Q3
(a) Let V = M₂ₓ₂(ℝ) denote a vector space over the field of real numbers. Find the matrix of the linear mapping φ: V → V given by φ(v) = $\begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}$v with respect to standard basis of M₂ₓ₂(ℝ), and hence find the rank of φ. Is φ invertible? Justify your answer. (15 marks) (b) Find the volume of the greatest cylinder which can be inscribed in a cone of height h and semi-vertical angle α. (20 marks) (c) Find the vertex of the cone 4x² - y² + 2z² + 2xy - 3yz + 12x - 11y + 6z + 4 = 0. (15 marks)
हिंदी में प्रश्न पढ़ें
(a) माना V = M₂ₓ₂(ℝ) वास्तविक संख्याओं के क्षेत्र पर एक सदिश समष्टि दर्शाता है। M₂ₓ₂(ℝ) के मानक आधार के सन्दर्भ में φ(v) = $\begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}$v द्वारा दिए गए रैखिक प्रतिचित्रण φ: V → V का आव्यूह ज्ञात कीजिए और तब φ की कोटि (रैंक) ज्ञात कीजिए। क्या φ व्युत्क्रमणीय है? अपने उत्तर का तर्क प्रस्तुत कीजिए। (15 अंक) (b) ऊँचाई h तथा अर्ध-शीर्ष कोण α वाले एक शंकु के अंतर्गत सबसे बड़े बेलन का आयतन ज्ञात कीजिए। (20 अंक) (c) शंकु 4x² - y² + 2z² + 2xy - 3yz + 12x - 11y + 6z + 4 = 0 का शीर्ष ज्ञात कीजिए। (15 अंक)
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How this answer will be evaluated
Approach
Solve all three parts systematically, allocating approximately 30% time to part (a) on linear mapping matrix representation, 40% to part (b) on optimization using calculus for the inscribed cylinder, and 30% to part (c) on finding the cone vertex through partial derivatives. Begin with clear identification of basis elements for (a), set up coordinate geometry for (b) with proper diagrammatic visualization, and use the condition for singular points for (c). Present each part with distinct sub-headings and conclude with boxed final answers.
Key points expected
- For (a): Correctly identify standard basis {E₁₁, E₁₂, E₂₁, E₂₂} of M₂ₓ₂(ℝ), compute φ(Eᵢⱼ) for each basis element, and construct the 4×4 matrix representation; determine rank via row reduction or determinant, and conclude non-invertibility due to zero determinant
- For (a): Explicitly state that φ is not invertible because det(φ) = 0 (the representing matrix is singular), connecting to the original 2×2 matrix having determinant -7 ≠ 0 but the induced map on 4×4 space being singular
- For (b): Set up coordinate system with cone vertex at origin, axis along z-axis, derive cone equation z = r cot(α), express cylinder volume V = πr²(h - r cot(α)) or equivalent, apply dV/dr = 0 to find optimal r = (2h/3)tan(α), and compute maximum volume V_max = (4πh³/27)tan²(α)
- For (b): Verify second derivative test confirms maximum, and express final volume in terms of given parameters h and α with proper dimensional analysis
- For (c): Apply condition that vertex satisfies ∂F/∂x = ∂F/∂y = ∂F/∂z = 0 for F(x,y,z) = 0, solve the resulting linear system: 8x + 2y + 12 = 0, -2y + 2x - 3z - 11 = 0, 4z - 3y + 6 = 0 to obtain vertex coordinates
- For (c): Verify the solution satisfies original cone equation and confirm the point is indeed a singular point (vertex) by checking all partial derivatives vanish simultaneously
Evaluation rubric
| Dimension | Weight | Max marks | Excellent | Average | Poor |
|---|---|---|---|---|---|
| Setup correctness | 20% | 10 | For (a): correctly identifies ordered basis E₁₁, E₁₂, E₂₁, E₂₂ and computes all four φ(Eᵢⱼ) images; for (b): establishes proper coordinate system with cone vertex at origin and derives correct geometric constraints; for (c): correctly sets up partial derivative equations for singular point | Identifies basis elements but with minor ordering errors; sets up coordinate system for (b) but with inverted height relationship; attempts partial derivatives for (c) but with algebraic errors in differentiation | Wrong basis identification (e.g., confuses matrix entries with basis ordering); incorrect cone orientation in (b) leading to wrong constraint equation; fails to recognize vertex condition requires partial derivatives to vanish |
| Method choice | 20% | 10 | For (a): uses systematic column-wise or row-wise vectorization to build 4×4 matrix; for (b): employs single-variable calculus optimization with clear substitution; for (c): applies Lagrange multiplier-free approach using direct singular point conditions, or recognizes this as quadratic form analysis | Builds matrix representation but with inefficient method; uses calculus for (b) but with suboptimal variable choice (e.g., height instead of radius); attempts vertex finding but with roundabout algebraic elimination | Attempts to find matrix representation by ad-hoc guessing without basis computation; uses geometric intuition without calculus for (b); tries to complete squares for (c) without recognizing general cone vertex method |
| Computation accuracy | 20% | 10 | For (a): correct 4×4 matrix with entries involving 1,2,3,-1 in proper positions, accurate determinant calculation showing singularity; for (b): correct derivative dV/dr, exact solution r = (2h/3)tan(α), final volume (4πh³/27)tan²(α); for (c): correct linear system solution (-2, -2, -3) or equivalent after verification | Correct matrix structure but arithmetic errors in entries; correct optimization setup but algebraic slip in solving critical point; correct method for (c) but arithmetic error in solving 3×3 system | Major computational errors in matrix entries (e.g., wrong placement of 2,3 coefficients); critical point calculation errors leading to r = h/2 or similar; fails to solve linear system or obtains coordinates not satisfying original equation |
| Step justification | 20% | 10 | For (a): justifies matrix construction via linearity and basis action, proves rank deficiency via determinant or row reduction; for (b): justifies maximum via second derivative test or physical reasoning; for (c): verifies vertex by confirming all partials vanish and point lies on surface | Shows some intermediate steps but skips key justifications (e.g., why chosen basis ordering); states maximum without verification; finds point but doesn't explicitly verify it satisfies all conditions | Jumps to final answers without showing work; no justification for why critical point is maximum; no verification that found point is actually on the cone |
| Final answer & units | 20% | 10 | Clear boxed/concluded answers: for (a) explicit 4×4 matrix, rank = 3, 'not invertible' with reason; for (b) volume = 4πh³tan²α/27 with dimensional consistency [L³]; for (c) vertex at (-2, -2, -3) or equivalent verified coordinates, with all three parts distinctly labeled | Correct answers present but poorly formatted or mixed between parts; missing units in (b); correct numerical answer for (c) but coordinates not clearly identified as vertex | Missing final answers for one or more parts; answers without proper labeling which part they belong to; dimensional inconsistency (e.g., volume with units of length) |
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