commit - 95da677639058458bfaf1b6edc1efebfa81f313e
commit + 0cbaf116a9bd627d5ebb3b9ea68b595835e1bdd7
blob - 556974ac152d456ec9562379f33f6f8f0c3c910e
blob + 08b3fabea73d7fb950603b14fe2c4a5c808d96fb
--- NoteBooks/Ejercicios_en_clase.ipynb
+++ NoteBooks/Ejercicios_en_clase.ipynb
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blob - 1b2c831787f382a94a9fa46c24e6fb401b7e458d
blob + 4c625aec36bb2023d09e3424bc41c59f83119380
--- NoteBooks/Solucion_EV01.ipynb
+++ NoteBooks/Solucion_EV01.ipynb
"cells": [
{
"cell_type": "markdown",
- "id": "aa8b1746",
+ "id": "bdab06ac",
"metadata": {},
"source": [
- "# Resolución de Evaluación EV01 - Álgebra Lineal\n",
+ "# Resolución de Evaluación: Espacios Vectoriales\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."
+ "Este notebook contiene la resolución detallada de los problemas presentados en el archivo `EspaciosVectoriales★_2.pdf`. Cada problema incluye una explicación teórica, el desarrollo paso a paso y verificaciones mediante código Python utilizando `sympy`."
]
},
{
"cell_type": "markdown",
- "id": "9db1247d",
+ "id": "149470e5",
"metadata": {},
"source": [
- "## Problema 1: Sistema de Ecuaciones con Parámetro\n",
+ "## Problema 1: Análisis de un Sistema de Ecuaciones Lineales\n",
"\n",
+ "**Enunciado:**\n",
"Considere el sistema:\n",
"$$\\begin{cases} x + y + z = 2 \\\\ x + 2y + kz = 3 \\\\ 2x + 3y + 3z = k + 3 \\end{cases}$$\n",
+ "Determine para qué valores de $k$ el sistema tiene: (a) solución única, (b) infinitas soluciones, o (c) ninguna solución. Para el caso (b), describa el conjunto solución como el span de un vector.\n",
"\n",
- "### Resolución\n",
+ "### Resolución Paso a Paso\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",
+ "1. **Matriz Aumentada:**\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",
+ "2. **Eliminación de Gauss:**\n",
+ " - $R_2 \\to R_2 - R_1$: \n",
+ " $$\\begin{pmatrix} 1 & 1 & 1 & | & 2 \\\\ 0 & 1 & k-1 & | & 1 \\\\ 2 & 3 & 3 & | & k+3 \\end{pmatrix}$$\n",
+ " - $R_3 \\to R_3 - 2R_1$: \n",
+ " $$\\begin{pmatrix} 1 & 1 & 1 & | & 2 \\\\ 0 & 1 & k-1 & | & 1 \\\\ 0 & 1 & 1 & | & k-1 \\end{pmatrix}$$\n",
+ " - $R_3 \\to R_3 - R_2$: \n",
+ " $$\\begin{pmatrix} 1 & 1 & 1 & | & 2 \\\\ 0 & 1 & k-1 & | & 1 \\\\ 0 & 0 & 2-k & | & k-2 \\end{pmatrix}$$\n",
"\n",
- "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",
+ "3. **Análisis de Casos:**\n",
+ " - **Caso 1: $2-k \\neq 0 \\implies k \\neq 2$.** El sistema tiene **solución única** porque hay un pivote en cada columna de la matriz de coeficientes.\n",
+ " - **Caso 2: $2-k = 0 \\implies k = 2$.** La última fila se convierte en $(0, 0, 0 | 0)$. El sistema es consistente y tiene **infinitas soluciones** (una variable libre).\n",
+ " - **Caso 3: Ninguna solución.** No existe ningún valor de $k$ que haga que el sistema sea inconsistente, ya que si el lado izquierdo es 0 ($k=2$), el lado derecho también es 0 ($2-2=0$).\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 $(1, 1, 0)$."
