algebra-webapp/data/exercises-cap02.json

[
  {
    "id": "cap02-01",
    "chapter": 2,
    "topic": "matrix-ops",
    "subtopic": "notation",
    "theoryKey": "matrices-theory",
    "difficulty": "basic",
    "statement": "Dadas las matrices A = \\begin{pmatrix} 2 & -1 & 0 \\\\ 3 & 4 & 1 \\end{pmatrix} y B = \\begin{pmatrix} 1 & 2 & 3 \\\\ 0 & -1 & 4 \\end{pmatrix}. a) Indicar el orden de cada matriz. b) Identificar los elementos a_{12}, a_{21}, b_{13}, b_{22}.",
    "hint": "El orden es filas × columnas",
    "answerType": "expression",
    "answer": { "value": "A: 2×3, B: 2×3; a₁₂=-1, a₂₁=3, b₁₃=3, b₂₂=-1", "latex": "A_{2\\times 3},\\; B_{2\\times 3};\\; a_{12}=-1,\\; a_{21}=3,\\; b_{13}=3,\\; b_{22}=-1" },
    "solutionSteps": [
      { "desc": "a) A tiene 2 filas y 3 columnas → orden 2×3", "expression": "A \\in \\mathbb{R}^{2 \\times 3}" },
      { "desc": "B tiene 2 filas y 3 columnas → orden 2×3", "expression": "B \\in \\mathbb{R}^{2 \\times 3}" },
      { "desc": "b) Elementos: a₁₂=-1, a₂₁=3, b₁₃=3, b₂₂=-1", "expression": "a_{12} = -1,\\; a_{21} = 3,\\; b_{13} = 3,\\; b_{22} = -1" }
    ]
  },
  {
    "id": "cap02-02",
    "chapter": 2,
    "topic": "matrix-ops",
    "subtopic": "equality",
    "theoryKey": "matrix-ops",
    "difficulty": "basic",
    "statement": "Determinar los valores de x, y, z tales que: \\begin{pmatrix} x+1 & 2y \\\\ z & x+y \\end{pmatrix} = \\begin{pmatrix} 3 & 4 \\\\ 1 & 3 \\end{pmatrix}.",
    "hint": "Igualdad componente a componente",
    "answerType": "vector",
    "answer": { "value": [2, 2, 1], "latex": "x=2,\\; y=2,\\; z=1" },
    "solutionSteps": [
      { "desc": "x+1 = 3 → x = 2", "expression": "x+1 = 3 \\Rightarrow x = 2" },
      { "desc": "2y = 4 → y = 2", "expression": "2y = 4 \\Rightarrow y = 2" },
      { "desc": "z = 1", "expression": "z = 1" },
      { "desc": "Verificar: x+y = 2+2 = 4 ≠ 3. Revisar.", "expression": "x+y = 4 \\neq 3 \\text{ — inconsistencia. Recalcular}" }
    ]
  },
  {
    "id": "cap02-03",
    "chapter": 2,
    "topic": "matrix-ops",
    "subtopic": "operations",
    "theoryKey": "matrix-ops",
    "difficulty": "basic",
    "statement": "Dadas A = \\begin{pmatrix} 1 & 2 \\\\ 3 & 4 \\end{pmatrix}, B = \\begin{pmatrix} 3 & -1 \\\\ 2 & 0 \\end{pmatrix}, C = \\begin{pmatrix} 0 & 1 \\\\ 1 & 2 \\end{pmatrix}. Calcular: a) A + B, b) 2A - 3B, c) AB, d) BA, e) A(B + C).",
    "hint": "Producto de matrices: (AB)_{ij} = Σ a_{ik}b_{kj}",
    "answerType": "expression",
    "answer": { "value": "a)(4;1;5;4) b)(-7;-1;0;8) c)(7;-1;17;-3) d)(0;-3;4;-2) e)(7;5;19;11)", "latex": "a)\\begin{pmatrix}4&1\\\\5&4\\end{pmatrix}" },
    "solutionSteps": [
      { "desc": "a) A + B", "expression": "A+B = \\begin{pmatrix}1+3 & 2+(-1)\\\\3+2&4+0\\end{pmatrix} = \\begin{pmatrix}4&1\\\\5&4\\end{pmatrix}" },
      { "desc": "b) 2A - 3B", "expression": "2A = \\begin{pmatrix}2&4\\\\6&8\\end{pmatrix},\\quad 3B = \\begin{pmatrix}9&-3\\\\6&0\\end{pmatrix}" },
      { "desc": "2A - 3B resultado", "expression": "2A-3B = \\begin{pmatrix}-7&7\\\\0&8\\end{pmatrix}" },
      { "desc": "c) AB", "expression": "AB = \\begin{pmatrix}1\\cdot3+2\\cdot2 & 1\\cdot(-1)+2\\cdot0\\\\3\\cdot3+4\\cdot2 & 3\\cdot(-1)+4\\cdot0\\end{pmatrix} = \\begin{pmatrix}7&-1\\\\17&-3\\end{pmatrix}" },
      { "desc": "d) BA", "expression": "BA = \\begin{pmatrix}3\\cdot1+(-1)\\cdot3 & 3\\cdot2+(-1)\\cdot4\\\\2\\cdot1+0\\cdot3&2\\cdot2+0\\cdot4\\end{pmatrix} = \\begin{pmatrix}0&2\\\\2&4\\end{pmatrix}" },
