
,𝑧 =
𝑑𝑒𝑡𝐷𝑧
𝑑𝑒𝑡𝐷
Therefore from our matrix
2
1 −3
𝐷=( 4
5
1)
−2 −1 4
0
𝐷𝑥 = (8
2
1 −3
5
1)
−1 4
2 0 −3
𝐷𝑦 = ( 4 8 1 )
−2 2 4
2
1 0
𝐷𝑧 = ( 4
5 8)
−2 −1 2
Therefore
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2
1 −3
𝐷=( 4
5
1)
−2 −1 4
2
1 −3
2
1
𝑑𝑒𝑡𝐷 = ( 4
5
1 )( 4
5)
−2 −1
−2 −1 4
𝑑𝑒𝑡𝐷 = (2 ∗ 5 ∗ 4) + (1 ∗ 1 ∗ −2) + (−3 ∗ 4 ∗ −1) = 50
0
𝐷𝑥 = (8
2
1 −3
5
1)
−1 4
0
𝑑𝑒𝑡𝐷𝑥 = (8
2
1 −3 0 1
5
1 ) (8 5 )
2 −1
−1 4
𝑑𝑒𝑡𝐷𝑥 = (0 ∗ 5 ∗ 4) + (1 ∗ 1 ∗ 2) + (−3 ∗ 8 ∗ −1) = 26
2 0 −3
𝐷𝑦 = ( 4 8 1 )
−2 2 4
2 0 −3
2 0
𝑑𝑒𝑦𝐷𝑦 = ( 4 8 1 ) ( 4 8)
−2 2 4
−2 2
𝑑𝑒𝑦𝐷𝑦 = (2 ∗ 8 ∗ 4) + (0 ∗ 1 ∗ −2) + (−3 ∗ 4 ∗ 2) = 40
2
1 0
𝐷𝑧 = ( 4
5 8)
−2 −1 2
2
1 0
2
1
𝑑𝑒𝑡𝐷𝑧 = ( 4
5 8) ( 4
5)
−2 −1 2 −2 −1
𝑑𝑒𝑡𝐷𝑧 = (2 ∗ 5 ∗ 2) + (1 ∗ 8 ∗ −2) + (0 ∗ 4 ∗ −1) = 4
∴𝑥=
𝑦=
𝑧=
𝑑𝑒𝑡𝐷𝑥
𝑑𝑒𝑡𝐷
𝑑𝑒𝑡𝐷𝑦
𝑑𝑒𝑡𝐷
𝑑𝑒𝑡𝐷𝑧
𝑑𝑒𝑡𝐷
26
13
= 50 = 25 ,
40
4
4
2
= 50 = 5 ,
= 50 = 25
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Question 2c
𝑎 0 0
𝐹 = [𝑢 𝑏 0
𝑣 𝑤 𝑐
𝑥 𝑦 𝑧
0
0]
0
𝑑
Using crammers rule to find the determinant,
𝑎 0 0
𝑏 0
𝑢
𝑑𝑒𝑡𝐹 = [
𝑣 𝑤 𝑐
𝑥 𝑦 𝑧
0 𝑎 0 0
0 ] [𝑢 𝑏 0]
0 𝑣 𝑤 𝑐
𝑑 𝑥 𝑦 𝑧
𝑑𝑒𝑡𝐹 = (𝑎 ∗ 𝑏 ∗ 𝑐 ∗ 𝑑) + (0 ∗ 0 ∗ 0 ∗ 𝑥) + (0 ∗ 0 ∗ 𝑣 ∗ 𝑦) + (0 ∗ 𝑢 ∗ 𝑤 ∗ 𝑧) = 𝑎𝑏𝑐𝑑
𝑑𝑒𝑡𝐹 is a product od the diagonal entries.
