Paper I — Q2
(a) Consider a linear operator T on R³ over R defined by T(x, y, z) = (2x, 4x – y, 2x + 3y – z). Is T invertible? If yes, justify…
Consider a linear operator T on R³ over R defined by T(x, y, z) = (2x, 4x – y, 2x + 3y – z). Is T invertible? If yes, justify your answer and find T⁻¹. 15 marks
If u = (x+y)/(1-xy) and v = tan⁻¹x + tan⁻¹y, then find ∂(u, v)/∂(x, y). Are u and v functionally related? If yes, find the relationship. 15 marks
Find the image of the line x = 3-6t, y = 2t, z = 3+2t in the plane 3x+4y-5z+26 = 0. 20 marks
हिंदी में प्रश्न पढ़ें
माना R के ऊपर R³ पर एक रैखिक संकारक T, T(x, y, z) = (2x, 4x – y, 2x + 3y – z) द्वारा परिभाषित है। क्या T व्युत्क्रमणीय है? यदि हाँ, तो अपने उत्तर का तर्क प्रस्तुत कीजिए तथा T⁻¹ ज्ञात कीजिए। (15 अंक)
यदि u = (x+y)/(1-xy) तथा v = tan⁻¹x + tan⁻¹y है, तब ∂(u, v)/∂(x, y) ज्ञात कीजिए। क्या u तथा v फलनतः सम्बन्धित हैं? यदि हाँ, तो सम्बन्ध ज्ञात कीजिए। (15 अंक)
रेखा x = 3-6t, y = 2t, z = 3+2t का समतल 3x+4y-5z+26 = 0 में प्रतिबिम्ब ज्ञात कीजिए। (20 अंक)
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) T(x, y, z) = (2x, 4x − y, 2x + 3y − z). Its standard matrix is
A = [[2, 0, 0], [4, −1, 0], [2, 3, −1]].
Using the determinant of a lower-triangular matrix,
det A = 2 × (−1) × (−1) = 2 ≠ 0.
Hence T is invertible.
Let T(x, y, z) = (X, Y, Z). Then
X = 2x ⇒ x = X/2.
Y = 4x − y ⇒ y = 4x − Y = 2X − Y.
Z = 2x + 3y − z ⇒ z = 2x + 3y − Z.
Substituting x and y:
z = X + 3(2X − Y) − Z = 7X − 3Y − Z.
Therefore
T⁻¹(X, Y, Z) = (X/2, 2X − Y, 7X − 3Y − Z).
Final answer (a): T is invertible since det A = 2 ≠ 0, and T⁻¹(X, Y, Z) = (X/2, 2X − Y, 7X − 3Y − Z).
(b) Given
u = (x + y)/(1 − xy), v = tan⁻¹ x + tan⁻¹ y.
By the quotient rule,
∂u/∂x = [(1)(1 − xy) − (x + y)(−y)]/(1 − xy)² = (1 + y²)/(1 − xy)².
Similarly,
∂u/∂y = [(1)(1 − xy) − (x + y)(−x)]/(1 − xy)² = (1 + x²)/(1 − xy)².
Also,
∂v/∂x = 1/(1 + x²), ∂v/∂y = 1/(1 + y²).
The Jacobian is
∂(u, v)/∂(x, y) = determinant of [[∂u/∂x, ∂u/∂y], [∂v/∂x, ∂v/∂y]]
= (∂u/∂x)(∂v/∂y) − (∂u/∂y)(∂v/∂x)
= [(1 + y²)/(1 − xy)²]·[1/(1 + y²)] − [(1 + x²)/(1 − xy)²]·[1/(1 + x²)]
= 1/(1 − xy)² − 1/(1 − xy)²
= 0.
Since the Jacobian vanishes identically wherever the derivatives exist, u and v are functionally related.
Now let x = tan α and y = tan β. Then
v = α + β,
and
u = (tan α + tan β)/(1 − tan α tan β) = tan(α + β) = tan v.
