Question

A chemical species can take on three different forms: A, B and C. After some amount of time passes, the chemical can change between each form in a predictable way. If it is found in form A, it has a 10% chance to convert to B, 80% chance to convert to C and 10% chance to remain as A. From form B, it has a 20% chance to become A, 30% chance to convert to C and a 50% chance to remain as B. Finally, if it is in form C, it converts to A 60% of the time, to B 20% of the time and stays in the form of C 20% of the time. (a - 2 points) If you start with 100 kg of pure form A and 50 kg of pure form B and allow the conversion above to take place, how many kg of form C are you likely to have? [kg of A] (b- 2 points) Create a matrix, M, that can be multiplied by any vector, v = kg of B kg of c] (c-2 points) Find the Eigenvalues of M using Newton's Method to solve the roots of the characteristic equation. (d - 2 points) There is a composition of v such that Mỹ = v. What will be the proportion of forms A, B and C at this composition? (Hint: Use one of the Eigenvectors to answer this)

Answer 1

0 chemical species can take on three different forms A,B,C : By writing given data into mathematical eq"'s. A = 0.1 A + 0.13 +0.80 -0. [found in form a] B = 0.2A + 0.5B+0.30 [Round in form B] C =0-6A +0.2 B +0.20 - [found in form c] a) given 100kg of Pure form A and so kg of Pare form B. = 100, B=50 substitute A, B values in ear 6 & @ 100 = 0.1 (100) + 0.1150)+ 0.80 50 =+0.2 (100) +.0.5(50)+ 0.30 A- + 50 - 10 - 20 +0.5c 0.50 = 50+ 30 80 C = 160 => 0.5 .: C = 160 kg b) From the equations 0,® ¢ we can write in matrix form. А 0.1 0.1 y = 0.5 0.3 0.6 0.2 0.2 0.8 - All

0.8 0.2 0.5 0.3 0.6 0.2 0.2 c) To find the eigen values , (A-11) = 0 ) Here, a= eigen value A = maatrix = M 딧 1M-HI ; charateristic equation 0.1-1 0.1 0.8 0.2 0.5-1 0.3 0.6 0.2 0.2-1 (0.1-1){(0-5-1)(0:2-4) - (0-24 0:)] – 0.1[02(02-4) -10.3 x 016)] + 0.8[(0.2) (0-2) – (0.5-4) (0-6)] by solving the above equation we get = -0.547, 0.347,1 d). composition of ✓ such that mū= ý. Find the Properties of ABC at this point let 1=1 (first guess) 0.1-1 0.8 А 0.2 0.5-1 0.3 0.6 0.2 0.2-1

VE -0.9A +0:1 B +0.86-0 0.2A -0.5B + 0.30 = 0 let A=1 substitute in above eana OIB + 0.8C = 0.9* -0.5B + 0.30 = -0.2x! by solving above 2 equations we get, which satisfies the above equations B=L, C=1 Eigen vector = (1,1,1) :: The Properties of A,B,C is lilil at MV = V