Test 5 - Sol

Pure Mathematics – Matrix and Determinants

p.1

King’s College 2007 – 2008 F.6 Pure Mathematics Revision Test 5 Suggested Solution 1.

(a) (E) has a unique solution 1 1 1 ⇔ k 1 k ≠0 k2 1 k 1 ⇔ k k2 −k

1 1 1 k ≠0 0 0

⇔ k (k − 1) ≠ 0 ⇔ k ≠ 0, 1 2

By Cramer’s Rule,

a 1 1

x=

y=

z=

b 1 k c 1 k k (k − 1) 1 k k2

=

2

a 1 b k c k

k (k − 1)

1

a

2

1 1 a k 1 b k2 1 c

1

b − a 0 k −1 c − a 0 k −1 k (k − 1)

=

0 0 2 k −k

2

a 1 b k c k

k (k − 1)

2

− (k − 1) =

k (k − 1) k (k − 1)

=

1 1 a k −1 0 b − a k 2 −1 0 c − a

b−a 1 c−a 1

a 1 b k

k (k − 1)

−

2

2

=

=

− (b − a − c + a ) c−b = k (k − 1) k (k − 1)

ak − b k −1

k −1 b − a k 2 −1 c − a

− (k − 1)

1 b−a k +1 c − a

= = = 2 2 2 2 k (k − 1) k (k − 1) k (k − 1) k (k − 1) − [c − a − (k + 1)(b − a )] − (c − a − bk + ak − b + a ) b(k + 1) − c − ak = = = k (k − 1) k (k − 1) k (k − 1)

(b) If k = 0, the system (E) becomes ⎧x + y + z = a (E ) : ⎪⎨ y = b ⎪y = c ⎩

Therefore, (E) is consistent iff b = c. The solution set is {(t, b, a – b – t); t ∈ R}

Pure Mathematics – Matrix and Determinants

2.

p.2

5π ⎛ ⎜ cos 6 (a) The transformation matrix is ⎜ 5 ⎜ sin π ⎜ 6 ⎝

5π 6 5π cos 6

− sin

⎞ ⎟ 1 ⎛− 3 −1 ⎞ ⎟ ⎟= ⎜ − 3 ⎟⎠ ⎟ 2 ⎜⎝ 1 ⎟ ⎠

Let (x’, y’) be the image of (x, y) under the rotation.

⎛ x' ⎞ 1 ⎛ − 3 − 1 ⎞⎛ x ⎞ ⎟⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ − 3 ⎟⎠⎝ y ⎠ ⎝ y' ⎠ 2 ⎝ 1 ⎛− 3 −1 ⎞ ⎛ x⎞ ⎟ ⎜⎜ ⎟⎟ = 2⎜⎜ ⎟ 1 3 − ⎝ y⎠ ⎝ ⎠

−1

( (

) )

⎛1 ⎞ ⎛ x' ⎞ 1 ⎛ − 3 1 ⎞⎛ x' ⎞ ⎜ 2 − 3 x'+ y ' ⎟ ⎟⎜⎜ ⎟⎟ = ⎜ ⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟ ⎝ y ' ⎠ 2 ⎝ − 1 − 3 ⎠⎝ y ' ⎠ ⎜⎜ 1 − x'− 3 y ' ⎟⎟ ⎝2 ⎠

Equation of the image of (Γ) is

(

)

2

(

) (

)

(

3 x' 2 +2 x' y '− 3 y ' 2 +

(

)

2

⎡1 ⎤ ⎡1 ⎤⎡1 ⎤ ⎡1 ⎤ 5⎢ − 3 x'+ y ' ⎥ − 2 3 ⎢ − 3 x'+ y ' ⎥ ⎢ − x'− 3 y ' ⎥ + 7 ⎢ − x'− 3 y ' ⎥ − 4 = 0 ⎣2 ⎦ ⎣2 ⎦⎣ 2 ⎦ ⎣2 ⎦

(

)

3 5 3 x' 2 −2 3 x' y '+ y ' 2 − 2 4 2 2 16 x' +32 y ' −16 = 0

) 74 (x' +2

)

3 x' y '+3 y ' 2 − 4 = 0

2

x' 2 +2 y ' 2 −1 = 0 2π ⎛ ⎜ cos 3 (b) The transformation matrix is ⎜ ⎜ sin 2π ⎜ 3 ⎝

2π 3 2π − cos 3 sin

⎞ ⎟ 1 ⎛ −1 ⎟= ⎜ ⎟ 2 ⎜⎝ 3 ⎟ ⎠

3⎞ ⎟ 1 ⎟⎠

Let (x’, y’) be the image of (x, y) under the rotation.

