Advanced Vector Calculus and Matrix Algebra Problems

Posted on Feb 3, 2026 in Mathematics

Cayley-Hamilton Theorem: Verification and Inverse

Problem 1: Verify Cayley-Hamilton theorem for A and find A^-1

Given matrix: A = [[2, 1, 1], [0, 1, 0], [1, 1, 2]]

Step 1: Find the Characteristic Equation |A – λI| = 0

A – λI = [[2-λ, 1, 1], [0, 1-λ, 0], [1, 1, 2-λ]]

Expand along the 2nd row (which has two zeros):

|A – λI| = (1-λ) × |[2-λ, 1], [1, 2-λ]|

= (1-λ) × [(2-λ)(2-λ) – 1×1]

= (1-λ) × [(2-λ)² – 1]

= (1-λ) × [4 – 4λ + λ² – 1]

= (1-λ)(λ² – 4λ + 3)

= (1-λ)(λ-1)(λ-3)

= -(λ-1)²(λ-3) = 0

Characteristic Equation: λ³ – 5λ² + 7λ – 3 = 0

Step 2: Verify Cayley-Hamilton (A satisfies its own characteristic equation)

We must show: A³ – 5A² + 7A – 3I = O (Zero matrix)

Calculate A²:

A² = A × A = [[5, 4, 4], [0, 1, 0], [4, 4, 5]]

Calculate A³:

A³ = A² × A = [[14, 13, 14], [0, 1, 0], [13, 13, 14]]

Now compute A³ – 5A² + 7A – 3I:

= [[14, 13, 14],[0, 1, 0],[13, 13, 14]] – 5[[5, 4, 4],[0, 1, 0],[4, 4, 5]] + 7[[2, 1, 1],[0, 1, 0],[1, 1, 2]] – 3[[1, 0, 0],[0, 1, 0],[0, 0, 1]]

= [[0, 0, 0], [0, 0, 0], [0, 0, 0]] = O. ✓ Verified!

Step 3: Find A^-1 using Cayley-Hamilton

From A³ – 5A² + 7A – 3I = O

Multiply both sides by A^-1:

A² – 5A + 7I – 3A^-1 = O

3A^-1 = A² – 5A + 7I

A^-1 = (1/3)[A² – 5A + 7I]

Calculate A² – 5A + 7I:

= [[5, 4, 4],[0, 1, 0],[4, 4, 5]] – 5[[2, 1, 1],[0, 1, 0],[1, 1, 2]] + 7[[1, 0, 0],[0, 1, 0],[0, 0, 1]]

= [[5-10+7, 4-5+0, 4-5+0], [0, 1-5+7, 0], [4-5+0, 4-5+0, 5-10+7]]

= [[2, -1, -1], [0, 3, 0], [-1, -1, 2]]

A^-1 = (1/3) × [[2, -1, -1], [0, 3, 0], [-1, -1, 2]]

Finding Eigenvalues and Eigenvectors for Matrix A

Problem 2: Find eigenvalues and eigenvectors of A

Given matrix: A = [[2, 1, 1], [2, 3, 2], [3, 3, 4]]

Step 1: Find Eigenvalues from |A – λI| = 0

|A – λI| = |[2-λ, 1, 1], [2, 3-λ, 2], [3, 3, 4-λ]| = 0

Expand (using R1):

= (2-λ)[(3-λ)(4-λ) – 6] – 1[2(4-λ) – 6] + 1[6 – 3(3-λ)]

= (2-λ)[λ² – 7λ + 6] – [2 – 2λ] + [3λ – 3]

= (λ-1)[(2-λ)(λ-6) + 5]

= -(λ-1)(λ² – 8λ + 7)

= -(λ-1)²(λ-7) = 0

Eigenvalues: λ₁ = 1 (repeated), λ₂ = 7

Step 2: Find Eigenvectors for λ = 1

Solve (A – 1·I)X = 0

[A – I] = [[1, 1, 1], [2, 2, 2], [3, 3, 3]]

Row reduce: All rows are proportional, leading to the equation x + y + z = 0.

