Paper I — Q3
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…
Let A = 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0
Verify the Cayley-Hamilton theorem for the matrix A.
Show that Aⁿ = Aⁿ⁻² + A² - I for n ≥ 3, where I is the identity matrix of order 3. Hence, find A⁴⁰. 10+10
Justify whether (0, 0) is an extreme point for the function f(x, y) = 2x⁴ - 3x^2y + y². 15 marks
Find the equation of the sphere through the circle
x² + y² + z² - 4x - 6y + 2z - 16 = 0; 3x + y + 3z - 4 = 0
in the following two cases.
the point (1, 0, -3) lies on the sphere.
the given circle is a great circle of the sphere. 15 marks
हिंदी में प्रश्न पढ़ें
दिया गया है A= 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0
आव्यूह A के लिये कैले-हैमिल्टन प्रमेय को सत्यापित कीजिए।
दर्शाइए कि n ≥ 3 के लिये Aⁿ = Aⁿ⁻² + A² – I; जहाँ I कोटि 3 का तत्समक आव्यूह है।
अतः A⁴⁰ ज्ञात कीजिए। 10+10
तर्क सहित दर्शाइये कि (0, 0), फलन f(x, y) = 2x⁴ - 3x^2y + y² का चरम-बिन्दु है अथवा नहीं। 15
वृत्त x² + y² + z² - 4x - 6y + 2z - 16 = 0; 3x + y + 3z - 4 = 0 से होकर गुजरने वाले गोले का समीकरण निम्न दो स्थितियों में ज्ञात कीजिए।
बिन्दु (1, 0, -3) गोले पर हो।
दिया गया वृत्त गोले का एक बृहत् वृत्त हो। 15
Model answer
Written by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key.
(a)(i) Cayley-Hamilton theorem states that every square matrix satisfies its 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]. Thus det(λI - A) = (λ-1)(λ² - 1) = λ³ - λ² - λ + 1. So the characteristic equation is A³ - A² - A + I = 0.
Now A² = [1 0 0; 1 1 0; 1 0 1], and A³ = [1 0 0; 2 0 1; 1 1 0]. Hence A³ - A² - A + I = [-1 0 0; 0 -1 0; 0 0 -1] + I = 0. Thus A satisfies its characteristic equation, verifying Cayley-Hamilton.
(a)(ii) From Cayley-Hamilton, A³ = A² + A - I, so A³ - A = A² - I. Let B = A² - I. Then AB = A³ - A = B. For n = 3, A³ = A + A² - I = A + B, so the formula 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.
For even n = 2m, 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) f(x, y) = 2x⁴ - 3x²y + y² = (y - x²)(y - 2x²). At (0, 0), f = 0. Also f_x = 8x³ - 6xy and f_y = -3x² + 2y, so both first partial derivatives vanish at (0, 0).
Take x = t, y = 0. Then f = 2t⁴ > 0 for t ≠ 0. Take x = t, y = 3t²/2. Then f = (3t²/2 - t²)(3t²/2 - 2t²) = (t²/2)(-t²/2) = -t⁴/4 < 0 for t ≠ 0. Both points approach (0, 0) as t → 0. Hence every neighbourhood of (0, 0) contains points where f is positive and points where f is negative. So (0, 0) is neither a local maximum nor a local minimum. Therefore (0, 0) is not an extreme point of f.
(c) Let S = x² + y² + z² - 4x - 6y + 2z - 16 and L = 3x + y + 3z - 4. Every sphere through the given circle has equation S + λL = 0, that is, x² + y² + z² + (-4 + 3λ)x + (-6 + λ)y + (2 + 3λ)z + (-16 - 4λ) = 0.
(c)(i) Since (1, 0, -3) lies on the sphere, S(1, 0, -3) + λL(1, 0, -3) = 0. Here S = -16 and L = -10. Hence -16 - 10λ = 0, so λ = -8/5. The required sphere is x² + y² + z² - 44/5 x - 38/5 y - 14/5 z - 48/5 = 0, or equivalently 5x² + 5y² + 5z² - 44x - 38y - 14z - 48 = 0.
(c)(ii) For the given circle to be a great circle of the sphere, the plane L = 0 must pass 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. This gives (12 - 19λ)/2 - 4 = 0, hence λ = 4/19. Therefore the sphere is x² + y² + z² - 64/19 x - 110/19 y + 50/19 z - 320/19 = 0, or equivalently 19x² + 19y² + 19z² - 64x - 110y + 50z - 320 = 0.
What "Solve" is asking you to do
Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.
Structure that answers it
Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity
Where marks are lost
Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.
How this answer will be evaluated
Approach
(a(i)) derive: given > assumptions > stepwise derivation > result > check | (a(ii)) derive: given > assumptions > stepwise derivation > result > check | (b) justify: claim > 3-4 reasons > evidence > conclusion | (c(i)) calculate: given > formula > substitution > result with units > interpretation | (c(ii)) calculate: given > formula > substitution > result with units > interpretation Full marks: Complete derivations with all steps shown, correct final answers, and clear justification of methods.
Key points expected
- Compute characteristic polynomial p(λ) of A
- Substitute matrix A into p(A)
- Show p(A) equals zero matrix
- Derive Aⁿ = Aⁿ⁻² + A² - I for n ≥ 3
- Apply recurrence to find A⁴⁰
- Present final matrix for A⁴⁰
- Find critical points via partial derivatives
- Compute Hessian matrix at (0,0)
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a(i)) Confirm Cayley-Hamilton theorem for matrix A.
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Compute characteristic polynomial p(λ) of A
- Substitute matrix A into p(A)
- Show p(A) equals zero matrix
Loses marks
- Skipping determinant expansion steps
- Arithmetic errors in matrix multiplication
Earns more
- Explicit calculation of determinant |A - λI|
- (a(ii)) Prove recurrence relation and calculate A⁴⁰.
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Derive Aⁿ = Aⁿ⁻² + A² - I for n ≥ 3
- Apply recurrence to find A⁴⁰
- Present final matrix for A⁴⁰
Loses marks
- Failing to justify the recurrence step
- Incorrect final matrix entries
Earns more
- Showing base case for induction
- Alternative method for A⁴⁰
- (b) Determine if (0,0) is an extreme point for f(x,y). 15 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- Find critical points via partial derivatives
- Compute Hessian matrix at (0,0)
- Apply second derivative test (D > 0, fxx > 0)
Loses marks
- Omitting Hessian determinant calculation
- Confusing local max/min with global
Earns more
- Explicit calculation of fxx, fyy, fxy
- Checking for saddle point conditions
- (c(i)) Find sphere equation passing through circle and point (1,0,-3).
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Use general sphere equation S + λL = 0
- Substitute point (1,0,-3) to find λ
- Write final sphere equation
Loses marks
- Incorrect substitution of point coordinates
- Algebraic errors in solving for λ
Earns more
- Correct identification of S and L
- (c(ii)) Find sphere equation where circle is a great circle.
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Identify center of given circle
- Set sphere center equal to circle center
- Determine λ using center coordinates
Loses marks
- Failing to equate sphere and circle centers
- Incorrect center calculation
Earns more
- Explicit calculation of circle center
- Verification that radius matches
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Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.
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