{
  "index": "1984-A-3",
  "type": "ALG",
  "tag": [
    "ALG",
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "Problem A-3\nLet \\( n \\) be a positive integer. Let \\( a, b, x \\) be real numbers, with \\( a \\neq b \\), and let \\( M_{n} \\) denote the \\( 2 n \\times 2 n \\) matrix whose ( \\( i, j \\) ) entry \\( m_{i j} \\) is given by\n\\[\nm_{i j}=\\left\\{\\begin{array}{ll}\nx & \\text { if } i=j, \\\\\na & \\text { if } i \\neq j \\\\\nb & \\text { if } i \\neq j\n\\end{array} \\text { and } i+j \\text { is even }, ~ i+j\\right. \\text { is odd. }\n\\]\n\nThus, for example, \\( M_{2}=\\left(\\begin{array}{llll}x & b & a & b \\\\ b & x & b & a \\\\ a & b & x & b \\\\ b & a & b & x\\end{array}\\right) \\). Express \\( \\lim _{x \\rightarrow a} \\operatorname{det} M_{n} /(x-a)^{2 n-2} \\) as a polynomial in \\( a, b \\), and \\( n \\), where \\( \\operatorname{det} M_{n} \\) denotes the determinant of \\( M_{n} \\).",
  "solution": "A-3.\nLet \\( \\left.N=M_{n}\\right]_{x=a} . N \\) has rank 2 , so that 0 is an eigenvalue of multiplicity \\( 2 n-2 \\). Let e denote the \\( 2 n \\times 1 \\) column vector of l's. Notice that \\( N e=n(a+b) \\mathbf{e} \\), and therefore \\( n(a+b) \\) is an eigenvalue. The trace of \\( N \\) is \\( 2 n a \\), and therefore the remaining eigenvalue is \\( 2 n a-n(a+b)= \\) \\( n(a-b) \\). [Note: This corresponds to the eigenvector \\( \\mathbf{f} \\), where \\( f_{1.1}=(-1)^{1+1}, i=1, \\ldots, 2 n \\).]\n\nThe preceding analysis implies that the characteristic equation of \\( N \\) is\n\\[\n\\operatorname{det}(N-\\lambda I)=\\lambda^{2 n-2}(\\lambda-n(a+b))(\\lambda-n(a-b))\n\\]\n\nLet \\( \\lambda=a-x \\). Then\n\\[\n\\operatorname{det} M_{n}=\\operatorname{det}(N-(a-x) I)=(a-x)^{2 n-2}(a-x-n(a+b))(a-x-n(a-b))\n\\]\n\nIt follows that\n\\[\n\\lim _{x \\rightarrow a} \\frac{\\operatorname{det} M_{n}}{(x-a)^{2 n-2}}=\\lim _{x \\rightarrow a}(a-x-n(a+b))(a-x-n(a-b))=n^{2}\\left(a^{2}-b^{2}\\right)\n\\]",
  "vars": [
    "x",
    "i",
    "j",
    "\\\\lambda"
  ],
  "params": [
    "n",
    "a",
    "b",
    "M_n",
    "m_ij",
    "e",
    "N",
    "I",
    "f",
    "f_1.1"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvar",
        "i": "rowindex",
        "j": "colindex",
        "\\lambda": "eigenparm",
        "n": "sizen",
        "a": "firstconst",
        "b": "secondconst",
        "M_n": "basematrix",
        "m_ij": "matrixentry",
        "e": "onesvector",
        "N": "specializedmatrix",
        "I": "identitymat",
        "f": "altvector",
        "f_1.1": "notedcomponent"
      },
      "question": "Problem A-3\nLet \\( sizen \\) be a positive integer. Let \\( firstconst, secondconst, realvar \\) be real numbers, with \\( firstconst \\neq secondconst \\), and let \\( basematrix \\) denote the \\( 2 sizen \\times 2 sizen \\) matrix whose ( \\( rowindex, colindex \\) ) entry \\( matrixentry \\) is given by\n\\[\nmatrixentry=\\left\\{\\begin{array}{ll}\nrealvar & \\text { if } rowindex=colindex, \\\\\nfirstconst & \\text { if } rowindex \\neq