(a) Let F be a subfield of complex numbers and let T be the linear transformation from F³ into F³, defined by T(x, y, z) = (x –…

(a) Null space and range of T The matrix of T with respect to the standard basis is A = [1 −1 2; 2 1 0; −1 −2 2]. (i) (a, b, c) ∈ N(T) ⇔ T(a, b, c) = 0: a − b + 2c = 0, 2a + b = 0, −a − 2b + 2c = 0. From the second, b = −2a; the first gives 3a + 2c = 0, so c = −3a/2; the third is then satisfied: −a + 4a − 3a = 0. Condition: b = −2a, c = −3a/2, i.e. (a, b, c) is a multiple of (2, −4, −3). N(T) = span{(2, −4, −3)}, so nullity of T = 1. (ii) (a, b, c) ∈ R(T) ⇔ T(x, y, z) = (a, b, c) is solvable. Row-reduce the augmented matrix: [1 −1 2 | a; 2 1 0 | b; −1 −2 2 | c] R₂ → R₂ − 2R₁, R₃ → R₃ + R₁: [1 −1 2 | a; 0 3 −4 | b − 2a; 0 −3 4 | c + a] R₃ → R₃ + R₂: [1 −1 2 | a; 0 3 −4 | b − 2a; 0 0 0 | b + c − a]. The system is consistent iff b + c − a = 0. Condition: a = b + c (equivalently a − b − c = 0). Check: the columns (1, 2, −1), (−1, 1, −2), (2, 0, 2) all satisfy it. The echelon form has two non-zero rows, so rank of T = 2, and rank + nullity = 2 + 1 = 3 = dim F³.

(b) Tracing y²(a + x) = x²(3a − x), a > 0 Symmetry: only even powers of y occur, so the curve is symmetric about the x-axis. Origin: the curve passes through (0, 0); the lowest-degree terms a y² − 3a x² = 0 give the tangents y = ±√3 x, real and distinct, so the origin is a node. Intercepts: y = 0 ⇒ x = 0 or x = 3a. At (3a, 0), shifting x = 3a + X gives 4a y² ≈ −9a² X to first order, so the tangent there is vertical (x = 3a). Asymptotes: the coefficient of the highest power of y is (a + x), so x = −a is an asymptote parallel to the y-axis. The highest-degree terms x(x² + y²) have no other real linear factor, so there is no other real asymptote. Region: y² = x²(3a − x)/(a + x). x > 3a: y² < 0, no curve. x < −a: y² < 0, no curve. −a < x < 3a: y² ≥ 0, curve exists. Shape: d(y²)/dx = −2x(x² − 3a²)/(a + x)², which vanishes in (−a, 3a) only at x = 0 and x = √3 a. For 0 < x < 3a: y² rises from 0 to its maximum (6√3 − 9)a² at x = √3 a (|y| ≈ 1.18a), then falls to 0 at x = 3a, so there is a loop between x = 0 and x = 3a. For −a < x < 0: y² decreases steadily from +∞ (as x → −a⁺) to 0, so two infinite branches come in from the asymptote x = −a (one above, one below the x-axis) and pass through the origin. Conclusion: a strophoid-type curve: a loop on 0 ≤ x ≤ 3a, a node at the origin with tangents y = ±√3 x, two infinite branches in −a < x < 0 approaching the asymptote x = −a, and no part of the curve for x < −a or x > 3a.

(c)(i) Generators of the hyperboloid x²/4 + y²/9 − z²/16 = 1 ⇔ (x/2 − z/4)(x/2 + z/4) = (1 − y/3)(1 + y/3). λ-system: x/2 − z/4 = λ(1 − y/3), λ(x/2 + z/4) = 1 + y/3. μ-system: x/2 − z/4 = μ(1 + y/3), μ(x/2 + z/4) = 1 − y/3. Both points lie on the surface: 1 + 1 − 1 = 1 and 1 + 1/9 − 1/9 = 1. At (2, 3, −4): x/2 − z/4 = 2, x/2 + z/4 = 0, 1 − y/3 = 0, 1 + y/3 = 2. λ-system: 2 = λ·0 has no finite λ; λ → ∞ gives the generator 1 − y/3 = 0, x/2 + z/4 = 0, i.e. y = 3, 2x + z = 0. μ-system: 2 = 2μ ⇒ μ = 1 (second equation 0 = 0 holds): x/2 − z/4 = 1 + y/3, x/2 + z/4 = 1 − y/3; adding and subtracting, x = 2, 4y + 3z = 0. At (2, −1, 4/3): x/2 − z/4 = 2/3, x/2 + z/4 = 4/3, 1 − y/3 = 4/3, 1 + y/3 = 2/3. λ-system: 2/3 = (4/3)λ ⇒ λ = 1/2 (check: (1/2)(4/3) = 2/3): x/2 − z/4 = ½(1 − y/3), x/2 + z/4 = 2(1 + y/3), i.e. 6x + 2y − 3z = 6, 6x − 8y + 3z = 24, or (x − 2)/3 = (y + 1)/6 = (z − 4/3)/10. μ-system: 2/3 = (2/3)μ ⇒ μ = 1, the same generator x = 2, 4y + 3z = 0. Hence the generator through both points is x = 2, 4y + 3z = 0, i.e. (x − 2)/0 = (y − 3)/3 = (z + 4)/(−4) (the line joining the points lies on the surface). The other generators are y = 3, 2x + z = 0 through (2, 3, −4) and (x − 2)/3 = (y + 1)/6 = (z − 4/3)/10 through (2, −1, 4/3).

(c)(ii) Right circular cylinder and its section by z = 0 Axis x = 2y = −z: through the origin with direction u = (2, 1, −2), |u|² = 9. For a point r = (x, y, z), distance² from the axis = |r × u|²/|u|², with r × u = (−2y − z, 2x + 2z, x − 2y). Radius 4: [(2y + z)² + 4(x + z)² + (x − 2y)²]/9 = 16 ⇒ 4y² + 4yz + z² + 4x² + 8xz + 4z² + x² − 4xy + 4y² = 144 ⇒ 5x² + 8y² + 5z² − 4xy + 4yz + 8zx = 144. Section by z = 0: 5x² − 4xy + 8y² = 144. The quadratic-form matrix [5 −2; −2 8] has eigenvalues from λ² − 13λ + 36 = 0, i.e. λ = 4, 9. In principal axes, 4X² + 9Y² = 144, i.e. X²/36 + Y²/16 = 1: an ellipse with semi-axes 6 and 4. Area = π·6·4 = 24π. (Geometric check: the plane z = 0 makes angle θ with the normal section, where cos θ = |u·k|/|u| = 2/3, so area = π·4²/(2/3) = 24π.)