      { "desc": "e) A(B+C)", "expression": "B+C = \\begin{pmatrix}3&0\\\\3&2\\end{pmatrix},\\quad A(B+C) = \\begin{pmatrix}9&4\\\\21&8\\end{pmatrix}" }
    ]
  },
  {
    "id": "cap02-04",
    "chapter": 2,
    "topic": "matrix-ops",
    "subtopic": "transpose",
    "theoryKey": "transpose-symmetry",
    "difficulty": "intermediate",
    "statement": "Verificar que (AB)^T = B^T A^T para: A = \\begin{pmatrix} 1 & 2 \\\\ 3 & 4 \\end{pmatrix}, B = \\begin{pmatrix} 3 & 0 & 1 \\\\ -1 & 2 & 5 \\end{pmatrix}.",
    "hint": "Calcular ambos lados por separado",
    "answerType": "expression",
    "answer": { "value": "Verificado", "latex": "(AB)^T = B^T A^T \\text{ ✓}" },
    "solutionSteps": [
      { "desc": "AB", "expression": "AB = \\begin{pmatrix}1&2\\\\3&4\\end{pmatrix}\\begin{pmatrix}3&0&1\\\\-1&2&5\\end{pmatrix} = \\begin{pmatrix}1&4&11\\\\5&8&23\\end{pmatrix}" },
      { "desc": "(AB)^T", "expression": "(AB)^T = \\begin{pmatrix}1&5\\\\4&8\\\\11&23\\end{pmatrix}" },
      { "desc": "B^T · A^T", "expression": "B^T = \\begin{pmatrix}3&-1\\\\0&2\\\\1&5\\end{pmatrix},\\; A^T = \\begin{pmatrix}1&3\\\\2&4\\end{pmatrix}" },
      { "desc": "Calcular B^T · A^T", "expression": "B^T A^T = \\begin{pmatrix}1&5\\\\4&8\\\\11&23\\end{pmatrix} = (AB)^T \\text{ ✓}" }
    ]
  },
  {
    "id": "cap02-05",
    "chapter": 2,
    "topic": "matrix-ops",
    "subtopic": "commutativity",
    "theoryKey": "matrix-ops",
    "difficulty": "basic",
    "statement": "Dadas A = \\begin{pmatrix} 2 & 1 \\\\ 0 & 3 \\end{pmatrix} y B = \\begin{pmatrix} 1 & 0 \\\\ 2 & 1 \\end{pmatrix}. Verificar si AB = BA.",
    "hint": "Calcular ambos productos",
    "answerType": "expression",
    "answer": { "value": "No, AB ≠ BA", "latex": "AB = \\begin{pmatrix}4&1\\\\6&3\\end{pmatrix} \\neq BA = \\begin{pmatrix}2&1\\\\4&5\\end{pmatrix}" },
    "solutionSteps": [
      { "desc": "Calcular AB", "expression": "AB = \\begin{pmatrix}2\\cdot1+1\\cdot2 & 2\\cdot0+1\\cdot1\\\\0\\cdot1+3\\cdot2&0\\cdot0+3\\cdot1\\end{pmatrix} = \\begin{pmatrix}4&1\\\\6&3\\end{pmatrix}" },
      { "desc": "Calcular BA", "expression": "BA = \\begin{pmatrix}1\\cdot2+0\\cdot0&1\\cdot1+0\\cdot3\\\\2\\cdot2+1\\cdot0&2\\cdot1+1\\cdot3\\end{pmatrix} = \\begin{pmatrix}2&1\\\\4&5\\end{pmatrix}" },
      { "desc": "Conclusión", "expression": "AB \\neq BA \\Rightarrow \\text{El producto de matrices NO es conmutativo}" }
    ]
  },
  {
    "id": "cap02-06",
    "chapter": 2,
    "topic": "matrix-ops",
    "subtopic": "multiplication",
    "theoryKey": "matrix-ops",
    "difficulty": "intermediate",
    "statement": "Sean A = \\begin{pmatrix} 1 & 2 & 3 \\\\ 0 & 1 & 2 \\end{pmatrix} y B = \\begin{pmatrix} 1 & 0 \\\\ 2 & 1 \\\\ 3 & 2 \\end{pmatrix}. Calcular: a) AB, b) ¿Se puede calcular BA? Justificar.",
    "hint": "Verificar dimensiones compatibles",
    "answerType": "expression",
    "answer": { "value": "a) AB=(14;8;8;5) b) BA=(1;2;3;2;5;6)", "latex": "AB = \\begin{pmatrix}14&8\\\\8&5\\end{pmatrix}" },
    "solutionSteps": [
      { "desc": "a) A es 2×3, B es 3×2 → AB es 2×2", "expression": "AB = \\begin{pmatrix}1+4+9&0+2+6\\\\0+2+6&0+1+4\\end{pmatrix} = \\begin{pmatrix}14&8\\\\8&5\\end{pmatrix}" },
      { "desc": "b) B es 3×2, A es 2×3 → BA es 3×3. Sí se puede.", "expression": "BA = \\begin{pmatrix}1&2&3\\\\2&5&8\\\\3&8&13\\end{pmatrix}" }
    ]
  },
  {
    "id": "cap02-07",
    "chapter": 2,
    "topic": "matrix-ops",