colindex \\\\\nsecondconst & \\text { if } rowindex \\neq colindex\n\\end{array} \\text { and } rowindex+colindex \\text { is even }, ~ rowindex+colindex\\right. \\text { is odd. }\n\\]\n\nThus, for example, \\( M_{2}=\\left(\\begin{array}{llll}realvar & secondconst & firstconst & secondconst \\\\ secondconst & realvar & secondconst & firstconst \\\\ firstconst & secondconst & realvar & secondconst \\\\ secondconst & firstconst & secondconst & realvar\\end{array}\\right) \\). Express \\( \\lim _{realvar \\rightarrow firstconst} \\operatorname{det} basematrix /(realvar-firstconst)^{2 sizen-2} \\) as a polynomial in \\( firstconst, secondconst \\), and \\( sizen \\), where \\( \\operatorname{det} basematrix \\) denotes the determinant of \\( basematrix \\).",
      "solution": "A-3.\nLet \\( \\left.specializedmatrix=basematrix\\right]_{realvar=firstconst} . specializedmatrix \\) has rank 2 , so that 0 is an eigenvalue of multiplicity \\( 2 sizen-2 \\). Let onesvector denote the \\( 2 sizen \\times 1 \\) column vector of l's. Notice that \\( specializedmatrix \\, onesvector=sizen(firstconst+secondconst) \\mathbf{onesvector} \\), and therefore \\( sizen(firstconst+secondconst) \\) is an eigenvalue. The trace of \\( specializedmatrix \\) is \\( 2 sizen firstconst \\), and therefore the remaining eigenvalue is \\( 2 sizen firstconst-sizen(firstconst+secondconst)= \\) \\( sizen(firstconst-secondconst) \\). [Note: This corresponds to the eigenvector \\( \\mathbf{altvector} \\), where \\( notedcomponent=(-1)^{1+1}, rowindex=1, \\ldots, 2 sizen \\).]\n\nThe preceding analysis implies that the characteristic equation of \\( specializedmatrix \\) is\n\\[\n\\operatorname{det}(specializedmatrix-eigenparm \\, identitymat)=eigenparm^{2 sizen-2}(eigenparm-sizen(firstconst+secondconst))(eigenparm-sizen(firstconst-secondconst))\n\\]\n\nLet \\( eigenparm=firstconst-realvar \\). Then\n\\[\n\\operatorname{det} basematrix=\\operatorname{det}(specializedmatrix-(firstconst-realvar) identitymat)=(firstconst-realvar)^{2 sizen-2}(firstconst-realvar-sizen(firstconst+secondconst))(firstconst-realvar-sizen(firstconst-secondconst))\n\\]\n\nIt follows that\n\\[\n\\lim _{realvar \\rightarrow firstconst} \\frac{\\operatorname{det} basematrix}{(realvar-firstconst)^{2 sizen-2}}=\\lim _{realvar \\rightarrow firstconst}(firstconst-realvar-sizen(firstconst+secondconst))(firstconst-realvar-sizen(firstconst-secondconst))=sizen^{2}\\left(firstconst^{2}-secondconst^{2}\\right)\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "coffeepot",
        "i": "sailorhat",
        "j": "lampshade",
        "\\lambda": "windshield",
        "n": "watermelon",
        "a": "rhinoceros",
        "b": "platypus",
        "M_n": "chocolate",
        "m_ij": "buttercup",
        "e": "tangerine",
        "N": "strawberry",
        "I": "blackboard",
        "f": "orangutan",
        "f_1.1": "caterpillar"
      },
