(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 φ(v) = 1 & 2
3 & -1 v with respect to standard basis of M₂ₓ₂(ℝ), and hence find the rank of φ. Is φ invertible? Justify

Question

(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 φ(v) = 1 & 2
 3 & -1 v with respect to standard basis of M₂ₓ₂(ℝ), and hence find the rank of φ. Is φ invertible? Justify your answer. (15 marks)

(b) Find the volume of the greatest cylinder which can be inscribed in a cone of height h and semi-vertical angle α. (20 marks)

(c) Find the vertex of the cone 4x² - y² + 2z² + 2xy - 3yz + 12x - 11y + 6z + 4 = 0. (15 marks)

UPSC Answer Check · Accepted Answer

(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.

(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…

Model answer

What "Solve" is asking you to do

Structure that answers it

Where marks are lost

How this answer will be evaluated

Approach

Key points expected

Evaluation rubric

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Practice this exact question

More from Mathematics 2024 Paper I