Let y = 1, z = 0 → x = -1: X₁ = [-1, 1, 0]^T
Let y = 0, z = 1 → x = -1: X₂ = [-1, 0, 1]^T

Step 3: Find Eigenvector for λ = 7

Solve (A – 7I)X = 0

[A – 7I] = [[-5, 1, 1], [2, -4, 2], [3, 3, -3]]

Row reduce and solve: X₃ = [1, 2, 3]^T

Answer: For λ=1, X₁=[-1, 1, 0]^T, X₂=[-1, 0, 1]^T | For λ=7, X₃=[1, 2, 3]^T

Leibnitz Theorem Proof for Higher Order Derivatives

Problem 3: Prove the relation for y = e^tan⁻¹x

If y = e^tan⁻¹x, prove that (1+x²)y_n+2 + [(2n+2)x – 1]y_n+1 + n(n+1)y_n = 0

Step 1: Find y₁ (First Derivative)

y = e^tan⁻¹x

y₁ = e^tan⁻¹x × 1/(1+x²) = y/(1+x²)

(1 + x²)y₁ = y … (1)

Step 2: Find y₂ (Differentiate equation 1)

Differentiating (1) using the product rule:

(1+x²)y₂ + 2xy₁ = y₁

(1+x²)y₂ + (2x-1)y₁ = 0 … (2)

Step 3: Apply Leibnitz Theorem to equation (2)

Leibnitz Formula for the n^th derivative of a product (uv)_n: (uv)_n = Σ nC_r u_n-r v_r

Differentiate equation (2) n times using Leibnitz:

For (1+x²)y₂:

[(1+x²)y₂]_n = (1+x²)y_n+2 + nC₁(2x)y_n+1 + nC₂(2)y_n

= (1+x²)y_n+2 + 2nx y_n+1 + n(n-1)y_n

For (2x-1)y₁:

[(2x-1)y₁]_n = (2x-1)y_n+1 + nC₁(2)y_n

= (2x-1)y_n+1 + 2n y_n

Step 4: Combine and Simplify

Summing the results:

(1+x²)y_n+2 + 2nx y_n+1 + n(n-1)y_n + (2x-1)y_n+1 + 2n y_n = 0

Group terms by derivative order:

(1+x²)y_n+2 + (2nx + 2x – 1)y_n+1 + [n(n-1) + 2n]y_n = 0

(1+x²)y_n+2 + [(2n+2)x – 1]y_n+1 + [n² – n + 2n]y_n = 0

(1+x²)y_n+2 + [(2n+2)x – 1]y_n+1 + n(n+1)y_n = 0

✓ Hence Proved!

Taylor Series Expansion of f(x,y) = x^y about (1,1)

Problem 4: Expand f(x,y) and approximate (1.1)^1.02

Function: f(x,y) = x^y. Expansion point: (a,b) = (1, 1). We expand up to second degree terms.

Step 1: Taylor’s Formula for Two Variables

f(x,y) = f(a,b) + [(x-a)f_x + (y-b)f_y] + (1/2!)[(x-a)²f_xx + 2(x-a)(y-b)f_xy + (y-b)²f_yy] + …

We expand in powers of (x-1) and (y-1).

Step 2: Calculate Required Partial Derivatives at (1,1)

f(x,y) = x^y → f(1,1) = 1¹ = 1
f_x = y x^y-1 → f_x(1,1) = 1 × 1⁰ = 1
f_y = x^y ln(x) → f_y(1,1) = 1¹ × ln(1) = 0
f_xx = y(y-1)x^y-2 → f_xx(1,1) = 1 × 0 × 1 = 0
f_yy = x^y (ln x)² → f_yy(1,1) = 1 × 0 = 0
f_xy = x^y-1 + y x^y-1 ln(x) → f_xy(1,1) = 1 + 0 = 1

Step 3: Substitute into Taylor’s Formula

x^y = 1 + [(x-1)×1 + (y-1)×0] + (1/2)[0 + 2(x-1)(y-1)×1 + 0] + …

x^y = 1 + (x-1) + (x-1)(y-1) + …

Approximation: x^y ≈ 1 + (x-1) + (x-1)(y-1)

Step 4: Find (1.1)^1.02

Here x = 1.1, y = 1.02. Thus, (x-1) = 0.1 and (y-1) = 0.02.