Thus the functional relation is
u = tan v, provided 1 − xy ≠ 0.
Equivalently, on branches,
v = tan⁻¹ u + kπ, k ∈ Z.
Final answer (b): ∂(u, v)/∂(x, y) = 0. u and v are functionally related, and the relation is u = tan v.
(c) The given line is
r(t) = (3 − 6t, 2t, 3 + 2t),
so its direction vector is
d = (−6, 2, 2).
The plane is
3x + 4y − 5z + 26 = 0,
so its normal vector is
n = (3, 4, −5).
For reflection in the plane, the component of d along n reverses sign. First find the intersection of the line with the plane. Substitute the line into the plane:
3(3 − 6t) + 4(2t) − 5(3 + 2t) + 26 = 0
= 9 − 18t + 8t − 15 − 10t + 26
= 20 − 20t.
Thus t = 1. The point of intersection is
A = (3 − 6, 2, 3 + 2) = (−3, 2, 5).
Now reflect the direction vector d across the plane. Use
d′ = d − [2(d·n)/(n·n)]n.
Compute
d·n = (−6)(3) + (2)(4) + (2)(−5) = −18 + 8 − 10 = −20,
n·n = 3² + 4² + (−5)² = 9 + 16 + 25 = 50.
Therefore,
d′ = (−6, 2, 2) − [2(−20)/50](3, 4, −5)
= (−6, 2, 2) + (4/5)(3, 4, −5)
= (−6 + 12/5, 2 + 16/5, 2 − 4)
= (−18/5, 26/5, −2).
This is proportional to (−9, 13, −5).
Hence the reflected line passes through A = (−3, 2, 5) and has direction (−9, 13, −5).
Its parametric form is
x = −3 − 9s, y = 2 + 13s, z = 5 − 5s, s ∈ R.
Its Cartesian form is
(x + 3)/(−9) = (y − 2)/13 = (z − 5)/(−5).
Equivalently,
(x + 3)/9 = (y − 2)/(−13) = (z − 5)/5.
Final answer (c): The image line is
x = −3 − 9s, y = 2 + 13s, z = 5 − 5s, s ∈ R,
or in Cartesian form,
(x + 3)/9 = (y − 2)/(−13) = (z − 5)/5.
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) justify: claim > 3-4 reasons > evidence > conclusion | (b) calculate: given > formula > substitution > result with units > interpretation | (c) calculate: given > formula > substitution > result with units > interpretation Full marks: Complete working with all steps, correct results, and verification.
Key points expected
- Matrix representation of T
- Calculation of determinant
- Invertibility condition stated
- Inverse matrix T⁻¹ derived
- Partial derivatives of u and v
- Jacobian determinant calculation
- Functional relationship derived
- Relationship explicitly stated
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Determine invertibility of T and find T⁻¹. 15 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- Matrix representation of T
- Calculation of determinant
- Invertibility condition stated
- Inverse matrix T⁻¹ derived
Loses marks
- Arithmetic error in determinant
- Missing inverse matrix
Earns more
- Verification T T⁻¹ = I
- Alternative method noted
Extra mark
- Neat matrix layout
- (b) Compute Jacobian ∂(u,v)/∂(x,y) and find functional relationship. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Partial derivatives of u and v
- Jacobian determinant calculation
- Functional relationship derived
- Relationship explicitly stated
Loses marks
- Incorrect partial derivatives
- Missing functional relationship
Earns more
- Trigonometric identity used
- Step-by-step differentiation
Extra mark
- Alternative derivation method
- (c) Find image of line in given plane. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Direction vector of line
- Normal vector of plane
- Reflection formula applied
- Image line equation derived
Loses marks
- Incorrect reflection formula
- Missing image line equation
Earns more
- Point of intersection found
- Verification of image line
Extra mark
- Geometric diagram included
Practice this exact question
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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