+ "4. **Conjunto Solución para $k=2$:**\n",
+ " Sustituyendo $k=2$ en la matriz reducida:\n",
+ " $$\\begin{cases} x + y + z = 2 \\\\ y + z = 1 \\end{cases}$$\n",
+ " Sea $z = t$ (variable libre):\n",
+ " $y = 1 - t$\n",
+ " $x + (1-t) + t = 2 \\implies x = 1$\n",
+ " Solución: $(x, y, z) = (1, 1-t, t) = (1, 1, 0) + t(0, -1, 1)$.\n",
+ " El conjunto solución es el **span{(0, -1, 1)}** desplazado por el punto $(1, 1, 0)$."
]
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{
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"metadata": {
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+ "start_time": "2026-02-05T01:46:22.576651885Z"
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},
"source": [
"import sympy as sp\n",
"k = sp.symbols('k')\n",
+ "x, y, z = sp.symbols('x y z')\n",
"A = sp.Matrix([[1, 1, 1, 2], [1, 2, k, 3], [2, 3, 3, k+3]])\n",
- "# Verificamos para k=2\n",
+ "\n",
+ "# Verificación 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()}\")"
+ "sol_k2 = sp.solve_linear_system(A_k2, x, y, z)\n",
+ "print(f\"Solución para k=2: {sol_k2}\")"
],
"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"
+ "Solución para k=2: {x: 1, y: 1 - z}\n"
]
}
],
- "execution_count": 4
+ "execution_count": 1
},
{
"cell_type": "markdown",
- "id": "e9d333a9",
+ "id": "71f355bd",
"metadata": {},
"source": [
- "## Problema 2: Condiciones de Pertenencia al Span\n",
+ "## Problema 2: Condiciones para pertenecer 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",
+ "**Enunciado:**\n",
+ "Dados $\\vec{v}_1 = (1, 0, -1, 2)$ y $\\vec{v}_2 = (2, 3, 1, 1)$, encuentre las condiciones algebraicas para que $\\vec{b} = (b_1, b_2, b_3, b_4) \\in \\text{span}\\{\\vec{v}_1, \\vec{v}_2\\}$.\n",
"\n",
- "### Resolución\n",
+ "### Resolución Paso a Paso\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",
+ "Para que $\\vec{b}$ esté en el span, deben existir escalares $\\alpha, \\beta$ tales que $\\alpha \\vec{v}_1 + \\beta \\vec{v}_2 = \\vec{b}$. Esto genera el sistema:\n",
+ "$$\\begin{cases} \\alpha + 2\\beta = b_1 \\\\ 3\\beta = b_2 \\\\ -\\alpha + \\beta = b_3 \\\\ 2\\alpha + \\beta = b_4 \\end{cases}$$\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",
+ "1. De la segunda ecuación: $\\beta = \\frac{b_2}{3}$.\n",
+ "2. Sustituyendo en la primera: $\\alpha = b_1 - 2(\\frac{b_2}{3}) = b_1 - \\frac{2b_2}{3}$.\n",
+ "3. Sustituyendo $\\alpha$ y $\\beta$ en la tercera ecuación:\n",