(1.1)^1.02 ≈ 1 + 0.1 + (0.1)(0.02)

= 1 + 0.1 + 0.002

= 1.102

Answer: (1.1)^1.02 ≈ 1.102

Euler’s Theorem Application: Proving a Partial Derivative Relation

Problem 5: Prove x(∂u/∂x) + y(∂u/∂y) = 2tan(u) for u = sin⁻¹[(x³+y³)/(x+y)]

Step 1: Recognize the Homogeneous Function

Let v = sin(u) = (x³+y³)/(x+y)

Check the degree of homogeneity for v:

v(tx, ty) = (t³x³ + t³y³)/(tx + ty) = t³(x³+y³)/t(x+y) = t²v

v is homogeneous of degree n = 2.

Step 2: Apply Euler’s Theorem

Euler’s Theorem states: If v is homogeneous of degree n, then x(∂v/∂x) + y(∂v/∂y) = nv

For v = sin(u) and n = 2:

x(∂v/∂x) + y(∂v/∂y) = 2v = 2sin(u) … (1)

Step 3: Express in terms of u

Since v = sin(u), we use the chain rule:

∂v/∂x = cos(u) × ∂u/∂x
∂v/∂y = cos(u) × ∂u/∂y

Substitute these into equation (1):

x [cos(u) × ∂u/∂x] + y [cos(u) × ∂u/∂y] = 2sin(u)

Factor out cos(u):

cos(u) [x(∂u/∂x) + y(∂u/∂y)] = 2sin(u)

Divide by cos(u):

x(∂u/∂x) + y(∂u/∂y) = 2sin(u)/cos(u)

x(∂u/∂x) + y(∂u/∂y) = 2tan(u)

✓ Proved!

Calculating the Jacobian ∂(x,y)/∂(u,v)

Problem 6: Find the Jacobian for x = e^vsec(u) and y = e^vtan(u)

Step 1: Jacobian Definition

The Jacobian J is defined as the determinant of the matrix of partial derivatives:

J = ∂(x,y)/∂(u,v) = | ∂x/∂u ∂x/∂v | | ∂y/∂u ∂y/∂v |

Step 2: Calculate Partial Derivatives

Given x = e^vsec(u):

∂x/∂u = e^v sec(u)tan(u)
∂x/∂v = sec(u) e^v

Given y = e^vtan(u):

∂y/∂u = e^v sec²(u)
∂y/∂v = tan(u) e^v

Step 3: Calculate the Determinant

J = (e^vsec(u)tan(u)) × (e^vtan(u)) – (e^vsec(u)) × (e^vsec²(u))

= e^2vsec(u)tan²(u) – e^2vsec³(u)

Factor out e^2vsec(u):

= e^2vsec(u) [tan²(u) – sec²(u)]

Using the identity sec²(u) – tan²(u) = 1:

= e^2vsec(u) [- (sec²(u) – tan²(u))]

= e^2vsec(u) [-1]

∂(x,y)/∂(u,v) = -e^2vsec(u)

Lagrange Multipliers: Minimizing Surface Area of a Box

Problem 7: Find dimensions for a rectangular box (open top) with capacity 32 ft³ for least material.

Step 1: Set up the Problem

Let length = x, width = y, height = z.

Constraint (Volume G): G(x, y, z) = xyz – 32 = 0

Objective (Surface Area S, open box): S(x, y, z) = xy + 2xz + 2yz

Step 2: Form the Lagrangian Function

F(x, y, z, λ) = S + λG = xy + 2xz + 2yz + λ(xyz – 32)

Step 3: Find Partial Derivatives and Set to Zero

∂F/∂x = y + 2z + λyz = 0 … (1)
∂F/∂y = x + 2z + λxz = 0 … (2)
∂F/∂z = 2x + 2y + λxy = 0 … (3)
∂F/∂λ = xyz – 32 = 0 … (4)

Step 4: Solve the System

From (1): λ = -(y + 2z)/(yz) = -1/z – 2/y

From (2): λ = -(x + 2z)/(xz) = -1/z – 2/x

Equating the expressions for λ:

-1/z – 2/y = -1/z – 2/x

Therefore: 2/y = 2/x → x = y

Substitute x = y into (3):

