Practice Test: Test III, Actual test is Wednesday, March 30, 2005 Monday, March 28, 2005 Math 307 Spring, 2005 Hentzel 1. For the Markov Chain, find the steady state. ______ .3 ______ | | ------------> | | | A | <---.2------ | B | |_____| __ __ |____| \ /\ /| / .4 .3 .1 .5 \ \ / / \ \ / / _\| \ / |/_ __________ | C | |________| If at the start (A,B,C) = {100,100,100} what happens after 3 instances of time? A B C A .3 .2 .3 B .3 .3 .1 C .4 .5 .6 | .3-1 .2 .3 | A - I = | .3 .3-1 .1 | | .4 .5 .6-1| | -.7 .2 .3 | | -7 2 3 | A - I = | .3 -.7 .1 | = 1/10 | 3 -7 1 | | .4 .5 -.4 | | 4 5 -4 | | -7 2 3 | | 1 12 -5 | | 1 12 -5 | A - I == | 3 -7 1 | == | 3 -7 1 | == | 0 -43 16 | | 1 12 -5 | | -7 2 3 | | 0 86 -32 | | 1 12 -5 | | 1 0 -23/43 | == | 0 1 -16/43 | == | 0 1 -16/43 | | 0 0 0 | | 0 0 0 | | 23 | The steady state is | 16 | | 43 | Check: | .3 .2 .3 | | 23 | | 6.9 + 3.2 +12.9 | | 23 | | .3 .3 .1 | | 16 | = | 6.9 + 4.8 + 4.3 | = | 16 | | .4 .5 .6 | | 43 | | 9.2 + 8.0 +25.8 | | 43 | inf |0.280488, 0.280488, 0.280488| Notice that all the columns A = |0.195122, 0.195122, 0.195122| are the same. This is the |0.52439, 0.52439, 0.52439 | This is the fraction of the total material that ends in each column. inf | 23. 23. 23. | (23+16+43) A = | 16. 16. 16. | | 43. 43. 43. | v0 v1 v2 v3 | .3 .2 .3 | | 100 | | 80 | |24+14+45| | 83| |24.9+12.0+47.1| | 84.0 | | .3 .3 .1 | | 100 |; | 70 |; |24+21+15|=| 60|;|24.9+18.0+15.7|= | 58.6 | | .4 .5 .6 | | 100 | |150 | |32+35+90| |157| |33.2+30.0+94.2| | 157.4 | 100 | | 84.0 | A.A.A.| 100 | = | 58.6 | | 100 | | 157.4 | ------------------------------------------------- A = {{.3,.2,.3},{.3,.3,.1},{.4,.5,.6}}; H = Table[0,{i,1,10}]; H[[1]] = {100,100,100}; Do[H[[i+1]] = A.H[[i]],{i,1,9}]; K = Graphics3D[ Line[H] ]; Show[K]; ------------------------------------------------- 2. Find the change of basis matrix P for the bases BC | 1 2 0 | |1 0 1 | B = | 0 1 1 | C = |0 1 0 | | 0 1 0 | |1 1 0 | V = B Vb = CVc -1 Vb = B C Vc Pbc -1 -1 | 1 2 0 | | 1 0 1 | B C = | 0 1 1 | | 0 1 0 | | 0 1 0 | | 1 1 0 | -1 | 1 0 -2 | | 1 0 1 | B C = | 0 0 1 | | 0 1 0 | | 0 1 -1 | | 1 1 0 | -1 |-1 -2 1 | Pbc B C = | 1 1 0 | |-1 0 0 | | x | Check. Vc = | y | | z | | -1 -2 1 | | x | | -x-2y+z | Vb = | 1 1 0 | | y | = | x+y | | -1 0 0 | | z | | -x | Pbc | 1 0 1 | | x | | x+z | V = C Vc = | 0 1 0 | | y | = | y | | 1 1 0 | | z | | x+y | | 1 2 0 | | -x-2y+z | | x + z | V = B Vb = | 0 1 1 | | x+y | = | y | | 0 1 0 | | -x | | x+y | It checks. --------------------------------------------------------- | 3 2 0 | 3. Suppose T = | 0 2 2 | BB | 1 1 1 | | 1 0 1 | where the basis B is | 0 1 1 | | 1 0 0 |. | 1 0 2 | Find the matrix T for the basis C is | 0 0 1 |. CC | 1 1 0 | Pcb Tbb Pbc Vc = Wc -1 -1 | 1 0 2 | | 1 0 1 | | 3 2 0 | | 1 0 1 | | 1 0 2 | | 0 0 1 | | 0 1 1 | | 0 2 2 | | 0 1 1 | | 0 0 1 | | 1 1 0 | | 1 0 0 | | 1 1 1 | | 1 0 0 | | 1 1 0 | -1 -1 C B Tbb B C | 1 -2 0| | 4 3 1 | | 0 0 1 | | 1 0 2 | |-1 2 1| | 1 3 3 | |-1 1 1 | | 0 0 1 | | 0 1 0| | 3 2 0 | | 1 0 -1 | | 