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+++ NoteBooks/Solucion_EV01.ipynb
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "id": "aa8b1746",
+ "metadata": {},
+ "source": [
+ "# Resolución de Evaluación EV01 - Álgebra Lineal\n",
+ "\n",
+ "Este notebook contiene la resolución detallada de los problemas presentados en el archivo EV01.pdf. Cada problema incluye una explicación teórica, el desarrollo paso a paso y, en algunos casos, una verificación mediante código Python."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "9db1247d",
+ "metadata": {},
+ "source": [
+ "## Problema 1: Sistema de Ecuaciones con Parámetro\n",
+ "\n",
+ "Considere el sistema:\n",
+ "$$\\begin{cases} x + y + z = 2 \\\\ x + 2y + kz = 3 \\\\ 2x + 3y + 3z = k + 3 \\end{cases}$$\n",
+ "\n",
+ "### Resolución\n",
+ "\n",
+ "Escribimos la matriz aumentada y aplicamos eliminación de Gauss:\n",
+ "$$\\begin{pmatrix} 1 & 1 & 1 & | & 2 \\\\ 1 & 2 & k & | & 3 \\\\ 2 & 3 & 3 & | & k+3 \\end{pmatrix}$$\n",
+ "\n",
+ "1. $R_2 \\to R_2 - R_1$ y $R_3 \\to R_3 - 2R_1$:\n",
+ "$$\\begin{pmatrix} 1 & 1 & 1 & | & 2 \\\\ 0 & 1 & k-1 & | & 1 \\\\ 0 & 1 & 1 & | & k-1 \\end{pmatrix}$$\n",
+ "\n",
+ "2. $R_3 \\to R_3 - R_2$:\n",
+ "$$\\begin{pmatrix} 1 & 1 & 1 & | & 2 \\\\ 0 & 1 & k-1 & | & 1 \\\\ 0 & 0 & 2-k & | & k-2 \\end{pmatrix}$$\n",
+ "\n",
+ "**Análisis:**\n",
+ "- **(a) Solución única:** Ocurre si el rango de la matriz es 3, es decir, $2-k \\neq 0 \\implies k \\neq 2$.\n",
+ "- **(b) Infinitas soluciones:** Si $k=2$, la última fila es $(0, 0, 0 | 0)$. El sistema es consistente con una variable libre.\n",
+ "- **(c) Ninguna solución:** No hay valores de $k$ que produzcan una contradicción del tipo $(0, 0, 0 | c)$ con $c \\neq 0$, ya que si $2-k=0$, entonces $k-2=0$.\n",
+ "\n",
+ "**Caso (b) k=2:**\n",
+ "El sistema queda:\n",
+ "$y + z = 1 \\implies y = 1 - z$\n",
+ "$x + y + z = 2 \\implies x + (1-z) + z = 2 \\implies x = 1$\n",
+ "Solución: $(1, 1-z, z) = (1, 1, 0) + z(0, -1, 1)$.\n",
+ "El conjunto solución es el **span{(0, -1, 1)}** desplazado por el punto $(1, 1, 0)$."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "id": "a41cdc7a",
+ "metadata": {
+ "ExecuteTime": {
+ "end_time": "2026-02-02T17:27:37.439236230Z",
+ "start_time": "2026-02-02T17:27:37.390643257Z"
+ }
+ },
+ "source": [
+ "import sympy as sp\n",
+ "k = sp.symbols('k')\n",
+ "A = sp.Matrix([[1, 1, 1, 2], [1, 2, k, 3], [2, 3, 3, k+3]])\n",
