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IIT M IMPROVEMENT AN EXAM QIA4 26 Oct - 26 Oct 2025

MLF Quiz 118 questions40 marks12 MCQ · 4 SA · 2 MSQImprovement / afternoon / QIA4
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Q1MCQ1 mark
A bank wants to decide whether a loan applicant should be given a loan or not, based on features like income, credit score and past records. Which of the following machine learning techniques is applicable for this scenario?
  1. Regression
  2. Classification
  3. Clustering
  4. Dimensionality Reduction
Q2MCQ3 marks
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Q3MSQ2 marksselect all that apply
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Q4SA3 marks
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Q5SA3 marks
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Q6SA2 marks
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Comprehension passage - the questions below it refer to this
Based on the above data, answer the given subquestions. question figure
Q7MCQ3 marks
Which of these models has the smallest squared loss on this training dataset?
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Q8MCQ2 marks
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  1. No
  2. Yes
  3. Insufficient information to answer the question
Comprehension passage - the questions below it refer to this
Based on the above data, answer the given subquestions. question figure
Q9MCQ3 marks
Which of the following sets forms a basis for the row space of A?
  1. {(1, 0, 2), (0, 1,−1)}
  2. {(1, 0, 2), (0, 1,−1), (0, 0,−1)}
  3. {(1, 0, 2), (0,−1, 1), (0,−1,−4)}
  4. {(0, 1,−1), (0, 0,−4)}
Q10MCQ2 marks
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Comprehension passage - the questions below it refer to this
Based on the above data, answer the given subquestions. question figure
Q11MSQ2 marksselect all that apply
Which of the following is/are true?
  1. A is an orthogonal matrix
  2. A is invertible
  3. A is symmetric
Q12MCQ3 marks
Which of the following is true?
  1. A does not have real eigenvalues
  2. A has two distinct real eigenvalues
  3. A has two real eigenvalues, not necessarily distinct
Comprehension passage - the questions below it refer to this
There is no common data for this question. There are three independent sub-questions for which you have to answer True/False.
Q13MCQ2 marks
If a square matrix A is diagonalizable, then A is invertible.
  1. True
  2. False
Q14MCQ1 mark
The eigenvalues of a real symmetric matrix are real.
  1. True
  2. False
Q15MCQ2 marks
A square matrix A of shape n x n is diagonalizable if and only if it has n distinct eigenvalues.
  1. True
  2. False
Comprehension passage - the questions below it refer to this
Based on the above data, answer the given subquestions. question figure
Q16MCQ2 marks
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Q17SA2 marks
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Q18MCQ2 marks
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  1. (A)
  2. (B)

Answer key

  1. Q1 (MCQ, 1 marks): B
  2. Q2 (MCQ, 3 marks): A
  3. Q3 (MSQ, 2 marks): B, C, D
  4. Q4 (SA, 3 marks): 72
  5. Q5 (SA, 3 marks): 1024
  6. Q6 (SA, 2 marks): 0
  7. Q7 (MCQ, 3 marks): A
  8. Q8 (MCQ, 2 marks): A
  9. Q9 (MCQ, 3 marks): B
  10. Q10 (MCQ, 2 marks): C
  11. Q11 (MSQ, 2 marks): A, B
  12. Q12 (MCQ, 3 marks): A
  13. Q13 (MCQ, 2 marks): B
  14. Q14 (MCQ, 1 marks): A
  15. Q15 (MCQ, 2 marks): B
  16. Q16 (MCQ, 2 marks): A
  17. Q17 (SA, 2 marks): 1
  18. Q18 (MCQ, 2 marks): A

Concepts this paper tests

Auto-generated summary from the source site. Handy as a checklist; the lessons above explain each idea from scratch.

Here’s a structured summary of core topics, concepts, principles, and equations essential for the exam, followed by Mermaid knowledge graphs to visualize relationships.


