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Maths for Electronics 2 Quiz 2: 16 March 2025 (January 2025 term)

Question 1

+40 marksOne correct option
  • Note: A vector [∙∙]\begin{bmatrix} \bullet \\ \bullet \end{bmatrix} could be represented as (∙,∙)(\bullet, \bullet) or [∙,∙][\bullet, \bullet].
  • u⋅vu \cdot v is the standard inner product of the two vectors uu and vv.
  • u‾\overline{u} is the complex conjugate of the vector uu.
  • ∥u∥=u⋅u\|u\| = \sqrt{u \cdot u}
  • S⊥S^{\perp} is the dual of the subspace SS.
  • T∗T^* is the adjoint of the operator TT.
  • AHA^H is the Hermitian of the matrix AA.

1 Objective Questions (16 marks)

  1. Consider a 5×75 \times 7 matrix AA and a 7×57 \times 5 matrix BB. Which of the following option(s) is(are) true? [4 Marks]

(a) Rank(AB)=Rank(A)+Rank(B)\text{Rank}(AB) = \text{Rank}(A) + \text{Rank}(B)

(b) Rank(5A)=Rank(A)\text{Rank}(5A) = \text{Rank}(A).

(c) If the nullity of the matrix ABAB is 2, then the rank of the matrix ABAB is 3.

(d) Rank(5A)≤5\text{Rank}(5A) \leq 5.

  1. Let VV be the subspace of R3\mathbb{R}^3 defined as follows:

V={(x,y,z)∣x=y−z, and x,y,z∈R}.V = \{(x,y,z) \mid x = y - z, \text{ and } x,y,z \in \mathbb{R}\}.

Which of the following option(s) is(are) true? [4 marks]

(a) The vectors (1,1,0),(1,0,−1)(1,1,0), (1,0,-1) in VV are linearly dependent.

(b) The vectors (1,1,0),(1,0,−1),(0,1,1)(1,1,0), (1,0,-1), (0,1,1) in VV are linearly independent.

(c) Span((0,1,1),(1,0,−1))=V\text{Span}((0,1,1), (1,0,-1)) = V.

(d) Span((1,1,0))\text{Span}((1,1,0)) is a subspace of VV.

  1. Consider the vectors a=(1,2)a = (1,2) and b=(2,2)b = (2,2). Find the value of ∣∣a+b∣∣−∣∣a−b∣∣||a + b|| - ||a - b||. [4 marks]

  2. Consider a linear operator T(x,y,z)=(2x+iy,y−5iz,x+(1−i)y+3z)T(x,y,z) = (2x + iy, y - 5iz, x + (1-i)y + 3z). Let AA be the matrix representation of TT. Which of the following option(s) is(are) correct? [4 marks]

(a) T∗=TT^* = T

(b) AH=[201−i11+i05i3]A^H = \begin{bmatrix} 2 & 0 & 1 \\ -i & 1 & 1+i \\ 0 & 5i & 3 \end{bmatrix}

(c) T∘T∗T \circ T^* is a self-adjoint operator.

(d) T∘T∗T \circ T^* is not diagonalizable.

2 Subjective Questions (24 marks)

  1. Consider a matrix

B=(1−333−536−64).B = \begin{pmatrix} 1 & -3 & 3 \\ 3 & -5 & 3 \\ 6 & -6 & 4 \end{pmatrix}.

(i) Find the inverse of the matrix BB. [4 marks]

(ii) If BB is diagonalizable, then find a diagonal matrix DD and an invertible matrix PP such that B=PDP−1B = PDP^{-1}. [6 marks]

(iii) What are the algebraic and geometric multiplicities of eigenvalues of B−1B^{-1}? [2 marks]

  1. Consider a matrix

A=[2−iii2−i−ii2].A = \begin{bmatrix} 2 & -i & i \\ i & 2 & -i \\ -i & i & 2 \end{bmatrix}.

(a) Find the left eigenvector for at least one eigenvalue of AA. [3 marks]

(b) Find the left null space of AA. [3 marks]

  1. Find the least squares solution to the system Ax=bAx = b, where

A=[011121] and b=[61−1].A = \begin{bmatrix} 0 & 1 \\ 1 & 1 \\ 2 & 1 \end{bmatrix} \text{ and } b = \begin{bmatrix} 6 \\ 1 \\ -1 \end{bmatrix}.

[6 marks]

  1. A

    I have written answers on the answer sheets

  2. B

    Not applicable

More on the Maths for Electronics 2 Quiz 2 16 Mar 2025 paper

The IIT Madras BS Mathematics for Electronics II (Maths for Electronics 2) Quiz 2 paper sat on 16 Mar 2025, in the January 2025 term: 1 question for 40 marks in 120 minutes. The first 1 question is below. Sign in with Google — it is free — to see the whole paper with its answers and explanations, in learning mode or as a timed mock test.

FeatureMaths for Electronics 2 Quiz 2 16 Mar 2025 at a glance
TermJanuary 2025 term
SubjectMathematics for Electronics II
Course codeMA2101
Questions1
Marks40
Duration120 min
MCQ1
Official paperIIT M ES FOUNDATION AN EXAM QEF2 16 Mar 2025
Negative markingNo negative marking.
Updated

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