+ "# Verificamos para k=2\n",
+ "A_k2 = A.subs(k, 2)\n",
+ "print(\"Matriz para k=2:\")\n",
+ "sp.pprint(A_k2)\n",
+ "print(\"\\nSolución para k=2:\")\n",
+ "sp.pprint(sp.solve_linear_system(A_k2, *sp.symbols('x y z')))\n",
+ "\n",
+ "# Verificamos determinante de la matriz de coeficientes\n",
+ "M = A[:, :3]\n",
+ "print(f\"\\nDeterminante de M: {M.det()}\")"
+ ],
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Matriz para k=2:\n",
+ "⎡1 1 1 2⎤\n",
+ "⎢ ⎥\n",
+ "⎢1 2 2 3⎥\n",
+ "⎢ ⎥\n",
+ "⎣2 3 3 5⎦\n",
+ "\n",
+ "Solución para k=2:\n",
+ "{x: 1, y: 1 - z}\n",
+ "\n",
+ "Determinante de M: 2 - k\n"
+ ]
+ }
+ ],
+ "execution_count": 4
+ },
+ {
+ "cell_type": "markdown",
+ "id": "e9d333a9",
+ "metadata": {},
+ "source": [
+ "## Problema 2: Condiciones de Pertenencia al Span\n",
+ "\n",
+ "Sean $v_1 = (1, 0, -1, 2)$ y $v_2 = (2, 3, 1, 1)$. Buscamos condiciones para $b = (b_1, b_2, b_3, b_4)$ tal que $b \\in \\text{span}\\{v_1, v_2\\}$.\n",
+ "\n",
+ "### Resolución\n",
+ "\n",
+ "El vector $b$ está en el span si existen escalares $c_1, c_2$ tales que $c_1 v_1 + c_2 v_2 = b$. Esto equivale a que el sistema sea consistente:\n",
+ "$$\\begin{pmatrix} 1 & 2 & | & b_1 \\\\ 0 & 3 & | & b_2 \\\\ -1 & 1 & | & b_3 \\\\ 2 & 1 & | & b_4 \\end{pmatrix}$$\n",
+ "\n",
+ "Reducimos la matriz:\n",
+ "1. $R_3 \\to R_3 + R_1$:\n",
+ "$$\\begin{pmatrix} 1 & 2 & | & b_1 \\\\ 0 & 3 & | & b_2 \\\\ 0 & 3 & | & b_3 + b_1 \\\\ 2 & 1 & | & b_4 \\end{pmatrix}$$\n",
+ "\n",
+ "2. $R_4 \\to R_4 - 2R_1$:\n",
+ "$$\\begin{pmatrix} 1 & 2 & | & b_1 \\\\ 0 & 3 & | & b_2 \\\\ 0 & 3 & | & b_3 + b_1 \\\\ 0 & -3 & | & b_4 - 2b_1 \\end{pmatrix}$$\n",
+ "\n",
+ "3. $R_3 \\to R_3 - R_2$ y $R_4 \\to R_4 + R_2$:\n",
+ "$$\\begin{pmatrix} 1 & 2 & | & b_1 \\\\ 0 & 3 & | & b_2 \\\\ 0 & 0 & | & b_3 + b_1 - b_2 \\\\ 0 & 0 & | & b_4 - 2b_1 + b_2 \\end{pmatrix}$$\n",
+ "\n",
+ "Para consistencia, las últimas dos entradas deben ser cero:\n",
+ "1. $b_1 - b_2 + b_3 = 0$\n",
+ "2. $-2b_1 + b_2 + b_4 = 0$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "47449e8a",
+ "metadata": {},
+ "source": [
+ "## Problema 3: Potencias de una Matriz\n",
+ "\n",
+ "Sea $A = \\begin{pmatrix} 1 & 1 & 0 \\\\ 0 & 1 & 1 \\\\ 0 & 0 & 1 \\end{pmatrix}$.\n",
+ "\n",
+ "### Resolución\n",
+ "\n",
+ "Podemos escribir $A = I + N$, donde $N = \\begin{pmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}$.\n",