⎛ x' ⎞ 1 ⎛ − 1 ⎜⎜ ⎟⎟ = ⎜⎜ ⎝ y' ⎠ 2 ⎝ 3

3 ⎞⎛ x ⎞ ⎟⎜⎜ ⎟⎟ 1 ⎟⎠⎝ y ⎠

⎛ −1 ⎛ x⎞ ⎜⎜ ⎟⎟ = 2⎜⎜ ⎝ y⎠ ⎝ 3

3⎞ ⎟ 1 ⎟⎠

−1

⎛ x' ⎞ 1 ⎛ − 1 ⎜⎜ ⎟⎟ = ⎜⎜ ⎝ y' ⎠ 2 ⎝ 3

(

)

⎛1 ⎞ 3 ⎞⎛ x' ⎞ ⎜ 2 − x'+ 3 y ' ⎟ ⎟⎜⎜ ⎟⎟ = ⎜ ⎟ 1 ⎟⎠⎝ y ' ⎠ ⎜⎜ 1 3 x'+ y ' ⎟⎟ ⎝ 2 ⎠

(

)

Equation of the image of (Γ) is ⎡1 ⎢2 ⎣

(

)

(

2

) (

⎤ ⎡1 ⎤⎡ 1 3 x'+ y ' ⎥ − 2 3 ⎢ − x'+ 3 y ' ⎥ ⎢ ⎦ ⎣2 ⎦⎣ 2

(

)

(

)

(

)

2

⎤ ⎡1 ⎤ 3 x'+ y ' ⎥ − ⎢ − x'+ 3 y ' ⎥ − 2 = 0 ⎦ ⎣2 ⎦

) (

)

1 3 1 − 3 x 2 '+2 x' y '+ 3 y ' 2 − x' 2 −2 3 x' y '+3 y ' 2 − 2 = 0 3 x' 2 +2 3 x' y '+ y ' 2 − 4 2 4 2 2 8 x' −8 y ' −8 = 0 x ' 2 − y ' 2 −1 = 0

Pure Mathematics – Matrix and Determinants

⎛0 (c) The reflection matrix is ⎜⎜ ⎝1 ⎛1 Let the shear matrix be ⎜⎜ ⎝μ

p.3

1⎞ ⎟ 0 ⎟⎠ 0⎞ ⎟ . Therefore, 1 ⎟⎠

⎛ 1 ⎞ ⎛ 1 0 ⎞⎛1⎞ ⎛ 1 ⎞ ⎟⎟ ⇒ μ = 2 ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎝ 3 ⎠ ⎝ μ 1 ⎠⎝1⎠ ⎝1 + μ ⎠ ⎛ 1 0⎞ ⎛ 1 0⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ μ 1⎠ ⎝ 2 1⎠

Let (x’, y’) be the image of (x, y) under the transformation.

⎛ x' ⎞ ⎛ 1 0 ⎞⎛ 0 1 ⎞⎛ x ⎞ ⎛ 0 1 ⎞⎛ x ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎝ y ' ⎠ ⎝ 2 1 ⎠⎝ 1 0 ⎠⎝ y ⎠ ⎝ 1 2 ⎠⎝ y ⎠ −1

⎛ x ⎞ ⎛ 0 1 ⎞ ⎛ x' ⎞ ⎛ − 2 1 ⎞⎛ x' ⎞ ⎛ − 2 x'+ y ' ⎞ ⎟⎟ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎜⎜ ⎟⎟ = ⎜⎜ y 1 2 y ' 1 0 y ' x ' ⎠ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎝ ⎠ ⎝ Equation of the image of (Γ) is

(− 2 x'+ y ')2 + (x')2 − 4(− 2 x'+ y ') = 0

(4 x'

2

)

−4 x' y '+ y ' 2 + x' 2 +8 x'−4 y ' = 0

5 x' −4 x' y '+ y ' +8 x'−4 y ' = 0 2

2