λ = -(2x + 2x)/(x²) = -4/x

Now equate this λ with the expression derived from (1) (using y=x):

-4/x = -1/z – 2/x → -2/x = -1/z → z = x/2

Step 5: Use Constraint to Find Values

Substitute x=y and z=x/2 into the volume constraint (4):

x × x × (x/2) = 32

x³/2 = 32

x³ = 64

x = 4

Since x=y=4 and z=x/2=2:

Answer: Length = 4 ft, Width = 4 ft, Height = 2 ft

Changing the Order of Integration and Evaluation

Problem 8: Change the order of integration and evaluate: ∫₀¹ ∫ₓ²²⁻ₓ xy dy dx

Step 1: Identify the Region of Integration

The original limits define the region R:

x: 0 ≤ x ≤ 1
y: x² ≤ y ≤ 2-x

The boundaries are the parabola y = x² and the line y = 2-x. The intersection is at x = 1, y = 1.

Step 2: Sketch and Divide Region for dx dy order

We express x as a function of y:

Lower boundary: y = x² → x = √y
Upper boundary: y = 2-x → x = 2-y

The region must be split at y = 1:

Region R₁ (0 ≤ y ≤ 1): x goes from 0 to √y.
Region R₂ (1 ≤ y ≤ 2): x goes from 0 to 2-y.

Step 3: Write the New Integral

I = ∫₀¹ ∫₀⁺√y xy dx dy + ∫₁² ∫₀²⁻ʸ xy dx dy

Step 4: Evaluate First Integral (I₁)

I₁ = ∫₀¹ [x²y/2]₀⁺√y dy = ∫₀¹ y²/2 dy = [y³/6]₀¹ = 1/6

Step 5: Evaluate Second Integral (I₂)

I₂ = ∫₁² [x²y/2]₀²⁻ʸ dy = ∫₁² y(2-y)²/2 dy

= (1/2)∫₁² (4y – 4y² + y³) dy

= (1/2)[2y² – 4y³/3 + y⁴/4]₁²

= (1/2)[(8 – 32/3 + 4) – (2 – 4/3 + 1/4)]

= (1/2) [ 4/3 – 11/12 ] = 5/24

Total Integral I = I₁ + I₂ = 1/6 + 5/24 = 9/24 = 3/8

Triple Integral: Calculating Volume of a Tetrahedron

Problem 9: Find the volume bounded by x=0, y=0, z=0 and x+y+z=1

Step 1: Set up the Triple Integral

Volume V = ∭∭∭ dV = ∭∭∭ dz dy dx.

The region is a tetrahedron with vertices (0,0,0), (1,0,0), (0,1,0), and (0,0,1).

Step 2: Determine Limits

x varies from 0 to 1.
For fixed x, y varies from 0 to 1-x.
For fixed x and y, z varies from 0 to 1-x-y.

Step 3: Evaluate the Integral

V = ∫₀¹ ∫₀¹⁻ₓ ∫₀¹⁻ₓ⁻ʸ dz dy dx

= ∫₀¹ ∫₀¹⁻ₓ (1 – x – y) dy dx

= ∫₀¹ [(1-x)y – y²/2]₀¹⁻ₓ dx

= ∫₀¹ [ (1-x)² – (1-x)²/2 ] dx

= ∫₀¹ (1-x)²/2 dx

= (1/2) × [-(1-x)³/3]₀¹

= (1/2) × [0 – (-1/3)] = 1/6

Volume = 1/6 cubic units

Directional Derivative of φ = xyz in Normal Direction

Problem 10: Find the directional derivative of φ = xyz at P(1,1,3) in the direction of the normal to the sphere x²+y²+z²=11 at P.

Step 1: Find the Gradient of φ

φ = xyz

∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

∇φ = yz i + xz j + xy k

At P(1, 1, 3): ∇φ = 3i + 3j + 1k

Step 2: Find the Normal Vector to the Sphere

The sphere is defined by F = x² + y² + z² – 11 = 0.

