Let A = 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 (i) Verify the Cayley-Hamilton theorem for the matrix A. (ii) Show that Aⁿ = Aⁿ⁻² + A²…

(a)(i) Verification of the Cayley-Hamilton theorem

The theorem states that every square matrix satisfies its own characteristic equation. For A = [1 0 0; 1 0 1; 0 1 0],

λI − A = [λ−1 0 0; −1 λ −1; 0 −1 λ],

so det(λI − A) = (λ−1)(λ² − 1) = λ³ − λ² − λ + 1.

The characteristic equation is therefore λ³ − λ² − λ + 1 = 0, and we must show A³ − A² − A + I = 0.

A² = [1 0 0; 1 1 0; 1 0 1] and A³ = A²·A = [1 0 0; 2 0 1; 1 1 0].

Then A³ − A² − A + I = [1 0 0; 2 0 1; 1 1 0] − [1 0 0; 1 1 0; 1 0 1] − [1 0 0; 1 0 1; 0 1 0] + [1 0 0; 0 1 0; 0 0 1] = [0 0 0; 0 0 0; 0 0 0].

Hence A satisfies its characteristic equation, and the theorem is verified.

(a)(ii) The relation Aⁿ = Aⁿ⁻² + A² − I and the value of A⁴⁰

From part (i), A³ = A² + A − I, so A³ − A = A² − I. Put B = A² − I. Then AB = A³ − A = B.

For n = 3, A³ = A + (A² − I) = A + B, so the relation holds.

Assume Aⁿ = Aⁿ⁻² + B for some n ≥ 3. Multiplying by A, Aⁿ⁺¹ = Aⁿ⁻¹ + AB = Aⁿ⁻¹ + B = A⁽ⁿ⁺¹⁾⁻² + A² − I. By induction, Aⁿ = Aⁿ⁻² + A² − I for all n ≥ 3.

Applying the relation repeatedly, A²ᵐ = A² + (m − 1)B. For n = 40, m = 20, so

A⁴⁰ = A² + 19B.

Here B = A² − I = [0 0 0; 1 0 0; 1 0 0]. Therefore

A⁴⁰ = [1 0 0; 20 1 0; 20 0 1].

(b) Whether (0, 0) is an extreme point of f(x, y) = 2x⁴ − 3x²y + y²

Factorise: f(x, y) = (y − x²)(y − 2x²). At (0, 0), f = 0, and f_x = 8x³ − 6xy, f_y = −3x² + 2y both vanish there, so (0, 0) is a critical point. The Hessian is singular there, so the second-derivative test fails and we examine the sign of f near the origin.

Along x = t, y = 0, f = 2t⁴ > 0 for t ≠ 0. Along x = t, y = 3t²/2, f = (t²/2)(−t²/2) = −t⁴/4 < 0 for t ≠ 0.

Both paths approach (0, 0) as t → 0. Every neighbourhood of (0, 0) thus contains points where f is positive and points where f is negative, while f(0, 0) = 0. So (0, 0) is neither a local maximum nor a local minimum.

Therefore (0, 0) is not an extreme point of f. (Geometrically, f is positive outside the parabolas y = x² and y = 2x² and negative between them, a saddle-type behaviour even though every straight line through the origin gives a minimum.)

(c) Sphere through the given circle

Let S = x² + y² + z² − 4x − 6y + 2z − 16 and L = 3x + y + 3z − 4. Every sphere through the circle S = 0, L = 0 has the form S + λL = 0, that is,

x² + y² + z² + (3λ − 4)x + (λ − 6)y + (3λ + 2)z − (16 + 4λ) = 0.

Case (i): the sphere passes through (1, 0, −3).

At (1, 0, −3), S = 1 + 0 + 9 − 4 − 0 − 6 − 16 = −16 and L = 3 + 0 − 9 − 4 = −10. So −16 − 10λ = 0, giving λ = −8/5.

The sphere is x² + y² + z² − (44/5)x − (38/5)y − (14/5)z − 48/5 = 0, or 5x² + 5y² + 5z² − 44x − 38y − 14z − 48 = 0.

Case (ii): the circle is a great circle.

The circle is a great circle when the plane L = 0 passes through the centre of the sphere. The centre is

((4 − 3λ)/2, (6 − λ)/2, (−2 − 3λ)/2).

Substituting in L = 0: 3(4 − 3λ)/2 + (6 − λ)/2 + 3(−2 − 3λ)/2 − 4 = 0 ⇒ (12 − 19λ)/2 − 4 = 0 ⇒ λ = 4/19.

The sphere is x² + y² + z² − (64/19)x − (110/19)y + (50/19)z − 320/19 = 0, or 19x² + 19y² + 19z² − 64x − 110y + 50z − 320 = 0.