(a) Let V₁ = (2, -1, 3, 2), V₂ = (-1, 1, 1, -3) and V₃ = (1, 1, 9, -5) be three vectors of the space ℝ⁴. Does (3, -1, 0, -1) ∈…

(a) Is (3, −1, 0, −1) in the span of V₁, V₂, V₃?

Let w = (3, −1, 0, −1). Suppose w ∈ span {V₁, V₂, V₃}. Then there exist real numbers α, β, γ such that αV₁ + βV₂ + γV₃ = w. Equating coordinates gives the system 2α − β + γ = 3 −α + β + γ = −1 3α + β + 9γ = 0 2α − 3β − 5γ = −1

From the second equation, β = α − γ − 1. Substituting in the first equation: 2α − (α − γ − 1) + γ = 3 ⇒ α + 2γ = 2. Substituting in the third equation: 3α + (α − γ − 1) + 9γ = 0 ⇒ 4α + 8γ − 1 = 0 ⇒ 4(α + 2γ) = 1. But α + 2γ = 2 gives 4(α + 2γ) = 8, not 1. Hence the system is inconsistent. Therefore no such α, β, γ exist. So (3, −1, 0, −1) is not in span {V₁, V₂, V₃}.

(b) Rank and nullity of T

The standard matrix of T : ℝ³ → ℝ³ is A = [1 0 1] [1 1 2] [2 1 3]

Use row reduction. R₂ → R₂ − R₁, R₃ → R₃ − 2R₁: [1 0 1] [0 1 1] [0 1 1]

Then R₃ → R₃ − R₂: [1 0 1] [0 1 1] [0 0 0]

There are two nonzero rows, so rank(T) = 2. By the rank-nullity theorem for T : ℝ³ → ℝ³, nullity(T) = dim ℝ³ − rank(T) = 3 − 2 = 1.

For the kernel, solve T(x, y, z) = (0, 0, 0): x + z = 0 x + y + 2z = 0 2x + y + 3z = 0

From x + z = 0, z = −x. Then x + y + 2(−x) = 0 gives y = x. The third equation is then automatically satisfied. Hence ker T = {t(1, 1, −1) : t ∈ ℝ}. Therefore rank = 2 and nullity = 1.

(c) Values of p and q

Let N(x) = x(1 + p cos x) − q sin x. Use the Taylor expansions cos x = 1 − x²/2 + x⁴/24 − ⋯ sin x = x − x³/6 + x⁵/120 − ⋯

Then x(1 + p cos x) = x + p x cos x = x + p(x − x³/2 + x⁵/24 − ⋯) = (1 + p)x − (p/2)x³ + (p/24)x⁵ − ⋯

Also q sin x = qx − (q/6)x³ + (q/120)x⁵ − ⋯

Therefore N(x) = (1 + p − q)x + (−p/2 + q/6)x³ + (p/24 − q/120)x⁵ + ⋯

For limₓ→₀ N(x)/x³ to exist finitely, the coefficient of x must vanish: 1 + p − q = 0 ⇒ q = 1 + p.

For the limit to equal 1, the coefficient of x³ must be 1: −p/2 + q/6 = 1.

Substitute q = 1 + p: −p/2 + (1 + p)/6 = 1 Multiply by 6: −3p + 1 + p = 6 ⇒ −2p = 5 ⇒ p = −5/2.

Then q = 1 + p = 1 − 5/2 = −3/2.

Thus p = −5/2 and q = −3/2.

(d) Convergence of the integral

Consider I = ∫₀¹ (log x)/(1 + x) dx. The integrand has a logarithmic singularity at x = 0. For 0 < x ≤ 1,

|(log x)/(1 + x)| = (−log x)/(1 + x) ≤ −log x.

Now ∫₀¹ (−log x) dx = [x − x log x]₀¹ = 1, since x log x → 0 as x → 0⁺. By the comparison test for improper integrals, ∫₀¹ |(log x)/(1 + x)| dx converges. Hence the given integral converges absolutely.

Its value may also be found by using 1/(1 + x) = 1 − x + x² − x³ + ⋯. For n ≥ 0, ∫₀¹ xⁿ log x dx = −1/(n + 1)². Therefore I = −Σₙ₌₀∞ (−1)ⁿ/(n + 1)² = −(1 − 1/2² + 1/3² − 1/4² + ⋯) = −π²/12.

So the integral converges absolutely, and its value is −π²/12.

(e) Locus of the centroid

Let the plane cut the coordinate axes at A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c). Its intercept form is x/a + y/b + z/c = 1.

The perpendicular distance from the origin O to this plane is d = 1 / √(1/a² + 1/b² + 1/c²). Given that this distance is 3p, 1 / √(1/a² + 1/b² + 1/c²) = 3p ⇒ 1/a² + 1/b² + 1/c² = 1/(9p²).

The centroid G of the tetrahedron OABC is G = (O + A + B + C)/4 = (a/4, b/4, c/4). Let G = (x, y, z). Then a = 4x, b = 4y, c = 4z.

Substitute these into the distance condition: 1/(4x)² + 1/(4y)² + 1/(4z)² = 1/(9p²) ⇒ (1/16)(1/x² + 1/y² + 1/z²) = 1/(9p²).

Multiplying by 16, 1/x² + 1/y² + 1/z² = 16/(9p²). Multiplying by 9, 9(1/x² + 1/y² + 1/z²) = 16/p².

Hence the locus of the centroid is exactly 9(1/x² + 1/y² + 1/z²) = 16/p². This holds for p > 0 and x, y, z ≠ 0.