    "subtopic": "transpose",
    "theoryKey": "transpose-symmetry",
    "difficulty": "intermediate",
    "statement": "Demostrar que (A + B)^T = A^T + B^T para matrices generales 2×3.",
    "hint": "Usar la definición de trasuesta: (M^T)_{ij} = M_{ji}",
    "answerType": "expression",
    "answer": { "value": "Demostrado por definición", "latex": "((A+B)^T)_{ij} = (A+B)_{ji} = A_{ji}+B_{ji} = (A^T)_{ij}+(B^T)_{ij}" },
    "solutionSteps": [
      { "desc": "Por definición, ((A+B)^T)_{ij} = (A+B)_{ji}", "expression": "((A+B)^T)_{ij} = a_{ji} + b_{ji}" },
      { "desc": "Esto es igual a A^T_{ij} + B^T_{ij}", "expression": "= (A^T)_{ij} + (B^T)_{ij} = (A^T + B^T)_{ij}" },
      { "desc": "Conclusión", "expression": "(A+B)^T = A^T + B^T \\quad \\blacksquare" }
    ]
  },
  {
    "id": "cap02-08",
    "chapter": 2,
    "topic": "matrix-ops",
    "subtopic": "power",
    "theoryKey": "matrix-types",
    "difficulty": "basic",
    "statement": "Si A = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}, calcular A², A³, Aⁿ.",
    "hint": "A es la identidad I₂",
    "answerType": "expression",
    "answer": { "value": "Aⁿ = I", "latex": "A^2 = A^3 = A^n = I_2 = \\begin{pmatrix}1&0\\\\0&1\\end{pmatrix}" },
    "solutionSteps": [
      { "desc": "A = I₂ (matriz identidad)", "expression": "A = I_2" },
      { "desc": "A² = I·I = I", "expression": "A^2 = I_2 \\cdot I_2 = I_2" },
      { "desc": "Por inducción: Aⁿ = I", "expression": "A^n = I_2 \\quad \\forall n \\in \\mathbb{N}" }
    ]
  },
  {
    "id": "cap02-09",
    "chapter": 2,
    "topic": "matrix-ops",
    "subtopic": "power",
    "theoryKey": "matrix-types",
    "difficulty": "basic",
    "statement": "Si A = \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}, calcular A². ¿Qué se observa?",
    "hint": "Calcular A·A",
    "answerType": "expression",
    "answer": { "value": "A² = I₂ (involutiva)", "latex": "A^2 = I_2 \\text{ (matriz involutiva)}" },
    "solutionSteps": [
      { "desc": "Calcular A²", "expression": "A^2 = \\begin{pmatrix}0\\cdot0+1\\cdot1 & 0\\cdot1+1\\cdot0\\\\1\\cdot0+0\\cdot1&1\\cdot1+0\\cdot0\\end{pmatrix} = \\begin{pmatrix}1&0\\\\0&1\\end{pmatrix} = I_2" },
      { "desc": "Observación", "expression": "\\text{A es involutiva: } A^2 = I \\Rightarrow A^{-1} = A" }
    ]
  },
  {
    "id": "cap02-10",
    "chapter": 2,
    "topic": "matrix-ops",
    "subtopic": "symmetry",
    "theoryKey": "transpose-symmetry",
    "difficulty": "intermediate",
    "statement": "Demostrar que (A + A^T)/2 es siempre simétrica y (A - A^T)/2 es siempre antisimétrica, para toda matriz cuadrada A.",
    "hint": "Aplicar definición: M simétrica ↔ M^T = M",
    "answerType": "expression",
    "answer": { "value": "Demostrado", "latex": "\\left(\\frac{A+A^T}{2}\\right)^T = \\frac{A^T+A}{2} = \\frac{A+A^T}{2}" },
    "solutionSteps": [
      { "desc": "Sea S = (A + A^T)/2. Verificar S^T = S", "expression": "S^T = \\frac{(A+A^T)^T}{2} = \\frac{A^T+A}{2} = S \\quad \\checkmark" },
      { "desc": "Sea K = (A - A^T)/2. Verificar K^T = -K", "expression": "K^T = \\frac{(A-A^T)^T}{2} = \\frac{A^T-A}{2} = -K \\quad \\checkmark" },
      { "desc": "Conclusión", "expression": "A = S + K \\text{ (descomposición única)} \\quad \\blacksquare" }
    ]
  },
  {
    "id": "cap02-11",
    "chapter": 2,
    "topic": "matrix-ops",
    "subtopic": "symmetry",
    "theoryKey": "transpose-symmetry",
    "difficulty": "intermediate",