+ "Notamos que:\n",
+ "$N^2 = \\begin{pmatrix} 0 & 0 & 1 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$, $N^3 = 0$.\n",
+ "\n",
+ "Usando el binomio de Newton (ya que $I$ y $N$ conmutan):\n",
+ "$A^n = (I + N)^n = \\sum_{k=0}^{n} \\binom{n}{k} I^{n-k} N^k = \\binom{n}{0}I + \\binom{n}{1}N + \\binom{n}{2}N^2$\n",
+ "$A^n = I + nN + \\frac{n(n-1)}{2}N^2$\n",
+ "\n",
+ "$$A^n = \\begin{pmatrix} 1 & n & \\frac{n(n-1)}{2} \\\\ 0 & 1 & n \\\\ 0 & 0 & 1 \\end{pmatrix}$$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "id": "c00bdb48",
+ "metadata": {
+ "ExecuteTime": {
+ "end_time": "2026-02-02T17:27:37.546444097Z",
+ "start_time": "2026-02-02T17:27:37.468342936Z"
+ }
+ },
+ "source": [
+ "A = sp.Matrix([[1, 1, 0], [0, 1, 1], [0, 0, 1]])\n",
+ "print(\"A^2:\")\n",
+ "sp.pprint(A**2)\n",
+ "print(\"\\nA^3:\")\n",
+ "sp.pprint(A**3)\n",
+ "print(\"\\nA^4:\")\n",
+ "sp.pprint(A**4)"
+ ],
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A^2:\n",
+ "⎡1 2 1⎤\n",
+ "⎢ ⎥\n",
+ "⎢0 1 2⎥\n",
+ "⎢ ⎥\n",
+ "⎣0 0 1⎦\n",
+ "\n",
+ "A^3:\n",
+ "⎡1 3 3⎤\n",
+ "⎢ ⎥\n",
+ "⎢0 1 3⎥\n",
+ "⎢ ⎥\n",
+ "⎣0 0 1⎦\n",
+ "\n",
+ "A^4:\n",
+ "⎡1 4 6⎤\n",
+ "⎢ ⎥\n",
+ "⎢0 1 4⎥\n",
+ "⎢ ⎥\n",
+ "⎣0 0 1⎦\n"
+ ]
+ }
+ ],
+ "execution_count": 5
+ },
+ {
+ "cell_type": "markdown",
+ "id": "a4e8703c",
+ "metadata": {},
+ "source": [
+ "## Problema 4: Matrices de Pauli y Conmutadores\n",
+ "\n",
+ "Demostrar $[\\sigma_x, \\sigma_y] = 2i\\sigma_z$.\n",
+ "\n",
+ "### Resolución\n",
+ "\n",
+ "$\\sigma_x = \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}, \\sigma_y = \\begin{pmatrix} 0 & -i \\\\ i & 0 \\end{pmatrix}, \\sigma_z = \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix}$.\n",
+ "\n",
+ "$\\sigma_x \\sigma_y = \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix} \\begin{pmatrix} 0 & -i \\\\ i & 0 \\end{pmatrix} = \\begin{pmatrix} i & 0 \\\\ 0 & -i \\end{pmatrix} = i\\sigma_z$\n",
+ "\n",
+ "$\\sigma_y \\sigma_x = \\begin{pmatrix} 0 & -i \\\\ i & 0 \\end{pmatrix} \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix} = \\begin{pmatrix} -i & 0 \\\\ 0 & i \\end{pmatrix} = -i\\sigma_z$\n",
+ "\n",
+ "$[\\sigma_x, \\sigma_y] = \\sigma_x \\sigma_y - \\sigma_y \\sigma_x = i\\sigma_z - (-i\\sigma_z) = 2i\\sigma_z$.\n",
+ "\n",
+ "**Implicación física:** En mecánica cuántica, si dos observables no conmutan ($[A, B] \\neq 0$), no pueden ser medidos simultáneamente con precisión infinita (Principio de Incertidumbre)."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "c8ad4688",
