(a) Find all solutions to the following system of equations by row-reduced method : x₁ + 2x₂ − x₃ = 2, 2x₁ + 3x₂ + 5x₃ = 5, − x₁…

(a) Solving the system by row reduction

The system is x₁+2x₂−x₃=2, 2x₁+3x₂+5x₃=5, −x₁−3x₂+8x₃=−1.

Augmented matrix: [1 2 −1 | 2] [2 3 5 | 5] [−1 −3 8 | −1]

Use row-reduced method. R₂→R₂−2R₁, R₃→R₃+R₁:

[1 2 −1 | 2] [0 −1 7 | 1] [0 −1 7 | 1]

R₃→R₃−R₂:

[1 2 −1 | 2] [0 −1 7 | 1] [0 0 0 | 0]

R₂→−R₂:

[1 2 −1 | 2] [0 1 −7 | −1] [0 0 0 | 0]

R₁→R₁−2R₂:

[1 0 13 | 4] [0 1 −7 | −1] [0 0 0 | 0]

The rank of the coefficient matrix and of the augmented matrix are both 2, which is less than 3, so the system is consistent with infinitely many solutions. Hence x₁+13x₃=4 and x₂−7x₃=−1. Let x₃=t, t∈ℝ. Then x₁=4−13t, x₂=−1+7t, x₃=t.

Final answer: (x₁,x₂,x₃)=(4−13t, −1+7t, t), t∈ℝ.

(b) Least total area of the square and circle

Let the length used for the square be x and for the circle be y. Then x+y=l. Area of square = (x/4)²=x²/16. If y is circumference of circle, radius = y/(2π), so area = y²/(4π). Minimise A=x²/16 + y²/(4π) subject to x+y=l.

Using Lagrange multiplier λ, take F=x²/16 + y²/(4π)+λ(x+y−l).

Then ∂F/∂x=x/8+λ=0 ⟹ λ=−x/8, ∂F/∂y=y/(2π)+λ=0 ⟹ λ=−y/(2π).

Thus x/8=y/(2π) ⟹ y=πx/4.

Using x+y=l: x+πx/4=l ⟹ x=4l/(π+4), y=πl/(π+4).

Minimum area: A=x²/16+y²/(4π) = [16l²/(π+4)²]/16 + [π²l²/(π+4)²]/(4π) = l²/(π+4)² + πl²/[4(π+4)²] = (π+4)l²/[4(π+4)²] = l²/[4(π+4)].

Final answer: minimum sum of areas = l²/[4(π+4)] square units.

(c) Plane through the other three feet of normals

Let A=a², B=b², C=c². Let the fixed point from which the six normals are drawn be (α,β,γ). At a foot (x,y,z), the normal direction is proportional to (x/A, y/B, z/C). Since (α,β,γ) lies on this normal, for a parameter μ, α=x(1+μ/A), β=y(1+μ/B), γ=z(1+μ/C), so x=αA/(A+μ), y=βB/(B+μ), z=γC/(C+μ).

The six normal feet correspond to the six roots μ₁,μ₂,…,μ₆ of α²A/(A+μ)² + β²B/(B+μ)² + γ²C/(C+μ)² = 1. Let P,Q,R correspond to μ₁,μ₂,μ₃. Their plane is lx+my+nz=p.

Substituting x,y,z in terms of μ and multiplying by (A+μ)(B+μ)(C+μ), define R(μ)=lαA(B+μ)(C+μ)+mβB(A+μ)(C+μ) +nγC(A+μ)(B+μ)−p(A+μ)(B+μ)(C+μ). Since μ₁,μ₂,μ₃ are roots, R(μ)=−p(μ−μ₁)(μ−μ₂)(μ−μ₃).

Putting μ=−A gives lαA(B−A)(C−A)=p(A+μ₁)(A+μ₂)(A+μ₃). (1)

Similarly, mβB(A−B)(C−B)=p(B+μ₁)(B+μ₂)(B+μ₃), (2) nγC(A−C)(B−C)=p(C+μ₁)(C+μ₂)(C+μ₃). (3)

Now consider the polynomial corresponding to the proposed plane x/(A l)+y/(B m)+z/(C n)+1/p=0:

S(μ)=α/l(B+μ)(C+μ)+β/m(A+μ)(C+μ) +γ/n(A+μ)(B+μ)+(1/p)(A+μ)(B+μ)(C+μ).

Its leading coefficient is 1/p.

From the normal equation, multiplying by (A+μ)²(B+μ)²(C+μ)², the resulting degree-six polynomial has roots μ₁,…,μ₆ and leading coefficient −1. Evaluating at μ=−A gives (A+μ₁)(A+μ₂)…(A+μ₆)=−α²A(B−A)²(C−A)². Using (1), (A+μ₄)(A+μ₅)(A+μ₆)=−pα(B−A)(C−A)/l.

Hence S(−A)=α(B−A)(C−A)/l =(1/p)(−A−μ₄)(−A−μ₅)(−A−μ₆).

Similarly, using (2) and (3), S(−B)=(1/p)(−B−μ₄)(−B−μ₅)(−B−μ₆), S(−C)=(1/p)(−C−μ₄)(−C−μ₅)(−C−μ₆).

Both S(μ) and T(μ)=(1/p)(μ−μ₄)(μ−μ₅)(μ−μ₆) are cubics with the same leading coefficient 1/p and agree at the three distinct points μ=−A,−B,−C (for the non-degenerate case; otherwise use continuity). Hence S(μ)=T(μ). Thus S(μ)=0 for μ=μ₄,μ₅,μ₆, so P′,Q′,R′ lie on x/(a²l)+y/(b²m)+z/(c²n)+1/p=0.

Final answer: P′Q′R′ is x/(a²l)+y/(b²m)+z/(c²n)+1/p=0, provided l,m,n,p≠0.