MATLAB Determinants and Invertibility Explained

Exercise 4.1 a. Use MATLAB to compute the determinants of the matrices A+B, A-B, AB, A- 1, and BT. (Recall that in MATLAB, BT is written as B'.) >> A=[8 11 2 8;0 -7 2 -1;-3 -7 2 1;1 1 2 4] A = 8 11 2 8 0 -7 2 -1 -3 -7 2 1 1 1 2 4 >> B=[1 -2 0 5;0 7 1 5;0 4 4 0;0 0 0 2] B = 1 -2 0 5 0 7 1 5 0 4 4 0 0 0 0 2 >> det(A) ans = 76 >> det(B) ans = 48 >> det(A+B) ans = 0

>> det(A-B) ans = 1038 >> det(A*B) ans = 3.6480e+03 >> det(A^-1) ans = 0.0132 >> det(inv(A)) ans = 0.0132 >> det(B') ans = 48

b. Which of the above matrices are not invertible? Explain your reasoning. A+B is not invertible because the determinant of a A+B is zero, that means the matrix is singular (or non-invertible). This is because a determinant of zero suggests that the rows (or columns) of the matrix are linearly dependent, i.e., one row (or column) can be written as a linear combination of other rows (or columns). This in turn implies that the matrix doesn't have a full rank and hence it is non-invertible. c. Suppose that you didn't know the entries of A and B, but you did know their determinants. Which of the above determinants would you still be able to compute from this information, even without having A or B at hand? Explain your reasoning. The determinant of a product of two matrices is equal to the product of their determinants (det(AB) = det(A)*det(B)). So, if you have the determinant of AB, and the determinant of either A or B, you could find the determinant of the other matrix. The determinant of the inverse of a matrix is the reciprocal of the determinant of the original matrix, provided that the original matrix is invertible (det(A^-1) = 1/det(A)). So, if you know the determinant of A^-1, you can compute the determinant of A. The determinants of (A+B) and (A-B) generally won't help you recover the determinants of A or B, because the determinant operation is not distributive over matrix addition or subtraction.

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Exercise 4.2 Use det() to calculate the determinant of N^100. Based on this information, do you think that N^100 is invertible? >> N=[0.003 0.02 0;0.1 1 0;0 0 0.015] N = 0.0030 0.0200 0 0.1000 1.0000 0 0 0 0.0150 >> det(N^100) ans = -1.7211e-201 In this case, the determinant of N^100 is very small (-1.7211e-201), but it is not exactly zero. Therefore, based on this information, N^100 is technically invertible. Now compute the determinant of N. Calculate the determinant of N^100 by hand. Now would you reconsider your answer to the previous question? Explain. >> det(N) ans = 1.5000e-05 I think even if the determinant of N^100 is technically non-zero, for all practical purposes we may consider the matrix as non-invertible if its determinant is very close to zero. This is because trying to invert such a matrix can lead to numerical instability and large errors due to the limitations of floating-point precision on computers.

Exercise 4.3 Enter the following matrix in MATLAB: >> V = [-8 6 -6 -30; 3 9 12 -10; 3 -6 -1 18; 3 0 4 7] a. Now use MATLAB to find the eigenvectors and corresponding eigenvalues of V. Assign them to matrices P and D, respectively. >> V = [-8 6 -6 -30; 3 9 12 -10; 3 -6 -1 18; 3 0 4 7] V = -8 6 -6 -30 3 9 12 -10 3 -6 -1 18 3 0 4 7 >> [P,D] = eig(V) P = -0.8874 -0.8660 -0.8706 0.7607 0.2662 0.2887 -0.0725 -0.5567 0.2662 0.2887 0.4353 -0.0874 0.2662 0.2887 0.2176 -0.3221 D = 1.0000 0 0 0 0 2.0000 0 0 0 0 3.0000 0 0 0 0 1.0000 b. Determine whether V is invertible by looking at the eigenvalues. Explain your reasoning.

From output, you can see that the eigenvalues of matrix V are the diagonal elements of matrix D: 1, 2, 3, 1. None of these eigenvalues is zero, which means the matrix V is invertible. The determinant of a matrix (which is the product of its eigenvalues) is zero if and only if the matrix is non-invertible. Since none of the eigenvalues is zero, the determinant is also not zero, hence V is invertible. c. Use MATLAB to evaluate P-1VP. What do you notice? >> (P^-1)*V*P ans = 1.0000 0.0000 0.0000 -0.0000 -0.0000 2.0000 -0.0000 0.0000 0.0000 0.0000 3.0000 0.0000 0.0000 0.0000 -0.0000 1.0000 I notice that the answer is equal to D but with some negative 0s.