+ "metadata": {},
+ "source": [
+ "## Problema 5: Matriz Nilpotente\n",
+ "\n",
+ "Sea $A = \\begin{pmatrix} 0 & a & b \\\\ 0 & 0 & c \\\\ 0 & 0 & 0 \\end{pmatrix}$.\n",
+ "\n",
+ "### Resolución\n",
+ "\n",
+ "Calculamos las potencias:\n",
+ "$A^2 = \\begin{pmatrix} 0 & 0 & ac \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$\n",
+ "$A^3 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix} = 0$\n",
+ "\n",
+ "Si $a, c \\neq 0$, entonces $ac \\neq 0$, por lo que $A^2 \\neq 0$. El índice de nilpotencia es **3**."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "7510dcbe",
+ "metadata": {},
+ "source": [
+ "## Problema 6: Matriz de Adyacencia\n",
+ "\n",
+ "$M = \\begin{pmatrix} 0 & 1 & 1 \\\\ 1 & 0 & 0 \\\\ 1 & 0 & 0 \\end{pmatrix}$.\n",
+ "\n",
+ "### Resolución\n",
+ "\n",
+ "$M^2 = \\begin{pmatrix} 0 & 1 & 1 \\\\ 1 & 0 & 0 \\\\ 1 & 0 & 0 \\end{pmatrix} \\begin{pmatrix} 0 & 1 & 1 \\\\ 1 & 0 & 0 \\\\ 1 & 0 & 0 \\end{pmatrix} = \\begin{pmatrix} 2 & 0 & 0 \\\\ 0 & 1 & 1 \\\\ 0 & 1 & 1 \\end{pmatrix}$.\n",
+ "\n",
+ "**Interpretación:** La entrada $(M^2)_{ij}$ representa el número de caminos de longitud 2 entre el nodo $i$ y el nodo $j$. Por ejemplo, $(M^2)_{11} = 2$ indica que hay dos caminos de ida y vuelta desde el nodo 1 (1-2-1 y 1-3-1)."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "fef32579",
+ "metadata": {},
+ "source": [
+ "## Problema 7: Interpolación Polinómica\n",
+ "\n",
+ "$p(x) = ax^2 + bx + c$ pasa por $(-1, 9), (1, 3), (2, 6)$.\n",
+ "\n",
+ "### Resolución\n",
+ "\n",
+ "Sustituimos los puntos:\n",
+ "1. $a(-1)^2 + b(-1) + c = 9 \\implies a - b + c = 9$\n",
+ "2. $a(1)^2 + b(1) + c = 3 \\implies a + b + c = 3$\n",
+ "3. $a(2)^2 + b(2) + c = 6 \\implies 4a + 2b + c = 6$\n",
+ "\n",
+ "Sistema matricial:\n",
+ "$$\\begin{pmatrix} 1 & -1 & 1 \\\\ 1 & 1 & 1 \\\\ 4 & 2 & 1 \\end{pmatrix} \\begin{pmatrix} a \\\\ b \\\\ c \\end{pmatrix} = \\begin{pmatrix} 9 \\\\ 3 \\\\ 6 \\end{pmatrix}$$\n",
+ "\n",
+ "Resolviendo:\n",
+ "$R_2 - R_1 \\implies 2b = -6 \\implies b = -3$\n",
+ "$R_1 \\implies a + 3 + c = 9 \\implies a + c = 6$\n",
+ "$R_3 \\implies 4a + 2(-3) + c = 6 \\implies 4a + c = 12$\n",
+ "$(4a + c) - (a + c) = 12 - 6 \\implies 3a = 6 \\implies a = 2$\n",
+ "$c = 6 - 2 = 4$\n",
+ "\n",