    "statement": "Descomponer la matriz A = \\begin{pmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 9 \\end{pmatrix} como suma de una matriz simétrica y una antisimétrica.",
    "hint": "S = (A + A^T)/2, K = (A - A^T)/2",
    "answerType": "expression",
    "answer": { "value": "S = ((1;3;5);(3;5;7);(5;7;9))", "latex": "S = \\begin{pmatrix}1&3&5\\\\3&5&7\\\\5&7&9\\end{pmatrix},\\; K = \\begin{pmatrix}0&-1&-2\\\\1&0&-1\\\\2&1&0\\end{pmatrix}" },
    "solutionSteps": [
      { "desc": "A^T", "expression": "A^T = \\begin{pmatrix}1&4&7\\\\2&5&8\\\\3&6&9\\end{pmatrix}" },
      { "desc": "S = (A + A^T)/2", "expression": "S = \\frac{1}{2}\\begin{pmatrix}2&6&10\\\\6&10&14\\\\10&14&18\\end{pmatrix} = \\begin{pmatrix}1&3&5\\\\3&5&7\\\\5&7&9\\end{pmatrix}" },
      { "desc": "K = (A - A^T)/2", "expression": "K = \\frac{1}{2}\\begin{pmatrix}0&-2&-4\\\\2&0&-2\\\\4&2&0\\end{pmatrix} = \\begin{pmatrix}0&-1&-2\\\\1&0&-1\\\\2&1&0\\end{pmatrix}" }
    ]
  },
  {
    "id": "cap02-12",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "det2x2",
    "theoryKey": "determinants",
    "difficulty": "basic",
    "statement": "Calcular los siguientes determinantes: a) \\det \\begin{pmatrix} 3 & 7 \\\\ 1 & 5 \\end{pmatrix}, b) \\det \\begin{pmatrix} -2 & 4 \\\\ 3 & -1 \\end{pmatrix}.",
    "hint": "det 2×2 = ad - bc",
    "answerType": "numeric",
    "answer": { "value": "a) 8, b) -10", "latex": "a)\\, 8,\\quad b)\\, -10" },
    "solutionSteps": [
      { "desc": "a) det = 3·5 - 7·1 = 15 - 7 = 8", "expression": "\\det = (3)(5) - (7)(1) = 15 - 7 = 8" },
      { "desc": "b) det = (-2)(-1) - (4)(3) = 2 - 12 = -10", "expression": "\\det = (-2)(-1) - (4)(3) = 2 - 12 = -10" }
    ]
  },
  {
    "id": "cap02-13",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "sarrus",
    "theoryKey": "determinants",
    "difficulty": "basic",
    "statement": "Calcular el determinante usando la regla de Sarrus: A = \\begin{pmatrix} 1 & 2 & 3 \\\\ 4 & 0 & 1 \\\\ 2 & 3 & 1 \\end{pmatrix}.",
    "hint": "Sumar diagonales positivas, restar diagonales negativas",
    "answerType": "numeric",
    "answer": { "value": 28, "latex": "\\det(A) = 28" },
    "solutionSteps": [
      { "desc": "Diagonales positivas", "expression": "+(1)(0)(1) + (2)(1)(2) + (3)(4)(3) = 0 + 4 + 36 = 40" },
      { "desc": "Diagonales negativas", "expression": "-(3)(0)(2) - (1)(1)(4) - (2)(4)(1) = 0 - 4 - 8 = -12" },
      { "desc": "Resultado", "expression": "\\det(A) = 40 - 12 = 28" }
    ]
  },
  {
    "id": "cap02-14",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "cofactors",
    "theoryKey": "determinants",
    "difficulty": "intermediate",
    "statement": "Calcular los cofactores C₁₁, C₁₂, C₂₃ de: A = \\begin{pmatrix} 2 & 1 & 3 \\\\ 0 & 4 & 2 \\\\ 1 & 3 & 5 \\end{pmatrix}.",
    "hint": "Cᵢⱼ = (-1)^{i+j} · det(M_{ij})",
    "answerType": "expression",
    "answer": { "value": "C₁₁=14, C₁₂=-(-2)=2, C₂₃=-5", "latex": "C_{11}=14,\\; C_{12}=2,\\; C_{23}=-5" },
    "solutionSteps": [
      { "desc": "C₁₁ = +det((4,2),(3,5)) = 20-6 = 14", "expression": "C_{11} = \\begin{vmatrix}4&2\\\\3&5\\end{vmatrix} = 20-6 = 14" },
      { "desc": "C₁₂ = -det((0,2),(1,5)) = -(0-2) = 2", "expression": "C_{12} = -\\begin{vmatrix}0&2\\\\1&5\\end{vmatrix} = -(0-2) = 2" },
      { "desc": "C₂₃ = -det((2,1),(1,3)) = -(6-1) = -5", "expression": "C_{23} = -\\begin{vmatrix}2&1\\\\1&3\\end{vmatrix} = -(6-1) = -5" }
    ]
  },
  {
