(a)(i) Reduce the following matrix to a row-reduced echelon form and hence also, find its rank: A = [1 3 2 4 1 0 0 2 2 0 2 6 2…

(a)(i) Row-reduced echelon form and rank

A = [1 3 2 4 1; 0 0 2 2 0; 2 6 2 6 2; 3 9 1 10 6].

R₃ → R₃ − 2R₁, R₄ → R₄ − 3R₁: [1 3 2 4 1; 0 0 2 2 0; 0 0 −2 −2 0; 0 0 −5 −2 3].

R₃ → R₃ + R₂ gives a zero row; R₂ → R₂/2: [1 3 2 4 1; 0 0 1 1 0; 0 0 0 0 0; 0 0 −5 −2 3].

R₁ → R₁ − 2R₂, R₄ → R₄ + 5R₂: [1 3 0 2 1; 0 0 1 1 0; 0 0 0 0 0; 0 0 0 3 3].

Interchange R₃ and R₄, then R₃ → R₃/3: [1 3 0 2 1; 0 0 1 1 0; 0 0 0 1 1; 0 0 0 0 0].

R₁ → R₁ − 2R₃, R₂ → R₂ − R₃ gives the row-reduced echelon form

[1 3 0 0 −1; 0 0 1 0 −1; 0 0 0 1 1; 0 0 0 0 0].

It has three non-zero rows, so rank A = 3.

(a)(ii) Eigenvalues and eigenvectors of [0 −i; i 0]

det(A − λI) = λ² − (−i)(i) = λ² − 1 = 0, so λ = 1 and λ = −1.

For λ = 1: −x − iy = 0, so x = −iy; taking y = i gives the eigenvector (1, i)ᵀ.

For λ = −1: x − iy = 0, so x = iy; taking y = 1 gives the eigenvector (i, 1)ᵀ.

(Any non-zero multiples are also eigenvectors.)

(b) Area of the astroid x^(2/3) + y^(2/3) = a^(2/3)

Put x = a cos³θ, y = a sin³θ, which satisfies the equation. The curve is symmetric in both axes, so the area is four times the first-quadrant area.

First-quadrant area = ∫ y dx with dx = −3a cos²θ sinθ dθ, θ from π/2 to 0:

= 3a² ∫₀^(π/2) sin⁴θ cos²θ dθ.

Using ∫₀^(π/2) sinᵐθ cosⁿθ dθ = Γ((m + 1)/2)Γ((n + 1)/2) / [2Γ((m + n + 2)/2)] with m = 4, n = 2:

= Γ(5/2)Γ(3/2) / (2Γ(4)) = (3√π/4)(√π/2) / 12 = π/32.

So the first-quadrant area = 3a²π/32, and the total area = 4 × 3πa²/32 = 3πa²/8.

(c) Plane containing the two lines, and their point of intersection

L₁: x = −1 + 3t, y = −3 + 5t, z = −5 + 7t. L₂: x = 2 + s, y = 4 + 3s, z = 6 + 5s.

Equating coordinates: 3t − s = 3, 5t − 3s = 7, 7t − 5s = 11. The first two give t = 1/2, s = −3/2, which also satisfy the third, so the lines intersect (and hence are coplanar) at (1/2, −1/2, −3/2).

The normal to the plane is d₁ × d₂ = (3, 5, 7) × (1, 3, 5) = (4, −8, 4), proportional to (1, −2, 1).

The plane through the point (1/2, −1/2, −3/2) is (x − 1/2) − 2(y + 1/2) + (z + 3/2) = 0, that is, x − 2y + z = 0.