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Exercise 4.4 a. Enter this matrix in MATLAB: >> F = [0 1; 1 1] Use MATLAB to find an invertible matrix P and a diagonal matrix D such that PDP-1 = F. >> F = [0 1; 1 1] F = 0 1 1 1 >> [P D]=eig(F); >> [P,D]=eig(F) P = -0.8507 0.5257 0.5257 0.8507 D = -0.6180 0 0 1.6180 >> P*D*P^-1 ans = -0.0000 1.0000 1.0000 1.0000

b. Use MATLAB to compare F10 and PD10P-1. >> F^10 ans = 34 55 55 89 >> P*D^10*P^-1 ans = 34.0000 55.0000 55.0000 89.0000 c. Let f = (1, 1)T. Compute Ff, F2f, F3f, F4f, and F5f. (You don't need to include the input and output for these.) Describe the pattern in your answers. (F^n)*f gives the n+1 and n+2 term of Fibonacci series. d. Consider the Fibonacci sequence {fn} = {1, 1, 2, 3, 5, 8, 13…}, where each term is formed by taking the sum of the previous two. If we start with a vector f = (f0, f1)T, then Ff = (f1, f0 + f1)T = (f1, f2)T, and in general Fnf = (fn, fn+1)T. Here, we're setting both f0 and f1 equal to 1. Given this, compute f30. >> F^30*f ans = 1346269 2178309

Exercise 4.5 a. Use MATLAB to find a matrix Q and a diagonal matrix D such that P = QDQ-1. >> P = [0.8100 0.0800 0.1600 0.1000; 0.0900 0.8400 0.0500 0.0800; 0.0600 0.0400 0.7400 0.0400; 0.0400 0.0400 0.0500 0.7800] P = 0.8100 0.0800 0.1600 0.1000 0.0900 0.8400 0.0500 0.0800 0.0600 0.0400 0.7400 0.0400 0.0400 0.0400 0.0500 0.7800 >> x0 = [0.5106; 0.4720; 0.0075; 0.0099] x0 = 0.5106 0.4720 0.0075 0.0099 >> [Q,D] = eig(P) Q = 0.6656 0.7676 0.5432 -0.4641

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0.6165 -0.2841 -0.8148 -0.1254 0.2946 -0.5682 0.1811 -0.2508 0.3001 0.0848 0.0905 0.8402 D = 1.0000 0 0 0 0 0.6730 0 0 0 0 0.7600 0 0 0 0 0.7370 b. Now recall that Pn = QDnQ-1. Find the limit as n tends to infinity of Dn by hand; we'll call the resulting matrix L. L = D^10000 L = 1.0000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 c. Using MATLAB, multiply L by Q and Q-1 (on the appropriate sides) to compute P∞, the limit as n tends to infinity of Pn. Store the answer in a variable called Pinf. >> Pinf = Q*L*(Q^-1) Pinf = 0.3547 0.3547 0.3547 0.3547 0.3285 0.3285 0.3285 0.3285 0.1570 0.1570 0.1570 0.1570 0.1599 0.1599 0.1599 0.1599 d. Now use MATLAB to compute P∞x0. How does your answer compare to the results you saw in the second part of the exercise from last lab? >> Pinf*x0

ans = 0.3547 0.3285 0.1570 0.1599 This answer is the same as the results from the last lab. e. Make up any vector y in R4 whose entries add up to 1. Compute P∞y, and compare your result to P∞x0. How does the initial distribution vector y of the electorate seem to affect the distribution in the long term? By looking at the matrix P∞, give a mathematical explanation. y = 0.1000 0.1500 0.2500 0.5000 >> Pinf*y ans = 0.3547 0.3285 0.1570 0.1599 Since P∞x0 and P∞y produces the same result, it means that regardless of the initial distribution of the electorate, the long-term distribution will converge to the same values given by the rows of P∞. This suggests that the initial preferences or proportions of the electorate do not play a significant role in determining the long-term distribution of the system. Instead, it is the transition probabilities represented by the matrix P that drive the dynamics of the system and determine its steady-state behavior.

Exercise 4.6 a. Let e0 = (1, 1, 1, 1, 1, 1, 1, 1)T, and define en+1 = Len (which is the same as saying en = Lne0). Use MATLAB to compute e10. How large must n be so that each entry of en changes by less than 1% when we increase n by 1? [Don't try to get an exact value, just try to get a value for n that's big enough.] >> L = [0,0,0,0,1,0,0,0; 0,0,0,0,0,0,0,1; 0,1/2,0,0,0,0,1,0; 1/2,0,1/2,0,0,0,0,0; 0,0,1/2,0,0,1,0,0; 1/2,0,0,0,0,0,0,0; 0,1/2,0,0,0,0,0,0; 0,0,0,1,0,0,0,0;] L = 0 0 0 0 1.0000 0 0 0 0 0 0 0 0 0 0 1.0000 0 0.5000 0 0 0 0 1.0000 0 0.5000 0 0.5000 0 0 0 0 0 0 0 0.5000 0 0 1.0000 0 0 0.5000 0 0 0 0 0 0 0 0 0.5000 0 0 0 0 0 0 0 0 0 1.0000 0 0 0 0

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>> e0 = [1 1 1 1 1 1 1 1]' e0 = 1 1 1 1 1 1 1 1 >> (L^10)*e0 ans = 1.0625 1.1250 1.1875 1.1250 1.1719 0.5781 0.5938 1.1562 To find a value for n that's big enough, we can continue raising L to higher powers, such as L^100, L^1000, etc., until we see that the changes in the entries of the resulting vector become negligible. b. In the graph of the network of webpages represented by L, which vertices have an edge going out and pointing toward website C? Which vertices do the edges coming out of C point to? (Here, by graph we mean a collection of vertices and edges.)

Vertices B and G have edges pointing towards C . C has edges pointing to vertices A, B, and G.

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