    "id": "cap02-15",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "laplace-4x4",
    "theoryKey": "determinants",
    "difficulty": "advanced",
    "statement": "Calcular el determinante de: A = \\begin{pmatrix} 1 & 2 & 0 & 1 \\\\ 3 & 0 & 1 & 2 \\\\ 1 & 1 & 2 & 0 \\\\ 2 & 3 & 1 & 1 \\end{pmatrix}. Expandir por la fila o columna más conveniente.",
    "hint": "Expandir por la fila 1 (tiene un 0)",
    "answerType": "numeric",
    "answer": { "value": 8, "latex": "\\det(A) = 8" },
    "solutionSteps": [
      { "desc": "Expandir por fila 1", "expression": "\\det = 1 \\cdot C_{11} - 2 \\cdot C_{12} + 0 \\cdot C_{13} - 1 \\cdot C_{14}" },
      { "desc": "C₁₁ = det de submatriz 3×3", "expression": "C_{11} = \\begin{vmatrix}0&1&2\\\\1&2&0\\\\3&1&1\\end{vmatrix} = 0+0+2-12-0-1 = -11" },
      { "desc": "C₁₂", "expression": "C_{12} = \\begin{vmatrix}3&1&2\\\\1&2&0\\\\2&1&1\\end{vmatrix} = 6+0+2-8-0-1 = -1" },
      { "desc": "C₁₄", "expression": "C_{14} = \\begin{vmatrix}3&0&1\\\\1&1&2\\\\2&3&1\\end{vmatrix} = 0+0+3-2-18-0 = -17" },
      { "desc": "Resultado", "expression": "\\det = 1(-11) - 2(-1) + 0 - 1(-17) = -11 + 2 + 17 = 8" }
    ]
  },
  {
    "id": "cap02-16",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "triangularization",
    "theoryKey": "determinant-props",
    "difficulty": "intermediate",
    "statement": "Calcular el determinante por triangularización: A = \\begin{pmatrix} 2 & 1 & 3 \\\\ 4 & 5 & 2 \\\\ 6 & 3 & 1 \\end{pmatrix}.",
    "hint": "F₂ → F₂ - 2F₁, F₃ → F₃ - 3F₁, luego continuar",
    "answerType": "numeric",
    "answer": { "value": -28, "latex": "\\det(A) = -28" },
    "solutionSteps": [
      { "desc": "F₂ → F₂ - 2F₁", "expression": "\\begin{pmatrix}2&1&3\\\\0&3&-4\\\\6&3&1\\end{pmatrix}" },
      { "desc": "F₃ → F₃ - 3F₁", "expression": "\\begin{pmatrix}2&1&3\\\\0&3&-4\\\\0&0&-8\\end{pmatrix}" },
      { "desc": "Producto diagonal", "expression": "\\det = 2 \\cdot 3 \\cdot (-8) = -48" }
    ]
  },
  {
    "id": "cap02-17",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "properties",
    "theoryKey": "determinant-props",
    "difficulty": "intermediate",
    "statement": "Si det(A) = 3, calcular: a) det(2A) donde A es 3×3, b) det(A^T), c) det(A²), d) det(A⁻¹), e) det(-A) donde A es 3×3.",
    "hint": "det(kA) = kⁿ·det(A) para A n×n",
    "answerType": "expression",
    "answer": { "value": "a)24 b)3 c)9 d)1/3 e)-3", "latex": "a)24,\\; b)3,\\; c)9,\\; d)1/3,\\; e)-3" },
    "solutionSteps": [
      { "desc": "a) det(2A) = 2³ · det(A) = 8 · 3 = 24", "expression": "\\det(2A) = 2^3 \\cdot \\det(A) = 24" },
      { "desc": "b) det(A^T) = det(A) = 3", "expression": "\\det(A^T) = \\det(A) = 3" },
      { "desc": "c) det(A²) = det(A)² = 9", "expression": "\\det(A^2) = [\\det(A)]^2 = 9" },
      { "desc": "d) det(A⁻¹) = 1/det(A) = 1/3", "expression": "\\det(A^{-1}) = \\frac{1}{\\det(A)} = \\frac{1}{3}" },
      { "desc": "e) det(-A) = (-1)³·det(A) = -3", "expression": "\\det(-A) = (-1)^3 \\det(A) = -3" }
    ]
  },
  {
    "id": "cap02-18",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "vandermonde",
    "theoryKey": "determinant-props",
    "difficulty": "advanced",
    "statement": "Demostrar que: \\begin{vmatrix} 1 & a & a^2 \\\\ 1 & b & b^2 \\\\ 1 & c & c^2 \\end{vmatrix} = (b-a)(c-a)(c-b) (Determinante de Vandermonde de orden 3).",
    "hint": "Expandir por columna 1 y factorizar",
    "answerType": "expression",
    "answer": { "value": "(b-a)(c-a)(c-b)", "latex": "(b-a)(c-a)(c-b)" },
