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MTH603 Midterm MCQs Set 2 by VU Topper RM

Numerical Analysis PDF 1.1 MBUpdated 23 Oct 2026
Midterm past paper MTH603 Midterm MCQs Set 2 by VU Topper RM PDF 1.1 MB
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Midterm past paper
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MTH603 Numerical Analysis
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PDF 1.1 MB
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23 Oct 2026
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Text from pages 2, 3, 4 of 46. Download the file for all 46 pages.

If n x n matrices A and B are similar, then they have the same eigenvalues (with the same multiplicities). A. False

B. True ok

In the context of Jacobi’s method for finding Eigen values and Eigen vectors of a real symmetric matrix of order 2*2, if |-5| be its largest offdiagonal then which of the following will be its corresponding offdiagonal values of Orthogonal Matrix?

A. -Sin(theta), Cos(theta)

B. Cos(theta), -Cos(theta) C. Sin(theta), -Sin(theta) D. Sin(theta), Cos(theta)

Q6. Let A be an n ×n matrix. The number x is an eigenvalue of A if there exists a non-zero vector v such that Av = xv

  1. True ok
  2. False
  3. λ=12.4817,(X)=⎛⎝⎜1.00.4365210.625431⎞⎠⎟
  4. λ=12.4817,(X)=⎛⎝⎜0.4365210.6254311.0⎞⎠⎟
  5. λ=0.436521,(X)=⎛⎝⎜12.48170.6254311.0⎞⎠⎟
  6. λ=0.625431,(X)=⎛⎝⎜0.43652112.48171.0⎞⎠⎟

which of the following will be 2nd order Backward difference at y2 = 5

A. -1

B. 1 ok

C. -2 D. 2

While using Jacobi method for the matrix A=⎡⎣⎢13–√23–√32–√22–√1⎤⎦⎥ and 'theta θ =pi/4',the orthogonal matrix S1 will be given by A. S1=⎡⎣⎢⎢cosπ40sinπ4010−sinπ40cosπ4⎤⎦⎥⎥ B. S1=⎡⎣⎢⎢1000cosπ4sinπ40−sinπ4cosπ4⎤⎦⎥⎥ C. S1=⎡⎣⎢⎢cosπ4sinπ40−sinπ4cosπ40001⎤⎦⎥⎥ D. S1=⎡⎣⎢⎢cosπ4000−sinπ40001⎤⎦⎥⎥

Let [A] be a 3x3 real symmetric matrix with |a23| be numerically the largest off-diagonal element of A, then we can construct orthogonal matrix S1 by Jacobi’s method as A. ⎡⎣⎢cosθ0sinθ010−sinθ0cosθ⎤⎦⎥ B. ⎡⎣⎢1000cosθsinθ0−cosθ−sinθ⎤⎦⎥ C. ⎡⎣⎢cosθsinθ0−sinθcosθ0001⎤⎦⎥ D. ⎡⎣⎢1000cosθsinθ0−sinθcosθ⎤⎦⎥

A. λ=1.123685,(X)=⎛⎝⎜0.6789581.09.85490⎞⎠⎟ B. λ=1.123685,(X)=⎛⎝⎜9.854901.00.678958⎞⎠⎟ C. λ=9.85490,(X)=⎛⎝⎜1.1236851.00.678958⎞⎠⎟

D. λ=9.85490,(X)=⎛⎝⎜0.6789581.01.123685⎞⎠⎟

A. u(2)=1.8⎛⎝⎜1−0.8/1.80.6/1.8⎞⎠⎟ B. u(2)=0.6⎛⎝⎜1.8/0.6−0.8/0.61⎞⎠⎟ C. u(2)=2.8⎛⎝⎜1.8/2.8−0.8/2.81⎞⎠⎟

D. u(2)=−0.8⎛⎝⎜−1.8/0.81−0.6/0.8⎞⎠⎟

A. u(1)=−2⎛⎝⎜−12/26/21⎞⎠⎟ B. u(1)=−12⎛⎝⎜−2/12−2/61⎞⎠⎟ C. u(1)=12⎛⎝⎜1−6/12−2/12⎞⎠⎟ D. u(1)=112⎛⎝⎜1−12/6−12/2⎞⎠⎟

following system 2x+2y+z=3x+3y+z=2x+y+z=2 with the initial guess

A. (32,−23,−2) B. (32,57,19)

C. (32,16,13)

D. (32,23,2)

Which of the following system of equation is diagonally dominant?

  1. 8x+y−z=5 2x+13y+5z=7 x+2y+9z=3
  2. 8x+y−z=5 x+2y+9z=3 2x+13y+5z=7
  3. 2x+13y+5z=7 8x+y−z=5 x+2y+9z=3
  4. x+2y+9z=3 8x+y−z=5 2x+13y+5z=7

Correct answer: 8x+y−z=5 2x+13y+5z=7 x+2y+9z=3

diagonally dominant coefficient matrix?

  1. −2x1+7x2+2x3=56x1−2x2+3x3=1x1+x2−5x3=−13
  2. −2x1+7x2+2x3=56x1−2x2+3x3=1x1+x2−5x3=−13
  3. 6x1−2x2+3x3=1−2x1+7x2+2x3=5x1+x2−5x3=−13
  4. x1+x2−5x3=−136x1−2x2+3x3=1−2x1+7x2+2x3=5

Correct answer: x1+x2−5x3=−136x1−2x2+3x3=1−2x1+7x2+2x3=5

While solving a system of three linear equations in three variables by Jacobi’s method, the first approximate solution is NECESSARY to be taken as (0,0,0). A. True

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