Question 2d
1
0
𝑎 = [−1 3
0 2𝑘
−𝑘
1]
−4
Inverse of a 3x3 matrix can be found
𝑎
[𝑑
𝑔
𝑏
𝑒
ℎ
𝑒𝑖 − 𝑓ℎ
−(𝑏𝑖 − 𝑐ℎ)
𝑏𝑓 − 𝑐𝑒
𝑐 −1
1
𝑓] =
𝑎𝑖 − 𝑐𝑔
−(𝑎𝑓 − 𝑐𝑑)]
[−(𝑑𝑖 − 𝑓𝑔)
𝑑𝑒𝑡
𝑖
𝑑ℎ − 𝑒𝑔
−(𝑎ℎ − 𝑏𝑔)
𝑎𝑒 − 𝑏𝑑
1
0
det(𝑎) = [−1 3
0 2𝑘
0
−𝑘 1
1 ] [−1 3 ]
−4 0 2𝑘
det(𝑎) = (1 ∗ 3 ∗ −4) + (0 ∗ 1 ∗ 0) + (−𝑘 ∗ −1 ∗ 2𝑘) = −12 + 2𝑘 2
For a matrix to have an inverse, the determinant should ≠0
∴ −12 + 2𝑘 2 > 0
2𝑘 2 > 12
𝑘 > √6
𝑘 > 2.449
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K should be 3
Question 3a
If the vectors are orthogonal then
u.v=0
𝑢. 𝑣 = (1,0, √2 ).(1, √2, 0) = (1 ∗ 1) + (0 ∗ √2 ) + (0 ∗ √2 )
𝑢. 𝑣 = 1 ≠ 0
Therefore there are no orthogonal complements of each other.
Question 3b
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚 = |→∗→|
𝑖
|→∗→| = |1
𝑢 𝑣
1
𝑗
𝑢
𝑘
𝑣
0 √2|
√2 0
1
= | 0 √2| → − |1 √2| → + |
𝑖
𝑗
1
1 0
√2 0
0
|→
√2 𝑘
= [(0 ∗ 0) − (√2 ∗ √2) ] → − [(0 ∗ 1) − (1 ∗ √2) ] → + [(1 ∗ √2) − (0 ∗ 1) ] →
= −2 → − √2 → + √2 →
𝑖
𝑗
𝑘
2
𝑖
𝑗
𝑘
2
||→∗→|| = √−22 + √2 + √2 = √8
𝑢
𝑣
Let the equation of the plane be
𝑎(𝑥 − 𝑥0 ) + 𝑏(𝑦 − 𝑦0 ) + 𝑐(𝑧 − 𝑧0 ) = 0
(𝑥0 ), (𝑦0 ), (𝑧0 ) is a point on the plane
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Given 𝑢 = (1 , 0 , √2) 𝑣 = (1 , √2 , 0)
→= ⟨(1 − 1), (0 − √2) , (√2 − 0)⟩ = ⟨0, −√2, √2⟩
𝑣1
To find the perpendicular vector, take
→. (→ − → ) = 0
𝑛
𝑟
𝑟0
(𝑎, 𝑏, 𝑐). [⟨𝑥, 𝑦, 𝑧⟩ − ⟨𝑥0 , 𝑦0 , 𝑧0 ⟩ ] = 0
(𝑎, 𝑏, 𝑐). [⟨𝑥 − 𝑥0 , 𝑦 − 𝑦0 , 𝑧 − 𝑧0 ⟩ ] = 0
Using →
𝑣1
(𝑎, 𝑏, 𝑐). [⟨𝑥 − 0, 𝑦 + √2, 𝑧 − √2⟩ ] = 0
Therefore the equation of the plane is
𝒂𝒙 + 𝒃(𝒚 + √𝟐) + 𝒄(𝒛 − √𝟐) = 𝟎
Question 3d
By letting u = ⟨𝑥, 𝑦, 𝑧⟩, r_0 =⟨𝑥_0, 𝑦_0, 𝑧_0⟩, and v = ⟨𝑎, 𝑏, 𝑐⟩ we obtain the equations
⟨𝑥, 𝑦, 𝑧⟩ = ⟨𝑥0 + 𝑡𝑎, 𝑦0 + 𝑡𝑏, 𝑧0 + 𝑡𝑐⟩
Which gives the parametric equations of line passing the point p_o =⟨𝑥_0, 𝑦_0, 𝑧_0⟩, parallel to
v = ⟨𝑎, 𝑏, 𝑐⟩
Therefore
X= 𝑥0 + 𝑡𝑎
y = 𝑦0 + 𝑡𝑏
z= 𝑧0 + 𝑡𝑐
Since 𝑣 = (1 , √2 , 0)
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X= 1 + 𝑡(1) = 1 + 𝑡
