(a) Can the set {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, –1, 0)} be extended to form a basis of the vector space ℝ⁴? Justify your…

(a) Let v₁ = (0, 0, 0, 3), v₂ = (1, 1, 0, 0), v₃ = (0, 1, –1, 0).

To check linear independence, suppose α v₁ + β v₂ + γ v₃ = (0, 0, 0, 0). Comparing coordinates:

• First coordinate: β = 0.
• Second coordinate: β + γ = 0, so γ = 0.
• Third coordinate: –γ = 0, so γ = 0.
• Fourth coordinate: 3α = 0, so α = 0.

Thus α = β = γ = 0, so the three given vectors are linearly independent. Since dim ℝ⁴ = 4, any linearly independent set of 3 vectors can be extended to a basis by adding one more vector outside its span.

Take e₁ = (1, 0, 0, 0). It is not in the span of the given three vectors, because if (1, 0, 0, 0) = α v₁ + β v₂ + γ v₃, then first coordinate gives β = 1, second gives β + γ = 0, so γ = –1, but third coordinate gives –γ = 0, so γ = 0, a contradiction.

Now check the four vectors v₁, v₂, v₃, e₁ are independent. Suppose α v₁ + β v₂ + γ v₃ + δ e₁ = (0, 0, 0, 0). Then coordinates give:

• x-coordinate: β + δ = 0.
• y-coordinate: β + γ = 0.
• z-coordinate: –γ = 0, so γ = 0.
• Then β = 0 from y-coordinate, δ = 0 from x-coordinate.
• w-coordinate: 3α = 0, so α = 0.

Hence the four vectors are linearly independent in ℝ⁴, so they form a basis.

Final answer: Yes. The set can be extended; for example, {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, –1, 0), (1, 0, 0, 0)} is a basis of ℝ⁴.

(b) The linear transformation is T(x, y, z, w) = (x – w, y + z, z – w).

Its matrix with respect to the standard bases of ℝ⁴ and ℝ³ is

A = [1 0 0 –1 0 1 1 0 0 0 1 –1]

The range of T is the column space of A. The columns are c₁ = (1, 0, 0), c₂ = (0, 1, 0), c₃ = (0, 1, 1), c₄ = (–1, 0, –1).

The first three columns c₁, c₂, c₃ are linearly independent. Indeed, if α c₁ + β c₂ + γ c₃ = (0, 0, 0), then first coordinate gives α = 0, third coordinate gives γ = 0, and second coordinate gives β + γ = 0, so β = 0.

Thus rank(T) = 3. Since the codomain ℝ³ has dimension 3, the range is the whole of ℝ³.

Now find the kernel. Put T(x, y, z, w) = (0, 0, 0). Then

x – w = 0 ⇒ x = w, y + z = 0 ⇒ y = –z, z – w = 0 ⇒ z = w.

Hence w = x = z and y = –z. Let z = t, where t ∈ ℝ. Then (x, y, z, w) = (t, –t, t, t).

Therefore kernel T = {(t, –t, t, t) : t ∈ ℝ} = span{(1, –1, 1, 1)}.

So nullity(T) = 1. The rank–nullity theorem checks this: rank + nullity = 3 + 1 = 4 = dim ℝ⁴.

Final answer: Range = ℝ³, rank = 3, kernel = span{(1, –1, 1, 1)}, nullity = 1.

(c) Let the side of each removed square be h meters. After folding, the height of the open box is h. The base length becomes 6 – 2h and the base width becomes 2 – 2h. For a positive base, 0 < h < 1.

Volume of the box is

V(h) = h(6 – 2h)(2 – 2h).

Simplify:

V(h) = h · 2(3 – h) · 2(1 – h) = 4h(3 – h)(1 – h) = 4h(3 – 4h + h²) = 12h – 16h² + 4h³.

Differentiate:

dV/dh = 12 – 32h + 12h².

Set dV/dh = 0:

12 – 32h + 12h² = 0 ⇒ 3h² – 8h + 3 = 0.

Using the quadratic formula,

h = [8 ± √(64 – 36)]/6 = [8 ± √28]/6 = [8 ± 2√7]/6 = (4 ± √7)/3.

The two values are approximately

(4 + √7)/3 ≈ 2.215 m, (4 – √7)/3 ≈ 0.451 m.

Since 0 < h < 1, the valid value is h = (4 – √7)/3 m.

Check maximum. The second derivative is

d²V/dh² = –32 + 24h.

At h = (4 – √7)/3,

d²V/dh² = –32 + 8(4 – √7) = –32 + 32 – 8√7 = –8√7 < 0.

So the volume is maximum at this value.

Final answer: h = (4 – √7)/3 m ≈ 0.451 m.

(d) Given f(x + y) = f(x) f(y) for all real x, y, and f(x) ≠ 0 for every real x.

First put x = y = 0:

f(0) = f(0) f(0) = f(0)².

Since f(0) ≠ 0, divide by f(0) to get f(0) = 1.

Now compute f′(x) from the definition of derivative:

f′(x) = lim_(h→0) [f(x + h) – f(x)]/h.

Using the functional equation,

f(x + h) = f(x) f(h).

Therefore,

f′(x) = lim_(h→0) [f(x)f(h) – f(x)]/h = f(x) lim_(h→0) [f(h) – 1]/h.

But f(0) = 1, so

lim_(h→0) [f(h) – 1]/h = lim_(h→0) [f(h) – f(0)]/h = f′(0) = 2.

Hence

f′(x) = 2f(x).

This is the required differential equation. Since f(x) ≠ 0, divide by f(x):

f′(x)/f(x) = 2.

Integrating,

ln|f(x)| = 2x + C.

Put x = 0. Since f(0) = 1,

ln 1 = 0 + C ⇒ C = 0.

Thus

ln|f(x)| = 2x ⇒ |f(x)| = e^(2x).

Because f is differentiable and f(0) = 1 > 0, f is positive for all real x. Hence

f(x) = e^(2x).

Final answer: f′(x) = 2f(x), and f(x) = e^(2x).

(e) The vertex of the cone is V = (1, 1, 0). The guiding curve is the circle y = 0, x² + z² = 4 in the plane y = 0.

Let P = (x, y, z) be any point on the cone. The generator through V and P is

(1, 1, 0) + t[(x, y, z) – (1, 1, 0)] = (1 + t(x – 1), 1 + t(y – 1), t z).

For y ≠ 1, this generator meets the plane y = 0 when

1 + t(y – 1) = 0 ⇒ t = 1/(1 – y).

At that intersection point Q, the coordinates are

X = 1 + t(x – 1) = 1 + (x – 1)/(1 – y) = (x – y)/(1 – y),

Y = 0,

Z = t z = z/(1 – y).

Since Q lies on the guiding curve y = 0, x² + z² = 4, we must have

X² + Z² = 4.

Substitute X and Z:

[(x – y)/(1 – y)]² + [z/(1 – y)]² = 4.

Multiplying by (1 – y)² gives

(x – y)² + z² = 4(1 – y)².

Equivalently, expanding,

x² – 2xy + y² + z² = 4 – 8y + 4y²,

so

x² + z² – 2xy – 3y² + 8y – 4 = 0.

This equation is satisfied by the vertex (1, 1, 0), as required.

Final answer: (x – y)² + z² = 4(1 – y)², or equivalently x² + z² – 2xy – 3y² + 8y – 4 = 0.