      "question": "Problem A-3\nLet \\( watermelon \\) be a positive integer. Let \\( rhinoceros, platypus, coffeepot \\) be real numbers, with \\( rhinoceros \\neq platypus \\), and let \\( chocolate_{watermelon} \\) denote the \\( 2 watermelon \\times 2 watermelon \\) matrix whose ( \\( sailorhat, lampshade \\) ) entry \\( buttercup_{sailorhat lampshade} \\) is given by\n\\[\nbuttercup_{sailorhat \\, lampshade}=\\left\\{\\begin{array}{ll}\ncoffeepot & \\text { if } sailorhat=lampshade, \\\\\nrhinoceros & \\text { if } sailorhat \\neq lampshade \\\\\nplatypus & \\text { if } sailorhat \\neq lampshade\n\\end{array} \\text { and } sailorhat+lampshade \\text { is even }, ~ sailorhat+lampshade\\right. \\text { is odd. }\n\\]\n\nThus, for example, \\( chocolate_{2}=\\left(\\begin{array}{llll}coffeepot & platypus & rhinoceros & platypus \\\\ platypus & coffeepot & platypus & rhinoceros \\\\ rhinoceros & platypus & coffeepot & platypus \\\\ platypus & rhinoceros & platypus & coffeepot\\end{array}\\right) \\). Express \\( \\lim _{coffeepot \\rightarrow rhinoceros} \\operatorname{det} chocolate_{watermelon} /(coffeepot-rhinoceros)^{2 watermelon-2} \\) as a polynomial in \\( rhinoceros, platypus \\), and \\( watermelon \\), where \\( \\operatorname{det} chocolate_{watermelon} \\) denotes the determinant of \\( chocolate_{watermelon} \\).",
      "solution": "A-3.\nLet \\( \\left.strawberry=chocolate_{watermelon}\\right]_{coffeepot=rhinoceros} . strawberry \\) has rank 2 , so that 0 is an eigenvalue of multiplicity \\( 2 watermelon-2 \\). Let tangerine denote the \\( 2 watermelon \\times 1 \\) column vector of l's. Notice that \\( strawberry\\,tangerine =watermelon(rhinoceros+platypus) \\mathbf{tangerine} \\), and therefore \\( watermelon(rhinoceros+platypus) \\) is an eigenvalue. The trace of \\( strawberry \\) is \\( 2 watermelon rhinoceros \\), and therefore the remaining eigenvalue is \\( 2 watermelon rhinoceros-watermelon(rhinoceros+platypus)= \\) \\( watermelon(rhinoceros-platypus) \\). [Note: This corresponds to the eigenvector \\( \\mathbf{orangutan} \\), where \\( caterpillar=(-1)^{1+1}, sailorhat=1, \\ldots, 2 watermelon \\).]\n\nThe preceding analysis implies that the characteristic equation of \\( strawberry \\) is\n\\[\n\\operatorname{det}(strawberry-windshield\\,blackboard)=windshield^{2 watermelon-2}(windshield-watermelon(rhinoceros+platypus))(windshield-watermelon(rhinoceros-platypus))\n\\]\n\nLet \\( windshield=rhinoceros-coffeepot \\). Then\n\\[\n\\operatorname{det} chocolate_{watermelon}=\\operatorname{det}(strawberry-(rhinoceros-coffeepot)\\,blackboard)=(rhinoceros-coffeepot)^{2 watermelon-2}(rhinoceros-coffeepot-watermelon(rhinoceros+platypus))(rhinoceros-coffeepot-watermelon(rhinoceros-platypus))\n\\]\n\nIt follows that\n\\[\n\\lim _{coffeepot \\rightarrow rhinoceros} \\frac{\\operatorname{det} chocolate_{watermelon}}{(coffeepot-rhinoceros)^{2 watermelon-2}}=\\lim _{coffeepot \\rightarrow rhinoceros}(rhinoceros-coffeepot-watermelon(rhinoceros+platypus))(rhinoceros-coffeepot-watermelon(rhinoceros-platypus))=watermelon^{2}\\left(rhinoceros^{2}-platypus^{2}\\right)\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixpoint",
        "i": "offsetindex",