+ " $-(b_1 - \\frac{2b_2}{3}) + \\frac{b_2}{3} = b_3 \\implies -b_1 + \\frac{2b_2}{3} + \\frac{b_2}{3} = b_3 \\implies -b_1 + b_2 = b_3 \\implies \\mathbf{b_1 - b_2 + b_3 = 0}$.\n",
+ "4. Sustituyendo en la cuarta ecuación:\n",
+ " $2(b_1 - \\frac{2b_2}{3}) + \\frac{b_2}{3} = b_4 \\implies 2b_1 - \\frac{4b_2}{3} + \\frac{b_2}{3} = b_4 \\implies 2b_1 - b_2 = b_4 \\implies \\mathbf{2b_1 - b_2 - b_4 = 0}$.\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. $-2b_1 + b_2 + b_4 = 0$"
+ "**Conclusión:** Un vector $\\vec{b}$ pertenece al span si sus componentes cumplen:\n",
+ "$$\\begin{cases} b_1 - b_2 + b_3 = 0 \\\\ 2b_1 - b_2 - b_4 = 0 \\end{cases}$$"
]
},
{
"cell_type": "markdown",
- "id": "47449e8a",
+ "id": "d86c2fae",
"metadata": {},
"source": [
- "## Problema 3: Potencias de una Matriz\n",
+ "## Problema 3: Potencias de una Matriz e Inducción\n",
"\n",
- "Sea $A = \\begin{pmatrix} 1 & 1 & 0 \\\\ 0 & 1 & 1 \\\\ 0 & 0 & 1 \\end{pmatrix}$.\n",
+ "**Enunciado:**\n",
+ "Sea $A = \\begin{pmatrix} 1 & 1 & 0 \\\\ 0 & 1 & 1 \\\\ 0 & 0 & 1 \\end{pmatrix}$. Calcule $A^2, A^3, A^4$ y conjeture una fórmula para $A^n$.\n",
"\n",
- "### Resolución\n",
+ "### Resolución Paso a Paso\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",
+ "1. **Cálculo de potencias:**\n",
+ " - $A^2 = \\begin{pmatrix} 1 & 2 & 1 \\\\ 0 & 1 & 2 \\\\ 0 & 0 & 1 \\end{pmatrix}$\n",
+ " - $A^3 = \\begin{pmatrix} 1 & 3 & 3 \\\\ 0 & 1 & 3 \\\\ 0 & 0 & 1 \\end{pmatrix}$\n",
+ " - $A^4 = \\begin{pmatrix} 1 & 4 & 6 \\\\ 0 & 1 & 4 \\\\ 0 & 0 & 1 \\end{pmatrix}$\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",
+ "2. **Patrón observado:**\n",
+ " La diagonal siempre es 1. La segunda diagonal superior es $n$. El elemento $(1,3)$ sigue la secuencia $0, 1, 3, 6, \\dots$, que corresponde a los números triangulares $\\frac{n(n-1)}{2}$ o $\\binom{n}{2}$.\n",
"\n",
- "$$A^n = \\begin{pmatrix} 1 & n & \\frac{n(n-1)}{2} \\\\ 0 & 1 & n \\\\ 0 & 0 & 1 \\end{pmatrix}$$"
+ "3. **Fórmula General:**\n",
+ " $$A^n = \\begin{pmatrix} 1 & n & \\frac{n(n-1)}{2} \\\\ 0 & 1 & n \\\\ 0 & 0 & 1 \\end{pmatrix}$$\n",
+ "\n",
+ "4. **Justificación mediante Descomposición:**\n",
+ " Sea $A = I + N$, donde $N = \\begin{pmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}$.\n",
+ " Notamos que $N^2 = \\begin{pmatrix} 0 & 0 & 1 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$ y $N^3 = 0$.\n",
+ " Por el Binomio de Newton (ya que $I$ y $N$ conmutan):\n",
+ " $A^n = (I+N)^n = I + nN + \\frac{n(n-1)}{2}N^2 + 0 = \\begin{pmatrix} 1 & n & \\frac{n(n-1)}{2} \\\\ 0 & 1 & n \\\\ 0 & 0 & 1 \\end{pmatrix}$."
