(a) If the matrix of a linear transformation T : IR³→IR³ relative to the basis (1, 0, 0), (0, 1, 0), (0, 0, 1) is 1 & 1 & 2 -1…

(a) Matrix of T in the new basis

Let A be the given matrix relative to the standard basis S. The new basis is {v₁, v₂, v₃}, where v₁=(1,1,1), v₂=(0,1,1), v₃=(0,0,1). The change-of-basis matrix is P = [v₁ v₂ v₃], so

P = [1 0 0] [1 1 0] [1 1 1]

Its inverse is

P⁻¹ = [1 0 0; -1 1 0; 0 -1 1].

The matrix relative to the new basis is M = P⁻¹ A P. First,

A P = [4 3 2; 2 3 1; 4 4 3].

Therefore,

M = P⁻¹(A P) = [1 0 0; -1 1 0; 0 -1 1] [4 3 2; 2 3 1; 4 4 3] = [4 3 2; -2 0 -1; 2 1 2].

Final answer (a): [4 3 2; -2 0 -1; 2 1 2].

(b) Volume between the two paraboloids

The two paraboloids are z = 5(x²+y²) and z = 6−7x²−y². Their intersection satisfies

5(x²+y²) = 6−7x²−y² ⇒ 12x²+6y² = 6 ⇒ 2x²+y² = 1.

Thus the projection region D is the ellipse 2x²+y²≤1. At the origin, the lower surface is z=0 and the upper surface is z=6. Hence

V = ∬_D [ ∫ from z = 5(x²+y²) to z = 6−7x²−y² of dz ] dA = ∬_D [(6−7x²−y²) − 5(x²+y²)] dA = ∬_D (6−12x²−6y²) dA.

Put x = u/√2, y = v. Then 2x²+y² = u²+v², so D becomes u²+v²≤1, and dx dy = du dv/√2. The integrand becomes

6−12(u²/2)−6v² = 6(1−u²−v²).

Therefore,

V = (6/√2) ∬ over u²+v²≤1 of (1−u²−v²) du dv.

Use polar coordinates u = r cosθ, v = r sinθ. Then

V = (6/√2) ∫₀²π ∫₀¹ (1−r²) r dr dθ = (6/√2)(2π)(1/4) = 3π√2/2.

Final answer (b): V = 3π√2/2 cubic units.

(c)(i) Elliptic paraboloid, axis and principal planes

The equation is

2x²+3y²−8x+6y−12z+11=0.

Complete squares:

2x²−8x = 2(x−2)²−8, 3y²+6y = 3(y+1)²−3.

Substituting,

2(x−2)²−8 + 3(y+1)²−3 −12z +11 = 0 ⇒ 2(x−2)²+3(y+1)²−12z = 0 ⇒ 2(x−2)²+3(y+1)² = 12z.

Put X=x−2, Y=y+1. Then

2X²+3Y² = 12z ⇒ z = X²/6 + Y²/4.

This is of the form z = X²/a² + Y²/b², hence it is an elliptic paraboloid opening along the positive z-axis. Its principal axis is the line X=0, Y=0, i.e.

x=2, y=−1.

The principal planes are the planes of symmetry X=0 and Y=0, i.e.

x=2 and y=−1.

Final answer (c)(i): Elliptic paraboloid; principal axis: x=2, y=−1; principal planes: x=2, y=−1.

(c)(ii) Cone through the circle ABC

Assume a,b,c ≠ 0. The plane

x/a + y/b + z/c = 1

meets the coordinate axes at A=(a,0,0), B=(0,b,0), C=(0,0,c). The circle ABC lies in this plane and also on the sphere

x²+y²+z²−ax−by−cz=0,

because this sphere passes through A, B, C; its intersection with the plane is the unique circle through A, B, C.

Let (x,y,z) be a point on the cone. The ray from O through (x,y,z) meets the base plane at (λx, λy, λz), where

λ(x/a + y/b + z/c)=1.

Let D = x/a + y/b + z/c. Then λ = 1/D. Since the base point lies on the sphere,

λ²(x²+y²+z²) − λ(ax+by+cz)=0.

For λ≠0, this gives

λ(x²+y²+z²) = ax+by+cz.

Substitute λ=1/D:

x²+y²+z² = D(ax+by+cz).

Now

D(ax+by+cz) = (x/a + y/b + z/c)(ax+by+cz) = x²+y²+z² + xy(b/a + a/b) + xz(c/a + a/c) + yz(c/b + b/c).

Cancelling x²+y²+z² from both sides,

xy(b/a + a/b) + xz(c/a + a/c) + yz(c/b + b/c)=0.

Rearranging,

yz(b/c + c/b) + zx(c/a + a/c) + xy(b/a + a/b)=0.

Final answer (c)(ii): yz(b/c + c/b) + zx(c/a + a/c) + xy(b/a + a/b)=0.