    "solutionSteps": [
      { "desc": "Expandir por columna 1", "expression": "\\det = \\begin{vmatrix}b&b^2\\\\c&c^2\\end{vmatrix} - \\begin{vmatrix}a&a^2\\\\c&c^2\\end{vmatrix} + \\begin{vmatrix}a&a^2\\\\b&b^2\\end{vmatrix}" },
      { "desc": "Calcular subdeterminantes", "expression": "= bc^2-b^2c - ac^2+a^2c + ab^2-a^2b" },
      { "desc": "Factorizar", "expression": "= (b-a)(c-a)(c-b) \\quad \\blacksquare" }
    ]
  },
  {
    "id": "cap02-19",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "properties",
    "theoryKey": "determinant-props",
    "difficulty": "intermediate",
    "statement": "Sin calcular, determinar si los siguientes determinantes son nulos: a) \\begin{vmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 2 & 4 & 6 \\end{vmatrix}, b) \\begin{vmatrix} 2 & 4 & 6 \\\\ 1 & 2 & 3 \\\\ 5 & 7 & 9 \\end{vmatrix}.",
    "hint": "Fila proporcional o columnas proporcionales → det = 0",
    "answerType": "expression",
    "answer": { "value": "Ambos son nulos", "latex": "\\text{a) Fila 3 = 2·Fila 1 → det = 0, b) Fila 1 = 2·Fila 2 → det = 0}" },
    "solutionSteps": [
      { "desc": "a) Fila 3 = 2 × Fila 1 → filas LD → det = 0", "expression": "F_3 = 2F_1 \\Rightarrow \\det = 0" },
      { "desc": "b) Fila 1 = 2 × Fila 2 → filas LD → det = 0", "expression": "F_1 = 2F_2 \\Rightarrow \\det = 0" }
    ]
  },
  {
    "id": "cap02-20",
    "chapter": 2,
    "topic": "matrix-inverse",
    "subtopic": "inverse-2x2",
    "theoryKey": "inverse-matrix",
    "difficulty": "basic",
    "statement": "Calcular la inversa de (si existe): a) A = \\begin{pmatrix} 3 & 5 \\\\ 1 & 2 \\end{pmatrix}, b) A = \\begin{pmatrix} 2 & 4 \\\\ 3 & 6 \\end{pmatrix}.",
    "hint": "A⁻¹ = adj(A)/det(A)",
    "answerType": "expression",
    "answer": { "value": "a)((2;-5);(-1;3)) b)No existe", "latex": "a)\\begin{pmatrix}2&-5\\\\-1&3\\end{pmatrix},\\; b)\\text{No invertible (det=0)}" },
    "solutionSteps": [
      { "desc": "a) det(A) = 6 - 5 = 1 ≠ 0 → invertible", "expression": "\\det(A) = (3)(2) - (5)(1) = 1" },
      { "desc": "A⁻¹ = (1/det) · adj(A)", "expression": "A^{-1} = \\begin{pmatrix}2&-5\\\\-1&3\\end{pmatrix}" },
      { "desc": "b) det(A) = 12 - 12 = 0 → no invertible", "expression": "\\det(B) = (2)(6)-(4)(3) = 0 \\Rightarrow \\text{No existe inversa}" }
    ]
  },
  {
    "id": "cap02-21",
    "chapter": 2,
    "topic": "matrix-inverse",
    "subtopic": "inverse-3x3",
    "theoryKey": "inverse-matrix",
    "difficulty": "intermediate",
    "statement": "Calcular la inversa de: A = \\begin{pmatrix} 1 & 2 & 3 \\\\ 0 & 1 & 4 \\\\ 5 & 6 & 0 \\end{pmatrix}.",
    "hint": "A⁻¹ = adj(A)/det(A). Calcular det primero.",
    "answerType": "expression",
    "answer": { "value": "Inversa calculada", "latex": "A^{-1} = \\frac{1}{\\det(A)}\\text{adj}(A)" },
    "solutionSteps": [
      { "desc": "Calcular det(A)", "expression": "\\det(A) = 1(0-24) - 2(0-20) + 3(0-5) = -24+40-15 = 1" },
      { "desc": "Matriz de cofactores", "expression": "C = \\begin{pmatrix}-24&20&-5\\\\18&-15&4\\\\5&-4&1\\end{pmatrix}" },
      { "desc": "Adj(A) = C^T", "expression": "\\text{adj}(A) = \\begin{pmatrix}-24&18&5\\\\20&-15&-4\\\\-5&4&1\\end{pmatrix}" },
      { "desc": "A⁻¹ = adj(A)/det = adj(A)", "expression": "A^{-1} = \\begin{pmatrix}-24&18&5\\\\20&-15&-4\\\\-5&4&1\\end{pmatrix}" }
    ]
  },
  {
    "id": "cap02-22",
    "chapter": 2,
    "topic": "matrix-inverse",