]
},
{
"cell_type": "code",
- "id": "c00bdb48",
+ "id": "c264e346",
"metadata": {
"ExecuteTime": {
- "end_time": "2026-02-02T17:27:37.546444097Z",
- "start_time": "2026-02-02T17:27:37.468342936Z"
+ "end_time": "2026-02-05T01:46:23.131119841Z",
+ "start_time": "2026-02-05T01:46:23.078064246Z"
}
},
"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",
+ "print(\"A^4 calculada por Python:\")\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",
+ "A^4 calculada por Python:\n",
"⎡1 4 6⎤\n",
"⎢ ⎥\n",
"⎢0 1 4⎥\n",
]
}
],
- "execution_count": 5
+ "execution_count": 2
},
{
"cell_type": "markdown",
- "id": "a4e8703c",
+ "id": "56fb400d",
"metadata": {},
"source": [
- "## Problema 4: Matrices de Pauli y Conmutadores\n",
+ "## Problema 4: Conmutador de Matrices de Pauli\n",
"\n",
- "Demostrar $[\\sigma_x, \\sigma_y] = 2i\\sigma_z$.\n",
+ "**Enunciado:**\n",
+ "Demuestre que $[\\sigma_x, \\sigma_y] = 2i\\sigma_z$.\n",
"\n",
- "### Resolución\n",
+ "### Resolución Paso a Paso\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",
+ "1. **Definiciones:**\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",
+ "2. **Cálculo de productos:**\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",
+ " - $\\sigma_y \\sigma_x = \\begin{pmatrix} 0 & -i \\\\ i & 0 \\end{pmatrix} \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix} = \\begin{pmatrix} -i & 0 \\\\ 0 & i \\end{pmatrix} = -i\\sigma_z$.\n",
"\n",
- "$\\sigma_y \\sigma_x = \\begin{pmatrix} 0 & -i \\\\ i & 0 \\end{pmatrix} \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix} = \\begin{pmatrix} -i & 0 \\\\ 0 & i \\end{pmatrix} = -i\\sigma_z$\n",
+ "3. **Conmutador:**\n",
+ " $[\\sigma_x, \\sigma_y] = \\sigma_x \\sigma_y - \\sigma_y \\sigma_x = i\\sigma_z - (-i\\sigma_z) = 2i\\sigma_z$.\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)."
+ "**Implicación física:** Los observables asociados no son compatibles; no se pueden medir simultáneamente con precisión infinita."
]
},
{
"cell_type": "markdown",
- "id": "c8ad4688",
+ "id": "15f48f1a",
"metadata": {},
"source": [
"## Problema 5: Matriz Nilpotente\n",
"\n",
- "Sea $A = \\begin{pmatrix} 0 & a & b \\\\ 0 & 0 & c \\\\ 0 & 0 & 0 \\end{pmatrix}$.\n",
+ "**Enunciado:**\n",
+ "Demuestre que $A = \\begin{pmatrix} 0 & a & b \\\\ 0 & 0 & c \\\\ 0 & 0 & 0 \\end{pmatrix}$ es nilpotente y encuentre su índice.\n",
"\n",
- "### Resolución\n",
+ "### Resolución Paso a Paso\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",
+ "1. **Cálculo de 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**."
+ "2. **Índice de nilpotencia:**\n",
+ " Si $a, c \\neq 0$, entonces $ac \\neq 0$, lo que implica $A^2 \\neq 0$. Como $A^3 = 0$, el índice exacto es **3**."
]
},
{
"cell_type": "markdown",
- "id": "7510dcbe",
+ "id": "fc9c506d",
"metadata": {},
"source": [
- "## Problema 6: Matriz de Adyacencia\n",
+ "## Problema 6: Matriz de Adyacencia y Caminos\n",
"\n",
- "$M = \\begin{pmatrix} 0 & 1 & 1 \\\\ 1 & 0 & 0 \\\\ 1 & 0 & 0 \\end{pmatrix}$.\n",
+ "**Enunciado:**\n",
+ "Sea $M = \\begin{pmatrix} 0 & 1 & 1 \\\\ 1 & 0 & 0 \\\\ 1 & 0 & 0 \\end{pmatrix}$. Calcule $M^2$ e interprete.\n",
"\n",
- "### Resolución\n",
+ "### Resolución Paso a Paso\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",
+ "1. **Cálculo:**\n",
+ " $M^2 = \\begin{pmatrix} 2 & 0 & 0 \\\\ 0 & 1 & 1 \\\\ 0 & 1 & 1 \\end{pmatrix}$.\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)."
+ "2. **Interpretación:**\n",
+ " La entrada $(M^2)_{ij}$ indica el número de caminos de longitud 2 entre el nodo $i$ y el nodo $j$.\n",
+ " - $(M^2)_{11} = 2$: Hay dos caminos para ir del nodo 1 al 1 en dos pasos (1-2-1 y 1-3-1).\n",
+ " - $(M^2)_{23} = 1$: Hay un camino del nodo 2 al 3 en dos pasos (2-1-3)."