1 1 0 | | 2 -3 -5 | | 1 1 0 | | 1 5 5 | | 0 1 -1 | | 1 3 3 | | 0 -1 2 | | 2 4 -7 | Tcc = | 1 1 5 | | 1 1 3 | ------------------------------- CC = {{1,0,2},{0,0,1},{1,1,0}}; B = {{1,0,1},{0,1,1},{1,0,0}}; Tbb = {{3,2,0},{0,2,2},{1,1,1}}; ans = Inverse[CC].B.Tbb.Inverse[B].CC; ------------------------------------------ 4. Solve the following differential equation. / |x| | 15 12 0 -12 | | x | |y| = | 4 0 -2 0 | | y | |z| | 24 25 3 -24 | | z | |w| | 20 13 -2 -13 | | w | A | 1 0 0 3 | | 0 1 2 0 | P = | 2 -1 -1 6 | | 1 1 2 4 | | 4 3 0 -3 | -1 | 4 -1 -2 0 | P = |-2 1 1 0 | |-1 -1 0 1 | -1 | 4 3 0 -3 | | 15 12 0 -12 | | 1 0 0 3 | P A P = | 4 -1 -2 0 | | 4 0 -2 0 | | 0 1 2 0 | |-2 1 1 0 | | 24 25 3 -24 | | 2 -1 -1 6 | |-1 -1 0 1 | | 20 13 -2 -13 | | 1 1 2 4 | -1 | 4 3 0 -3 | | 3 0 0 -3 | P A P = | 4 -1 -2 0 | | 0 2 2 0 | |-2 1 1 0 | | 6 -2 -1 -6 | |-1 -1 0 1 | | 3 2 2 -4 | -1 | 3 0 0 0 | P A P = | 0 2 0 0 | | 0 0 1 0 | | 0 0 0 -1 | P = { { 1, 0, 0, 3 }, { 0, 1, 2, 0 }, { 2,-1,-1, 6 }, { 1, 1, 2, 4 }}; A = { { 15, 12, 0, -12}, {4, 0, -2, 0}, {24,25,3, -24}, { 20, 13, -2, -13 }}; | x | | 1 0 0 3 | | c1 e^(3t) | | y | = | 0 1 2 0 | | c2 e^(2t) | | z | | 2 -1 -1 6 | | c3 e^t | | w | | 1 1 2 4 | | c4 e^(-t) | Check / | x | | 1 0 0 3 | | 3 c1 e^(3t) | | 3 0 0 -3 | | c1 e^(3t) | | y | = | 0 1 2 0 | | 2 c2 e^(2t) | = | 0 2 2 0 | | c2 e^(2t) | | z | | 2 -1 -1 6 | | c3 e^t | | 6 -2 -1 -6 | | c3 e^t | | w | | 1 1 2 4 | |-1 c4 e^(-t) | | 3 2 2 -4 | | c4 e^(-t | |x| | 15 12 0 -12 | | 1 0 0 3 | | c1 e^(3t) | A |y| = | 4 0 -2 0 | | 0 1 2 0 | | c2 e^(2t) | |z| | 24 25 3 -24 | | 2 -1 -1 6 | | c3 e^t | |w| | 20 13 -2 -13 | | 1 1 2 4 | | c4 e^(-t) | |x| | 3 0 0 -3 | | c1 e^(3t) | A |y| = | 0 2 2 0 | | c2 e^(2t) | |z| | 6 -2 -1 -6 | | c3 e^t | |w| | 3 2 2 -4 | | c4 e^(-t) | ----------------------------------------------------------------------- 5. Tidbits: (a) Fill in the bottom row to make this matrix stochastic. | .1 .2 .3 | | .4 .5 .7 | | .5 .3 0 | (b) If B and C are basis and T is a linear transformation, BB Express T in terms of B, C and T . CC BB -1 -1 Tcc = C B Tbb B C (c) What is the dimension of / | 1 | | 2 | | 3 | | 4 | \ / | 5 |, | 6 |, | 7 |, | 8 | \ \ | 9 | |10 | |11 | |12 | / \ |13 | |14 | |15 | |16 | / These columns are spanned by the first column and the all 1's vector. The rank is 2 (d What is the process to find the steady state of a stochastic matrix B? Solve (B-I)X = 0 (e) Is the product of two stochastic matrices also stochastic? Yes (f) Prove that if a set V1, V2, ..., Vn of vectors are dependent and V1 =/= 0, then some vector is a linear combination of the previous. If c1 V2 + c2 V2 + ... cn Vn = 0, let cm be the last non zero coefficient. Then -1/cm (c1 V1 + c2 V2 + ... + cm-1 Vm-1) = Vm (g) When is a matrix upper triangular matrix? When the entries below the diagonal are zero. (h) Give 5 criterion that imply that an nxn matrix has an inverse. 1. Det =/= 0 2. Rows are Linearly independent 3. Columns are Linearly independent 4. RCF = I 5. NullSpace = 0