Core Topics & Concepts

1. Machine Learning Foundations

A. Supervised Learning

y=θ0+θ1x+θ2x2+⋯+θnxny = \theta_0 + \theta_1 x + \theta_2 x^2 + \dots + \theta_n x^n
min⁡θ∑i=1m(yi−(θ0+θ1xi+θ2xi2))2\min_{\theta} \sum_{i=1}^m (y_i - (\theta_0 + \theta_1 x_i + \theta_2 x_i^2))^2

B. Model Evaluation

L=1n∑i=1n(yi−y^i)2L = \frac{1}{n} \sum_{i=1}^n (y_i - \hat{y}_i)^2
L=1n∑i=1n∥xi−g(f(xi))∥2L = \frac{1}{n} \sum_{i=1}^n \|x_i - g(f(x_i))\|^2

where ff = encoder, gg = decoder.

C. Linear Algebra for ML

projv2v1=(v1⋅v2v2⋅v2)v2\text{proj}_{v_2} v_1 = \left( \frac{v_1 \cdot v_2}{v_2 \cdot v_2} \right) v_2
det⁡(A)=∏i=1nλi\det(A) = \prod_{i=1}^n \lambda_i

D. Dimensionality Reduction

E. Basis & Row Space


2. Linear Algebra Deep Dive

A. Matrix Properties

A=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

B. Linear Approximation

La[f](x)=f(a)+∇f(a)⋅(x−a)L_a[f](x) = f(a) + \nabla f(a) \cdot (x - a)
L(0,0)[f](x)=f(0,0)+∂f∂x1(0,0)x1+∂f∂x2(0,0)x2=1+x1+x2L_{(0,0)}[f](x) = f(0,0) + \frac{\partial f}{\partial x_1}(0,0) x_1 + \frac{\partial f}{\partial x_2}(0,0) x_2 = 1 + x_1 + x_2
wTx+b=0wherew=∇f(a),b=−f(a)w^T x + b = 0 \quad \text{where} \quad w = \nabla f(a), \quad b = -f(a)

3. Key Equations Summary

TopicEquation
Polynomial Regressiony=θ0+θ1x+θ2x2y = \theta_0 + \theta_1 x + \theta_2 x^2
Least Squaresmin⁡θ∑(yi−y^i)2\min_{\theta} \sum (y_i - \hat{y}_i)^2
Reconstruction Loss1n∑∥xi−g(f(xi))∥2\frac{1}{n} \sum \|x_i - g(f(x_i))\|^2
Orthogonalityv1Tv2=0v_1^T v_2 = 0
Projectionprojv2v1=(v1⋅v2v2⋅v2)v2\text{proj}_{v_2} v_1 = \left( \frac{v_1 \cdot v_2}{v_2 \cdot v_2} \right) v_2
Eigenvalue DefinitionAv=λvAv = \lambda v
Determinantdet⁡(A)=∏λi\det(A) = \prod \lambda_i
Taylor ApproximationLa[f](x)=f(a)+∇f(a)⋅(x−a)L_a[f](x) = f(a) + \nabla f(a) \cdot (x - a)
Rotation Matrix[cos⁡θ−sin⁡θsin⁡θcos⁡θ]\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

Graph 2: Linear Algebra Concepts

Graph 3: Calculus for ML

Graph 4: Eigenvalues & Special Matrices


Exam Tips

  1. Polynomial Regression: Always use least squares to solve for θ\theta (normal equations or gradient descent).
  2. Orthogonality/Projection: Check dot product for orthogonality; projection formula is critical.
  3. Eigenvalues:

    • Sum of columns = zero vector ⇒λ=0\Rightarrow \lambda = 0.
    • Rotation matrices have complex eigenvalues (unless θ=0,π\theta = 0, \pi).
  4. Linear Approximation: Compute gradient and evaluate at the point.
  5. Basis/Row Space: Row reduce the matrix to find basis vectors.
  6. Autoencoders: Reconstruction loss is MSE between input and output.