(a) Let A = 3 & 2 & 4 2 & 0 & 2 4 & 2 & 3 be a 3×3 matrix. Find the eigenvalues and the corresponding eigenvectors of A. Hence…

(a) Let A = ((3,2,4),(2,0,2),(4,2,3)).

The characteristic equation is det(A − λI) = 0. Expanding,

det(A − λI) = −λ³ + 6λ² + 15λ + 8 = −(λ + 1)²(λ − 8).

Hence the eigenvalues are λ = −1, −1, 8.

For λ = 8, solve (A − 8I)X = 0:

−5x + 2y + 4z = 0, 2x − 8y + 2z = 0, 4x + 2y − 5z = 0.

From the second equation, x = 4y − z. Substituting in the first gives z = 2y, so x = 2y. Thus an eigenvector is v₃ = (2, 1, 2).

For λ = −1, solve (A + I)X = 0:

4x + 2y + 4z = 0, so 2x + y + 2z = 0.

Two independent eigenvectors are v₁ = (1, −2, 0), v₂ = (1, 0, −1).

Any nonzero linear combination of v₁ and v₂ is also an eigenvector for λ = −1.

Since A is symmetric, it is diagonalizable. Hence A⁻¹⁵ has the same eigenvectors, and its eigenvalues are λ⁻¹⁵. Therefore:

• For λ = −1, the eigenvalue of A⁻¹⁵ is (−1)⁻¹⁵ = −1, occurring twice. Corresponding eigenvectors: (1, −2, 0) and (1, 0, −1).
• For λ = 8, the eigenvalue of A⁻¹⁵ is 8⁻¹⁵ = 1/8¹⁵ = 1/35184372088832. Corresponding eigenvector: (2, 1, 2).

Final eigenvalues of A⁻¹⁵: −1, −1, 1/35184372088832.

(b) Let S be the required area. The region is inside the cardioid r = a(1 + cos θ) and outside the circle r = a. Thus

a ≤ r ≤ a(1 + cos θ).

For this interval to be nonempty, a(1 + cos θ) ≥ a, so cos θ ≥ 0. Hence θ ranges from −π/2 to π/2. By symmetry about the x-axis,

S = 2 ∫₀^π/2 ∫_a^a(1+cos θ) r dr dθ.

First integrate with respect to r:

∫_a^a(1+cos θ) r dr = (1/2)[r²]_a^a(1+cos θ) = (a²/2)[(1 + cos θ)² − 1] = (a²/2)(2cos θ + cos²θ) = a²(cos θ + (1/2)cos²θ).

Therefore,

S = 2a² ∫₀^π/2 (cos θ + (1/2)cos²θ) dθ = a² ∫₀^π/2 (2cos θ + cos²θ) dθ.

Now,

∫₀^π/2 2cos θ dθ = 2,

∫₀^π/2 cos²θ dθ = π/4.

Hence

S = a²(2 + π/4).

Final area: a²(2 + π/4) square units, with a > 0.

(c) Let the required sphere have centre C = (α, β, γ) and radius R. The given plane is 3x + 2y − z + 2 = 0, whose normal is n = (3, 2, −1). It touches the sphere at P = (1, −2, 1). Therefore CP is perpendicular to the plane, so C lies on the normal line through P:

C = P + t n = (1 + 3t, −2 + 2t, 1 − t).

Thus

R = |t|√(3² + 2² + (−1)²) = |t|√14.

The given sphere is

x² + y² + z² − 4x + 6y + 4 = 0.

Completing squares:

(x − 2)² + (y + 3)² + z² = 9.

So its centre is O = (2, −3, 0) and radius r = 3.

For two spheres cutting orthogonally, the distance d between their centres satisfies

d² = R² + r².

Now

OC² = (1 + 3t − 2)² + (−2 + 2t + 3)² + (1 − t − 0)² = (3t − 1)² + (2t + 1)² + (1 − t)² = 14t² − 4t + 3.

Also R² = 14t² and r² = 9. Hence

14t² − 4t + 3 = 14t² + 9,

so −4t = 6, giving t = −3/2.

Therefore

C = (1 + 3(−3/2), −2 + 2(−3/2), 1 − (−3/2)) = (−7/2, −5, 5/2).

Also

R² = 14(9/4) = 63/2.

Thus the required sphere is

(x + 7/2)² + (y + 5)² + (z − 5/2)² = 63/2.

Equivalently,

x² + y² + z² + 7x + 10y − 5z + 12 = 0.