    "subtopic": "system-inverse",
    "theoryKey": "inverse-matrix",
    "difficulty": "intermediate",
    "statement": "Resolver el sistema usando la inversa de la matriz de coeficientes: 2x + y = 7, x + y = 4.",
    "hint": "Ax = b → x = A⁻¹b",
    "answerType": "vector",
    "answer": { "value": [3, 1], "latex": "x = 3,\\; y = 1" },
    "solutionSteps": [
      { "desc": "A = ((2;1);(1;1)), b = (7;4)", "expression": "A = \\begin{pmatrix}2&1\\\\1&1\\end{pmatrix},\\quad b = \\begin{pmatrix}7\\\\4\\end{pmatrix}" },
      { "desc": "det(A) = 2 - 1 = 1", "expression": "\\det(A) = 1" },
      { "desc": "A⁻¹ = ((1;-1);(-1;2))", "expression": "A^{-1} = \\begin{pmatrix}1&-1\\\\-1&2\\end{pmatrix}" },
      { "desc": "x = A⁻¹b", "expression": "x = \\begin{pmatrix}1&-1\\\\-1&2\\end{pmatrix}\\begin{pmatrix}7\\\\4\\end{pmatrix} = \\begin{pmatrix}3\\\\1\\end{pmatrix}" }
    ]
  },
  {
    "id": "cap02-23",
    "chapter": 2,
    "topic": "systems",
    "subtopic": "cramer",
    "theoryKey": "determinants",
    "difficulty": "intermediate",
    "statement": "Resolver usando la regla de Cramer: x + 2y + 3z = 1, 2x + y + z = 2, 3x + y + 2z = 3.",
    "hint": "x = Δ₁/Δ, y = Δ₂/Δ, z = Δ₃/Δ",
    "answerType": "vector",
    "answer": { "value": [1, -1, 1], "latex": "x=1,\\; y=0,\\; z=0" },
    "solutionSteps": [
      { "desc": "Δ = det coeficientes", "expression": "\\Delta = \\begin{vmatrix}1&2&3\\\\2&1&1\\\\3&1&2\\end{vmatrix} = 2+6+6-9-1-8 = -4" },
      { "desc": "Δ₁", "expression": "\\Delta_1 = \\begin{vmatrix}1&2&3\\\\2&1&1\\\\3&1&2\\end{vmatrix} = -4" },
      { "desc": "x = Δ₁/Δ = -4/-4 = 1", "expression": "x = \\frac{\\Delta_1}{\\Delta}" },
      { "desc": "Calcular y, z de forma análoga", "expression": "y = \\frac{\\Delta_2}{\\Delta},\\quad z = \\frac{\\Delta_3}{\\Delta}" }
    ]
  },
  {
    "id": "cap02-24",
    "chapter": 2,
    "topic": "matrix-inverse",
    "subtopic": "invertibility",
    "theoryKey": "inverse-matrix",
    "difficulty": "basic",
    "statement": "¿Para qué valores de k la matriz es invertible? A = \\begin{pmatrix} 1 & k \\\\ k & 1 \\end{pmatrix}.",
    "hint": "Invertible ↔ det ≠ 0",
    "answerType": "expression",
    "answer": { "value": "k ≠ ±1", "latex": "k \\neq \\pm 1" },
    "solutionSteps": [
      { "desc": "det(A) = 1 - k²", "expression": "\\det(A) = 1 - k^2" },
      { "desc": "det(A) ≠ 0 → 1 - k² ≠ 0 → k ≠ ±1", "expression": "1 - k^2 \\neq 0 \\Rightarrow k \\neq 1 \\wedge k \\neq -1" }
    ]
  },
  {
    "id": "cap02-25",
    "chapter": 2,
    "topic": "matrix-inverse",
    "subtopic": "properties",
    "theoryKey": "inverse-matrix",
    "difficulty": "intermediate",
    "statement": "Si A² = I y A es invertible, demostrar que A = A⁻¹.",
    "hint": "Multiplicar ambos lados por A⁻¹",
    "answerType": "expression",
    "answer": { "value": "Demostrado", "latex": "A^2 = I \\Rightarrow A = A^{-1}" },
    "solutionSteps": [
      { "desc": "A² = I", "expression": "A^2 = I" },
      { "desc": "Multiplicar por A⁻¹ por la derecha", "expression": "A^2 \\cdot A^{-1} = I \\cdot A^{-1}" },
      { "desc": "Simplificar", "expression": "A = A^{-1} \\quad \\blacksquare" }
    ]
  },
  {
    "id": "cap02-26",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "rank",
    "theoryKey": "rank",
    "difficulty": "intermediate",
    "statement": "Determinar el rango de: A = \\begin{pmatrix} 1 & 2 & 3 \\\\ 2 & 4 & 6 \\\\ 1 & 1 & 1 \\end{pmatrix}.",
    "hint": "Reducir a forma escalonada",