        "j": "terminalindex",
        "\\lambda": "fixedscalar",
        "n": "singleton",
        "a": "variance",
        "b": "meanvalue",
        "M_n": "vectorscale",
        "m_ij": "vectorcomponent",
        "e": "zerosvector",
        "N": "highrankmatrix",
        "I": "zeromatrix",
        "f": "constantvector",
        "f_1.1": "constvectornode"
      },
      "question": "Problem A-3\nLet \\( singleton \\) be a positive integer. Let \\( variance, meanvalue, fixpoint \\) be real numbers, with \\( variance \\neq meanvalue \\), and let \\( vectorscale_{singleton} \\) denote the \\( 2 singleton \\times 2 singleton \\) matrix whose ( \\( offsetindex, terminalindex \\) ) entry \\( vectorcomponent_{offsetindex terminalindex} \\) is given by\n\\[\nvectorcomponent_{offsetindex terminalindex}=\\left\\{\\begin{array}{ll}\nfixpoint & \\text { if } offsetindex=terminalindex, \\\\\nvariance & \\text { if } offsetindex \\neq terminalindex \\\\\nmeanvalue & \\text { if } offsetindex \\neq terminalindex\n\\end{array} \\text { and } offsetindex+terminalindex \\text { is even }, ~ offsetindex+terminalindex\\right. \\text { is odd. }\n\\]\n\nThus, for example, \\( vectorscale_{2}=\\left(\\begin{array}{llll}fixpoint & meanvalue & variance & meanvalue \\\\ meanvalue & fixpoint & meanvalue & variance \\\\ variance & meanvalue & fixpoint & meanvalue \\\\ meanvalue & variance & meanvalue & fixpoint\\end{array}\\right) \\). Express \\( \\lim _{fixpoint \\rightarrow variance} \\operatorname{det} vectorscale_{singleton} /(fixpoint-variance)^{2 singleton-2} \\) as a polynomial in \\( variance, meanvalue \\), and \\( singleton \\), where \\( \\operatorname{det} vectorscale_{singleton} \\) denotes the determinant of \\( vectorscale_{singleton} \\).",
      "solution": "A-3.\nLet \\( \\left.highrankmatrix=vectorscale_{singleton}\\right]_{fixpoint=variance} . highrankmatrix \\) has rank 2 , so that 0 is an eigenvalue of multiplicity \\( 2 singleton-2 \\). Let zerosvector denote the \\( 2 singleton \\times 1 \\) column vector of l's. Notice that \\( highrankmatrix zerosvector=singleton(variance+meanvalue) \\mathbf{zerosvector} \\), and therefore \\( singleton(variance+meanvalue) \\) is an eigenvalue. The trace of \\( highrankmatrix \\) is \\( 2 singleton variance \\), and therefore the remaining eigenvalue is \\( 2 singleton variance-singleton(variance+meanvalue)= \\) \\( singleton(variance-meanvalue) \\). [Note: This corresponds to the eigenvector \\( \\mathbf{constantvector} \\), where \\( constvectornode=(-1)^{1+1}, offsetindex=1, \\ldots, 2 singleton \\).]\n\nThe preceding analysis implies that the characteristic equation of \\( highrankmatrix \\) is\n\\[\n\\operatorname{det}(highrankmatrix-fixedscalar zeromatrix)=fixedscalar^{2 singleton-2}(fixedscalar-singleton(variance+meanvalue))(fixedscalar-singleton(variance-meanvalue))\n\\]\n\nLet \\( fixedscalar=variance-fixpoint \\). Then\n\\[\n\\operatorname{det} vectorscale_{singleton}=\\operatorname{det}(highrankmatrix-(variance-fixpoint) zeromatrix)=(variance-fixpoint)^{2 