Preview question: Cadena de Markov estable de dimensión 3. POTENCIA DE LA MATRIZ, VECTOR DE TRANSICIÓN Y VECTOR DE ESTABILIDAD

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 = [[0.84,0.054,0.11],[0.82,0.099,0.08],[0.97,0.026,0.0037]] y el vector de condiciones iniciales V = [0.28,0.38,0.33] , calcula: a) La potencia 5 de la matriz M es [M to the power of # n end exponent]={[[0.85,0.053,0.098],[0.85,0.053,0.098],[0.85,0.053,0.098]]; [[0.85,0.053,0.095],[0.85,0.053,0.095],[0.85,0.053,0.095]]; [[0.86,0.0041,0.14],[0.86,0.0041,0.14],[0.86,0.0041,0.14]]} b) El vector que representa las condiciones para ese estado 5 es V = {[0.29,0.29,0.29]; [0.29,0.29,0.29]; [0.29,0.29,0.29]} c) El vector de estabilidad es W = {{x=0.86,y=0.0041,z=0.14}; {x=0.85,y=0.053,z=0.098}; {x=0.85,y=0.053,z=0.095}}

Right answer summary: part 1: [[0.85,0.053,0.098],[0.85,0.053,0.098],[0.85,0.053,0.098]]; part 2: [0.29,0.29,0.29]; part 3: {x=0.85,y=0.053,z=0.098}

Question state: todo