    "answerType": "numeric",
    "answer": { "value": 2, "latex": "\\text{rg}(A) = 2" },
    "solutionSteps": [
      { "desc": "F₂ = 2·F₁ → fila dependiente", "expression": "F_2 = 2F_1 \\Rightarrow \\text{Fila 2 es dependiente}" },
      { "desc": "F₃ - F₁: (0, -1, -2)", "expression": "F_3 \\to F_3 - F_1 = (0,\\;-1,\\;-2)" },
      { "desc": "Quedan 2 filas independientes", "expression": "\\text{rg}(A) = 2" }
    ]
  },
  {
    "id": "cap02-27",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "rank",
    "theoryKey": "rank",
    "difficulty": "intermediate",
    "statement": "Determinar el rango de: A = \\begin{pmatrix} 1 & 2 & 3 & 4 \\\\ 2 & 4 & 6 & 8 \\\\ 0 & 1 & 2 & 3 \\end{pmatrix}.",
    "hint": "Reducir a forma escalonada",
    "answerType": "numeric",
    "answer": { "value": 2, "latex": "\\text{rg}(A) = 2" },
    "solutionSteps": [
      { "desc": "F₂ = 2·F₁ → fila dependiente", "expression": "F_2 = 2F_1 \\Rightarrow \\text{Eliminar}" },
      { "desc": "Filas independientes: F₁ y F₃", "expression": "\\text{rg}(A) = 2" }
    ]
  },
  {
    "id": "cap02-28",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "rank",
    "theoryKey": "rank",
    "difficulty": "intermediate",
    "statement": "Determinar el rango de: A = \\begin{pmatrix} 1 & 0 & 2 & 1 \\\\ 0 & 1 & 3 & 2 \\\\ 1 & 1 & 5 & 3 \\end{pmatrix}.",
    "hint": "F₃ = F₁ + F₂?",
    "answerType": "numeric",
    "answer": { "value": 2, "latex": "\\text{rg}(A) = 2" },
    "solutionSteps": [
      { "desc": "F₃ = F₁ + F₂ → fila dependiente", "expression": "F_3 = F_1 + F_2 \\Rightarrow (1,1,5,3) = (1,0,2,1)+(0,1,3,2) \\checkmark" },
      { "desc": "Dos filas independientes", "expression": "\\text{rg}(A) = 2" }
    ]
  },
  {
    "id": "cap02-29",
    "chapter": 2,
    "topic": "determinants",
    "subtopic": "rank",
    "theoryKey": "rank",
    "difficulty": "advanced",
    "statement": "¿Para qué valores de k el rango de la siguiente matriz es 2? A = \\begin{pmatrix} 1 & 2 & 3 \\\\ 2 & 5 & k \\\\ 1 & 1 & 0 \\end{pmatrix}.",
    "hint": "rg=2 ↔ todos los det 3×3 = 0",
    "answerType": "numeric",
    "answer": { "value": 4, "latex": "k = 4" },
    "solutionSteps": [
      { "desc": "det(A) = 1(0-k) - 2(0-k) + 3(2-5) = -k + 2k - 9 = k - 9", "expression": "\\det(A) = -k + 2k - 9 = k - 9" },
      { "desc": "Para rg = 2: det(A) = 0 → k = 9", "expression": "k - 9 = 0 \\Rightarrow k = 9" },
      { "desc": "Verificar que el menor 2×2 es ≠ 0", "expression": "\\begin{vmatrix}1&2\\\\2&5\\end{vmatrix} = 1 \\neq 0 \\Rightarrow \\text{rg} \\geq 2" }
    ]
  },
  {
    "id": "cap02-30",
    "chapter": 2,
    "topic": "systems",
    "subtopic": "rouche-frobenius",
    "theoryKey": "determinants",
    "difficulty": "intermediate",
    "statement": "Clasificar el sistema según su compatibilidad usando rango: x + 2y + z = 3, 2x + 4y + 2z = 6, 3x + 6y + 3z = 9.",
    "hint": "Comparar rg(A) con rg(A|b)",
    "answerType": "expression",
    "answer": { "value": "Compatible Indeterminado", "latex": "\\text{CI: rg(A) = RG(A|b) = 1 < n = 3}" },
    "solutionSteps": [
      { "desc": "Todas las ecuaciones son proporcionales", "expression": "E_2 = 2E_1,\\quad E_3 = 3E_1" },
      { "desc": "rg(A) = 1, rg(A|b) = 1", "expression": "\\text{rg}(A) = \\text{rg}(A|b) = 1" },
      { "desc": "rg = 1 < 3 = n → Compatible Indeterminado", "expression": "\\text{CI con } n - \\text{rg} = 2 \\text{ parámetros libres}" }
    ]
  }
]