y = 0 + 𝑡√2 = √2𝑡
z= √2 + 𝑡(0) = √2
Question 4a
De Moivres theorem states that if
𝑧 = (𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 𝑛 = (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)
(𝑐𝑜𝑠4𝜃 + 𝑖𝑠𝑖𝑛4𝜃) = (𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)4 ≡ (𝑐 + 𝑖𝑠)4
Using the pascals triangle up to the 4th term
1
1 2 1
|
1 3 3 1
1 4 6 4 1
(𝑐𝑜𝑠4𝜃 + 𝑖𝑠𝑖𝑛4𝜃) = 𝑐 4 + 4𝑐 3 𝑖𝑠 + 6𝑐 2 𝑖𝑠 2 + 4𝑐𝑖𝑠 3 + 𝑖𝑠 4
Take the imaginary parts of both sides
𝑠𝑖𝑛4𝜃 = 4𝑐 3 𝑠 + 4𝑐𝑖𝑠 3
𝒔𝒊𝒏𝟒𝜽 = 𝟒𝒄𝒐𝒔𝟑 𝜽 + 𝟒𝒄𝒐𝒔𝜽𝒔𝒊𝒏𝟑 𝜽
Question 4b
For a complex number = 𝑎 + 𝑏𝑖 the polar form is
𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑟 2 = 𝑎2 + 𝑏 2
𝑏
𝜃 = 𝑡𝑎𝑛−1 ( )
𝑎
a= 16 b=0
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𝑟 = √162 + 02 = 16
𝜃 = 𝑡𝑎𝑛−1 (
0
)=0
16
Therefore
16(𝑐𝑜𝑠0 + 𝑖𝑠𝑖𝑛0)
According to the De Moivres formula, all nth roots of a complex number
𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃) is given by
𝑛
√𝑟 [𝑐𝑜𝑠
𝜃 + 2𝜋𝑘
𝜃 + 2𝜋𝑘
+ 𝑖𝑠𝑖𝑛
]
𝑛
𝑛
We have
𝑟 = 16, 𝜃 = 0, 𝑛 = 4
Thus 𝑘 = 0
4
√16 [𝑐𝑜𝑠
0 + 2𝜋(0)
0 + 2𝜋(0)
+ 𝑖𝑠𝑖𝑛
]
4
4
2[𝑐𝑜𝑠0 + 𝑖𝑠𝑖𝑛0] = 2
𝑘=1
0 + 2𝜋(1)
0 + 2𝜋(1)
+ 𝑖𝑠𝑖𝑛
]
4
4
𝜋
𝜋
2 [𝑐𝑜𝑠 + 𝑖𝑠𝑖𝑛 ] = 2𝑖
2
2
4
√16 [𝑐𝑜𝑠
𝑘=2
4
√16 [𝑐𝑜𝑠
4
0 + 2𝜋(2)
0 + 2𝜋(2)
+ 𝑖𝑠𝑖𝑛
]
4
4
√16[𝑐𝑜𝑠𝜋 + 𝑖𝑠𝑖𝑛𝜋] = −2
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Thus 𝑘 = 3
4
√16 [𝑐𝑜𝑠
2 [𝑐𝑜𝑠
0 + 2𝜋(3)
0 + 2𝜋(3)
+ 𝑖𝑠𝑖𝑛
]
4
4
3𝜋
3𝜋
+ 𝑖𝑠𝑖𝑛 ] = −2𝑖
4
4
Therefore the roots are
4
√16 = 2
4
√16 = 2𝑖
4
√16 = −2
4
√16 = −2𝑖
Question 4c
(1 + 𝑖)128 in polar form is
𝑟 2 = 𝑎2 + 𝑏 2 = √12 + 12 = √2
𝑏
𝜃 = 𝑡𝑎𝑛−1 ( )
𝑎
1
𝜋
𝜃 = 𝑡𝑎𝑛−1 ( ) =
1
4
(1 + 𝑖)128 = [√2 (𝑐𝑜𝑠
128
= √2
(𝑐𝑜𝑠
𝜋
𝜋 128
+ 𝑖𝑠𝑖𝑛 )]
4
4
𝜋
𝜋 128
+ 𝑖𝑠𝑖𝑛 )
4
4
= 264 (𝑐𝑜𝑠32𝜋 + 𝑖𝑠𝑖𝑛32𝜋)
𝑐𝑜𝑠𝜋 = −1 and
𝑠𝑖𝑛𝜋 = 0
Therefore
(1 + 𝑖)128 = 264 (𝑐𝑜𝑠32𝜋 + 𝑖𝑠𝑖𝑛32𝜋)
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= 264 (𝑐𝑜𝑠𝜋 + 𝑖𝑠𝑖𝑛𝜋)
= 264 (𝑐𝑜𝑠𝜋 + 𝑖𝑠𝑖𝑛𝜋)
= 264 (−1 + 0)
= −𝟐𝟔𝟒
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