singleton-2}(variance-fixpoint-singleton(variance+meanvalue))(variance-fixpoint-singleton(variance-meanvalue))\n\\]\n\nIt follows that\n\\[\n\\lim _{fixpoint \\rightarrow variance} \\frac{\\operatorname{det} vectorscale_{singleton}}{(fixpoint-variance)^{2 singleton-2}}=\\lim _{fixpoint \\rightarrow variance}(variance-fixpoint-singleton(variance+meanvalue))(variance-fixpoint-singleton(variance-meanvalue))=singleton^{2}\\left(variance^{2}-meanvalue^{2}\\right)\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "i": "hjgrksla",
        "j": "pkldweng",
        "\\\\lambda": "mcnvbqer",
        "n": "sldkfjwe",
        "a": "vncmzqwe",
        "b": "plkmqwer",
        "M_n": "gjrkslwe",
        "m_ij": "zxcvbnml",
        "e": "poiuytre",
        "N": "lkjhgfds",
        "I": "asdfghjk",
        "f": "qwertyui",
        "f_1.1": "cvbnmert"
      },
      "question": "Problem A-3\nLet \\( sldkfjwe \\) be a positive integer. Let \\( vncmzqwe, plkmqwer, qzxwvtnp \\) be real numbers, with \\( vncmzqwe \\neq plkmqwer \\), and let \\( gjrkslwe \\) denote the \\( 2 sldkfjwe \\times 2 sldkfjwe \\) matrix whose ( \\( hjgrksla, pkldweng \\) ) entry \\( zxcvbnml \\) is given by\n\\[\nzxcvbnml=\\left\\{\\begin{array}{ll}\nqzxwvtnp & \\text { if } hjgrksla=pkldweng, \\\\\nvncmzqwe & \\text { if } hjgrksla \\neq pkldweng \\\\\nplkmqwer & \\text { if } hjgrksla \\neq pkldweng\n\\end{array} \\text { and } hjgrksla+pkldweng \\text { is even }, ~ hjgrksla+pkldweng\\right. \\text { is odd. }\n\\]\n\nThus, for example, \\( gjrkslwe_{2}=\\left(\\begin{array}{llll}qzxwvtnp & plkmqwer & vncmzqwe & plkmqwer \\\\ plkmqwer & qzxwvtnp & plkmqwer & vncmzqwe \\\\ vncmzqwe & plkmqwer & qzxwvtnp & plkmqwer \\\\ plkmqwer & vncmzqwe & plkmqwer & qzxwvtnp\\end{array}\\right) \\). Express \\( \\lim _{qzxwvtnp \\rightarrow vncmzqwe} \\operatorname{det} gjrkslwe /(qzxwvtnp-vncmzqwe)^{2 sldkfjwe-2} \\) as a polynomial in \\( vncmzqwe, plkmqwer \\), and \\( sldkfjwe \\), where \\( \\operatorname{det} gjrkslwe \\) denotes the determinant of \\( gjrkslwe \\).",
      "solution": "A-3.\nLet \\( \\left.lkjhgfds=gjrkslwe\\right]_{qzxwvtnp=vncmzqwe} . lkjhgfds \\) has rank 2 , so that 0 is an eigenvalue of multiplicity \\( 2 sldkfjwe-2 \\). Let poiuytre denote the \\( 2 sldkfjwe \\times 1 \\) column vector of l's. Notice that \\( lkjhgfds \\, poiuytre = sldkfjwe(vncmzqwe+plkmqwer) \\mathbf{poiuytre} \\), and therefore \\( sldkfjwe(vncmzqwe+plkmqwer) \\) is an eigenvalue. The trace of \\( lkjhgfds \\) is \\( 2 sldkfjwe \\, vncmzqwe \\), and therefore the remaining eigenvalue is \\( 2 sldkfjwe \\, vncmzqwe - sldkfjwe(vncmzqwe+plkmqwer)=sldkfjwe(vncmzqwe-plkmqwer) \\). [Note: This corresponds to the eigenvector \\( \\mathbf{qwertyui} \\), where \\( cvbnmert=(-1)^{1+1}, hjgrksla=1, \\ldots, 2 sldkfjwe \\).]