Normal = ∇F = 2xi + 2yj + 2zk

At P(1, 1, 3): ∇F = 2i + 2j + 6k

Step 3: Find the Unit Normal Vector (n)

Magnitude: |∇F| = √(4 + 4 + 36) = √44 = 2√11

Unit vector: n = (2i + 2j + 6k) / (2√11) = (i + j + 3k) / √11

Step 4: Calculate the Directional Derivative

Directional Derivative = ∇φ · n

= (3i + 3j + 1k) · (i + j + 3k)/√11

= (3×1 + 3×1 + 1×3)/√11

= 9/√11

Directional Derivative = 9/√11 or (9√11)/11

Gauss Divergence Theorem Verification on a Cylinder

Problem 11: Evaluate ∭∭_S F·n dS where F = x²i + y²j + 4zk and S is the cylinder x²+y²=4, z=0 to z=3.

Step 1: State Gauss Divergence Theorem

The theorem relates the surface integral to the volume integral of the divergence:

∭∭_S F·n dS = ∭∭∭_V (∇·F) dV

Step 2: Calculate Divergence (∇·F)

F = x²i + y²j + 4zk

∇·F = ∂(x²)/∂x + ∂(y²)/∂y + ∂(4z)/∂z

∇·F = 2x + 2y + 4

Step 3: Set up the Volume Integral

The volume V is a cylinder defined by r ≤ 2 and 0 ≤ z ≤ 3. We use cylindrical coordinates (dV = r dr dθ dz).

∇·F = 2r cosθ + 2r sinθ + 4

∭∭∭_V (∇·F) dV = ∫₀³ ∫₀²π ∫₀² (2r cosθ + 2r sinθ + 4) r dr dθ dz

Step 4: Evaluate the Integral

Integrate r first:

∫₀² (2r² cosθ + 2r² sinθ + 4r) dr = [2r³ cosθ/3 + 2r³ sinθ/3 + 2r²]₀²

= 16 cosθ/3 + 16 sinθ/3 + 8

Integrate θ:

∫₀²π (16 cosθ/3 + 16 sinθ/3 + 8) dθ = [16 sinθ/3 – 16 cosθ/3 + 8θ]₀²π

= (0 – 16/3 + 16π) – (0 – 16/3 + 0) = 16π

Integrate z:

∫₀³ 16π dz = 16π × 3 = 48π

Answer: ∭∭_S F·n dS = 48π

Verifying Stoke’s Theorem for a Square Region

Problem 12: Verify Stoke’s theorem for F = x²i + xyj around the square x=0, y=0, x=a, y=a in the z=0 plane.

Step 1: State Stoke’s Theorem

The theorem relates the line integral around a closed curve C to the surface integral over the surface S bounded by C:

∮_C F·dr = ∭∭_S (∇×F)·n dS

Step 2: Evaluate LHS (Line Integral ∮_C F·dr)

F = x²i + xyj. Thus, F·dr = x²dx + xy dy.

The path C is the square OABCO (O(0,0) → A(a,0) → B(a,a) → C(0,a) → O(0,0)).

Along OA (y=0, dy=0, x: 0→a):
∫_OA = ∫₀ⁿ x² dx = [x³/3]₀ⁿ = a³/3
Along AB (x=a, dx=0, y: 0→a):
∫_AB = ∫₀ⁿ a y dy = a[y²/2]₀ⁿ = a³/2
Along BC (y=a, dy=0, x: a→0):
∫_BC = ∫ⁿ₀ x² dx = [x³/3]ⁿ₀ = -a³/3
Along CO (x=0, dx=0, y: a→0):
∫_CO = ∫ⁿ₀ 0 dy = 0

LHS = a³/3 + a³/2 – a³/3 + 0 = a³/2

Step 3: Evaluate RHS (Surface Integral ∭∭_S (∇×F)·n dS)

Calculate the Curl (∇×F):

∇×F = yk

The surface S is in the z=0 plane, so the unit normal vector is n = k.

(∇×F)·n = yk · k = y

RHS = ∭∭_S y dS = ∫₀ⁿ ∫₀ⁿ y dx dy

= ∫₀ⁿ y [x]₀ⁿ dy = ∫₀ⁿ a y dy

= a [y²/2]₀ⁿ = a³/2

Since LHS = RHS = a³/2, ✓ Stoke’s Theorem is Verified!