(a) Let V = M₂ₓ₂(ℝ) denote a vector space over the field of real numbers. Find the matrix of the linear mapping φ: V → V given by…

(a) Let A = (1 2; 3 −1). For v = (x₁₁ x₁₂; x₂₁ x₂₂),

φ(v) = A v = (x₁₁+2x₂₁, x₁₂+2x₂₂; 3x₁₁−x₂₁, 3x₁₂−x₂₂).

Take the standard ordered basis E₁₁, E₁₂, E₂₁, E₂₂ of M₂ₓ₂(ℝ). Then

φ(E₁₁) = E₁₁ + 3E₂₁, φ(E₁₂) = E₁₂ + 3E₂₂, φ(E₂₁) = 2E₁₁ − E₂₁, φ(E₂₂) = 2E₁₂ − E₂₂.

Hence the matrix of φ is

M = (1 0 2 0; 0 1 0 2; 3 0 −1 0; 0 3 0 −1).

Now det A = 1(−1) − 2·3 = −1 − 6 = −7 ≠ 0. Reordering the basis as E₁₁, E₂₁, E₁₂, E₂₂, the matrix M becomes block diagonal with two copies of A, so det M = (−7)² = 49 ≠ 0.

Therefore rank φ = 4. Since V has dimension 4 and rank φ = 4, φ is bijective. Thus φ is invertible. Its inverse is left multiplication by

A⁻¹ = (1/7 2/7; 3/7 −1/7).

Indeed A A⁻¹ = I, so φ⁻¹(w) = A⁻¹w.

(b) Let the cone have height h and semi-vertical angle α. Its base radius is

R = h tan α.

Let the inscribed cylinder have height y and radius r. At height y above the base, the cone cross-section radius is

R(1 − y/h) = (h − y) tan α.

For maximum volume at fixed y, the cylinder radius must equal this available radius, so

r = (h − y) tan α.

Thus the cylinder volume is

V(y) = π r² y = π tan²α · y(h − y)².

Differentiate:

dV/dy = π tan²α [(h − y)² − 2y(h − y)] = π tan²α (h − y)(h − 3y).

The critical points are y = h and y = h/3. The endpoints y = 0 and y = h give zero volume, so the maximum occurs at

y = h/3.

Then

r = (h − h/3) tan α = (2h/3) tan α.

Therefore

Vmax = π r² y = π (4h²/9 tan²α)(h/3)

= 4π h³ tan²α / 27.

Condition: h > 0 and 0 < α < π/2. The maximum volume is 4πh³ tan²α/27 cubic units.

(c) Let

F(x,y,z) = 4x² − y² + 2z² + 2xy − 3yz + 12x − 11y + 6z + 4.

The vertex of a cone is its unique singular point, so solve

Fₓ = 8x + 2y + 12 = 0, F_y = 2x − 2y − 3z − 11 = 0, F_z = −3y + 4z + 6 = 0.

From the first equation,

y = −4x − 6.

From the third equation,

4z = 3y − 6 = 3(−4x − 6) − 6 = −12x − 24,

so

z = −3x − 6.

Substitute into the second equation:

2x − 2(−4x − 6) − 3(−3x − 6) − 11 = 0

⇒ 2x + 8x + 12 + 9x + 18 − 11 = 0

⇒ 19x + 19 = 0

⇒ x = −1.

Then

y = −4(−1) − 6 = −2, z = −3(−1) − 6 = −3.

Thus the vertex is

(−1, −2, −3).

Indeed, putting X = x + 1, Y = y + 2, Z = z + 3 gives

4X² − Y² + 2Z² + 2XY − 3YZ = 0,

a homogeneous quadratic equation in X, Y, Z, confirming that (−1, −2, −3) is the vertex.