+ "El polinomio es **$p(x) = 2x^2 - 3x + 4$**."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "id": "892e5c9e",
+ "metadata": {
+ "ExecuteTime": {
+ "end_time": "2026-02-02T17:27:37.604801408Z",
+ "start_time": "2026-02-02T17:27:37.560845494Z"
+ }
+ },
+ "source": [
+ "A_poly = sp.Matrix([[1, -1, 1], [1, 1, 1], [4, 2, 1]])\n",
+ "b_poly = sp.Matrix([9, 3, 6])\n",
+ "sol_poly = A_poly.solve(b_poly)\n",
+ "print(\"Solución (a, b, c):\")\n",
+ "sp.pprint(sol_poly)"
+ ],
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solución (a, b, c):\n",
+ "⎡2 ⎤\n",
+ "⎢ ⎥\n",
+ "⎢-3⎥\n",
+ "⎢ ⎥\n",
+ "⎣4 ⎦\n"
+ ]
+ }
+ ],
+ "execution_count": 6
+ },
+ {
+ "cell_type": "markdown",
+ "id": "e0c4414b",
+ "metadata": {},
+ "source": [
+ "## Problema 8: Matriz por Bloques\n",
+ "\n",
+ "Demostrar que si $M = \\begin{pmatrix} A & 0 \\\\ 0 & B \\end{pmatrix}$, entonces $M^k = \\begin{pmatrix} A^k & 0 \\\\ 0 & B^k \\end{pmatrix}$.\n",
+ "\n",
+ "### Resolución\n",
+ "\n",
+ "Por inducción:\n",
+ "- Base $k=1$: Es trivial.\n",
+ "- Hipótesis: $M^k = \\begin{pmatrix} A^k & 0 \\\\ 0 & B^k \\end{pmatrix}$.\n",
+ "- Paso $k+1$:\n",
+ "$M^{k+1} = M^k M = \\begin{pmatrix} A^k & 0 \\\\ 0 & B^k \\end{pmatrix} \\begin{pmatrix} A & 0 \\\\ 0 & B \\end{pmatrix} = \\begin{pmatrix} A^k A + 0 & 0 + 0 \\\\ 0 + 0 & 0 + B^k B \\end{pmatrix} = \\begin{pmatrix} A^{k+1} & 0 \\\\ 0 & B^{k+1} \\end{pmatrix}$.\n",
+ "Queda demostrado."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "3995d78c",
+ "metadata": {},
+ "source": [
+ "## Problema 9: Propiedad de la Traza\n",
+ "\n",
+ "Demostrar $\\text{tr}(AB) = \\text{tr}(BA)$.\n",
+ "\n",
+ "### Resolución\n",
+ "\n",
+ "Sea $A$ de $n \\times n$ con elementos $a_{ij}$ y $B$ con elementos $b_{ij}$.\n",
+ "La entrada $(i, i)$ de $AB$ es $(AB)_{ii} = \\sum_{j=1}^n a_{ij} b_{ji}$.\n",
+ "La traza es:\n",
+ "$$\\text{tr}(AB) = \\sum_{i=1}^n (AB)_{ii} = \\sum_{i=1}^n \\sum_{j=1}^n a_{ij} b_{ji}$$\n",
+ "\n",
+ "Para $BA$, la entrada $(j, j)$ es $(BA)_{jj} = \\sum_{i=1}^n b_{ji} a_{ij}$.\n",
+ "La traza es:\n",
+ "$$\\text{tr}(BA) = \\sum_{j=1}^n (BA)_{jj} = \\sum_{j=1}^n \\sum_{i=1}^n b_{ji} a_{ij}$$\n",
+ "\n",
+ "Como la suma es finita, podemos intercambiar el orden:\n",
+ "$$\\sum_{i=1}^n \\sum_{j=1}^n a_{ij} b_{ji} = \\sum_{j=1}^n \\sum_{i=1}^n b_{ji} a_{ij}$$\n",
+ "Por lo tanto, $\\text{tr}(AB) = \\text{tr}(BA)$."
+ ]
+ }
+ ],
+ "metadata": {},
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
