Behaviour being used: Adaptive mode
Minimum fraction: 0
Question summary: CADENA DE MARKOV ESTABLE DE DIMENSIÓN 3. EJERCICIO DE APLICACIÓN Dada la matriz de transición M = [[0.72,0.12,0.16],[0.79,0.067,0.15],[0.61,0.18,0.21]] y el vector de condiciones iniciales V = [0.42,0.78,-0.2] , calcula: a) La potencia 3 de la matriz M es [M to the power of # n end exponent]={[[0.71,0.12,0.17],[0.71,0.12,0.17],[0.71,0.12,0.17]]; [[0.76,0.12,0.13],[0.76,0.12,0.12],[0.79,0.13,0.086]]; [[0.69,0.05,0.26],[0.69,0.049,0.26],[0.7,0.047,0.26]]} b) El vector que representa las condiciones para ese estado 3 es V = {[0.36,0.36,0.36]; [0.39,0.39,0.41]; [0.28,0.28,0.28]} c) El vector de estabilidad es W = {{x=0.71,y=0.12,z=0.17}; {x=0.69,y=0.049,z=0.26}; {x=0.76,y=0.12,z=0.12}}
Right answer summary: part 1: [[0.71,0.12,0.17],[0.71,0.12,0.17],[0.71,0.12,0.17]]; part 2: [0.36,0.36,0.36]; part 3: {x=0.71,y=0.12,z=0.17}
Question state: todo