Ncku Engineering Mathematics Super-mega Practice Final Exam Mk. II
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Consider the matrix shown. From the list below, choose the eigenvalues for this matrix. Choose as many correct answers as you wish.HINT: This is a 2x2. So choose 2 values below - Prof. Smith
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15
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2
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5
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3
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0
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Please use your student number in both fields below (I. E. First name AND Last name = Student number).
Have fun - this is just to give you an idea about what to expect in your REAL exam.
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- 2.
For the matrix shown, choose the correct eigenvectors from the list below.
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[1; -2]
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[2; -1]
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None of these options are correct. (DON'T PICK THIS ONE)
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[-2; 1]
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[1; 2]
Correct Answer(s)
A. [-2; 1]
A. [1; 2]Explanation
The correct eigenvectors for the given matrix are [-2; 1] and [1; 2]. These vectors satisfy the equation Av = λv, where A is the given matrix, v is the eigenvector, and λ is the eigenvalue. By substituting the values, we can see that both [-2; 1] and [1; 2] satisfy the equation, making them the correct eigenvectors.Rate this question:
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- 3.
For the 2x2 matrix shown, what are the eigenvalues? Choose the correct option(s) from the list below. You may choose one or more options.
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4
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5
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-3
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2
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None of these options are correct.
Correct Answer
A. None of these options are correct. -
- 4.
For the matrix shown, choose the correct Eigenvectors from the list below. You may choose more than one option.
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[-4; 1]
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[1; 1]
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[2; 2]
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[4; -1]
Correct Answer(s)
A. [-4; 1]
A. [1; 1]
A. [2; 2]
A. [4; -1]Explanation
The given matrix has eigenvalues of -2 and 3. To find the eigenvectors, we need to solve the equation (A - λI)v = 0, where A is the matrix, λ is the eigenvalue, I is the identity matrix, and v is the eigenvector. Plugging in the eigenvalue -2, we get the equation [-6, 1; 1, 1]v = 0. Solving this equation, we find that v can be any nonzero multiple of [-4, 1]. Similarly, plugging in the eigenvalue 3, we get the equation [-1, 1; 1, -2]v = 0. Solving this equation, we find that v can be any nonzero multiple of [1, 1]. Therefore, the correct eigenvectors are [-4, 1], [1, 1], [2, 2], and [4, -1].Rate this question:
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- 5.
For the 1D heat conduction problem shown here, with the initial condition shown (i.e. T(x) = 1 - x/L), what is a suitable temperature in the bar after t (time) equals infinity? (i.e. the steady solution)
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T(x) = 0
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T(x) = 0.5
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T(x) = x
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None of the above options
Correct Answer
A. T(x) = 0Explanation
The initial condition given is T(x) = 1 - x/L, which means that the temperature at any point x in the bar decreases linearly with x. As time goes to infinity, the system reaches a steady state where the temperature no longer changes with time. In this steady state, the temperature throughout the bar will be constant. Since the initial condition is a linear function that decreases with x, the suitable temperature in the bar after t equals infinity would be T(x) = 0, as it satisfies the condition of a constant temperature throughout the bar.Rate this question:
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- 6.
For the ODE shown, what is the correct Finite Difference (FD) discretization?
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Option 1
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Option 2
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Option 3
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Option 4
Correct Answer
A. Option 1 -
- 7.
Shown here is a finite difference discretization of a PDE. What options below would be correct choices for that PDE? (Choose one)
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None of these options are correct.
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Option 2
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Option 3
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Option 4
Correct Answer
A. Option 4 -
- 8.
Find the fourier series representation of the function shown.
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Option 1
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Option 2
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Option 3
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None of these options are correct.
Correct Answer
A. Option 1 -
- 9.
Consider the MATLAB code attached. Someone thinks it contains a bug, because it produces strange results. Where is the bug?
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Line 1 - The variable u is undefined.
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Line 8 - The command "total(j) = 0" is on the wrong line.
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Line 14 - The command "result(i,:) = total(:)" is not correct.
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There is no bug in this code.
Correct Answer
A. There is no bug in this code.Explanation
The given code does not contain any bugs.Rate this question:
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Apr 19, 2023Quiz Edited by
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Jun 14, 2016Quiz Created by
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