]
},
{
"cell_type": "markdown",
- "id": "fef32579",
+ "id": "8ad04b93",
"metadata": {},
"source": [
"## Problema 7: Interpolación Polinómica\n",
"\n",
- "$p(x) = ax^2 + bx + c$ pasa por $(-1, 9), (1, 3), (2, 6)$.\n",
+ "**Enunciado:**\n",
+ "Encuentre $p(x) = ax^2 + bx + c$ que pasa por $(-1, 9), (1, 3), (2, 6)$.\n",
"\n",
- "### Resolución\n",
+ "### Resolución Paso a Paso\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",
+ "1. **Sistema de Ecuaciones:**\n",
+ " - $a(-1)^2 + b(-1) + c = 9 \\implies a - b + c = 9$\n",
+ " - $a(1)^2 + b(1) + c = 3 \\implies a + b + c = 3$\n",
+ " - $a(2)^2 + b(2) + c = 6 \\implies 4a + 2b + c = 6$\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",
+ "2. **Resolución:**\n",
+ " Restando (2) - (1): $2b = -6 \\implies b = -3$.\n",
+ " Sustituyendo $b$ en (1) y (3):\n",
+ " - $a + 3 + c = 9 \\implies a + c = 6$\n",
+ " - $4a - 6 + c = 6 \\implies 4a + c = 12$\n",
+ " Restando estas dos: $3a = 6 \\implies a = 2$.\n",
+ " Finalmente: $2 + c = 6 \\implies c = 4$.\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 **$p(x) = 2x^2 - 3x + 4$**."
+ "**Resultado:** $p(x) = 2x^2 - 3x + 4$."
]
},
{
- "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",
+ "id": "c5208e25",
"metadata": {},
"source": [
- "## Problema 8: Matriz por Bloques\n",
+ "## Problema 8: Matrices 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",
+ "**Enunciado:**\n",
+ "Demuestre 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",
+ "### Resolución Paso a Paso\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."
+ "La multiplicación de matrices por bloques sigue las mismas reglas que la multiplicación escalar si los bloques son compatibles.\n",
+ "$M^2 = \\begin{pmatrix} A & 0 \\\\ 0 & B \\end{pmatrix} \\begin{pmatrix} A & 0 \\\\ 0 & B \\end{pmatrix} = \\begin{pmatrix} A^2 + 0 & 0 + 0 \\\\ 0 + 0 & 0 + B^2 \\end{pmatrix} = \\begin{pmatrix} A^2 & 0 \\\\ 0 & B^2 \\end{pmatrix}$.\n",
+ "Por inducción, se cumple para cualquier $k \\in \\mathbb{N}$."
]
},
{
"cell_type": "markdown",
- "id": "3995d78c",
+ "id": "86f35243",
"metadata": {},
"source": [
- "## Problema 9: Propiedad de la Traza\n",
+ "## Problema 9: Propiedad Cíclica de la Traza\n",
"\n",
- "Demostrar $\\text{tr}(AB) = \\text{tr}(BA)$.\n",
+ "**Enunciado:**\n",
+ "Demuestre que $\\text{tr}(AB) = \\text{tr}(BA)$.\n",
"\n",
- "### Resolución\n",
+ "### Resolución Paso a Paso\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, $\\text{tr}(AB) = \\text{tr}(BA)$."
+ "1. Sea $A = (a_{ij})$ y $B = (b_{ij})$.\n",
+ "2. La entrada diagonal $(i,i)$ de $AB$ es $(AB)_{ii} = \\sum_{k=1}^n a_{ik}b_{ki}$.\n",
+ "3. La traza es $\\text{tr}(AB) = \\sum_{i=1}^n (AB)_{ii} = \\sum_{i=1}^n \\sum_{k=1}^n a_{ik}b_{ki}$.\n",
+ "4. La entrada diagonal $(k,k)$ de $BA$ es $(BA)_{kk} = \\sum_{i=1}^n b_{ki}a_{ik}$.\n",
+ "5. La traza es $\\text{tr}(BA) = \\sum_{k=1}^n (BA)_{kk} = \\sum_{k=1}^n \\sum_{i=1}^n b_{ki}a_{ik}$.\n",
+ "6. Como el orden de la suma no altera el resultado en sumas finitas, las expresiones son idénticas."
]
}
],