\n\nThe preceding analysis implies that the characteristic equation of \\( lkjhgfds \\) is\n\\[\n\\operatorname{det}(lkjhgfds-mcnvbqer \\, asdfghjk)=mcnvbqer^{2 sldkfjwe-2}(mcnvbqer-sldkfjwe(vncmzqwe+plkmqwer))(mcnvbqer-sldkfjwe(vncmzqwe-plkmqwer))\n\\]\n\nLet \\( mcnvbqer=vncmzqwe-qzxwvtnp \\). Then\n\\[\n\\operatorname{det} gjrkslwe=\\operatorname{det}(lkjhgfds-(vncmzqwe-qzxwvtnp) \\, asdfghjk)=(vncmzqwe-qzxwvtnp)^{2 sldkfjwe-2}(vncmzqwe-qzxwvtnp-sldkfjwe(vncmzqwe+plkmqwer))(vncmzqwe-qzxwvtnp-sldkfjwe(vncmzqwe-plkmqwer))\n\\]\n\nIt follows that\n\\[\n\\lim _{qzxwvtnp \\rightarrow vncmzqwe} \\frac{\\operatorname{det} gjrkslwe}{(qzxwvtnp-vncmzqwe)^{2 sldkfjwe-2}}=\\lim _{qzxwvtnp \\rightarrow vncmzqwe}(vncmzqwe-qzxwvtnp-sldkfjwe(vncmzqwe+plkmqwer))(vncmzqwe-qzxwvtnp-sldkfjwe(vncmzqwe-plkmqwer))=sldkfjwe^{2}\\left(vncmzqwe^{2}-plkmqwer^{2}\\right)\n\\]"
    },
    "kernel_variant": {
      "question": "For a positive integer m ≥ 1 and three distinct real numbers a, b, c, define the 3m × 3m matrix  \n  Qₘ(x) = (q_{ij})_{1≤i,j≤3m}, where  \n\n  q_{ij} = ⎧  \n    x,       if i = j,  \n    a,       if i≠j and i+j≡0 (mod 3),  \n    b,       if i≠j and i+j≡1 (mod 3),  \n    c.       if i≠j and i+j≡2 (mod 3).  \n\nEvaluate the limit  \n\n  Lₘ = lim_{x→a} det Qₘ(x) ∕ (x − a)^{\\,m-1},  \n\nand give Lₘ explicitly as a polynomial in a, b, c and m.\n\n——————————————————————————————————————————",
      "solution": "Notation.  \nIₘ = identity_{m×m},   Jₘ = all-ones_{m×m},          \n1 = (1,…,1)ᵀ ∈ ℝᵐ.\n\n--------------------------------------------------------------------\n1. 3 × 3 block form.  \nWrite every index as i = 3s+r (s = 0,…,m–1, r = 0,1,2).  \nAfter cyclicly regrouping rows and columns (0-block first, then 1-block, then 2-block) the matrix becomes a 3 × 3 block matrix  \n\n                     Qₘ(x)=⎛    D₀   B₀₁ B₀₂  \n                                    B₁₀ D₁ B₁₂  \n                                    B₂₀ B₂₁ D₂ ⎞,  \n\nwith  \n\n D₀ = (x−a)Iₘ + aJₘ,             B₀₁ = B₁₀ᵀ = bJₘ,  \n D₁ = (x−c)Iₘ + cJₘ,             B₀₂ = B₂₀ᵀ = cJₘ,  \n D₂ = (x−b)Iₘ + bJₘ,             B₁₂ = B₂₁ᵀ = aJₘ.\n\n--------------------------------------------------------------------\n2. Orthogonal decomposition with respect to Jₘ.  \nLet  \n\n W = {v∈ℝᵐ | 1ᵀv = 0}  (dim W = m−1),  \n\nso Jₘ acts as 0 on W and as multiplication by m on the 1-dimensional complement ⟨1⟩.\n\n--------------------------------------------------------------------\n3. Eigenvalues coming from W.  \nFor each r ∈ {0,1,2} and v∈W insert v in block r and zeros elsewhere.  \nBecause Jₘv = 0 such vectors are eigenvectors of Qₘ(x) with eigenvalue  \n\n λ_r(x) = ⎧x−a if r=0; x−c if r=1; x−b if r=2⎫.\n\nTherefore  \n\n (x−a)^{m−1}(x−b)^{m−1}(x−c)^{m−1}  \n\nis a factor of det Qₘ(x).\n\n--------------------------------------------------------------------\n4. Compression to a 3-dimensional subspace.  \nDefine  \n\n 1_r (r=0,1,2) = vector that equals 1 in block r and 0 elsewhere.\n\nThe vectors 1_r ∕√m are orthonormal and span the orthogonal complement of the 3(m−1)-dimensional space from Step 3.  \nRestricting Qₘ(x) to this 3-space yields the 3×3 matrix  \n\n Sₘ(x)=⎛ x−a+ma   mb   mc  \n      mb x−c+mc  ma  \n      mc  ma x−b+mb ⎞.             (1)\n\nConsequently  \n\n det Qₘ(x) = (x−a)^{m−1}(x−b)^{m−1}(x−c)^{m−1} det Sₘ(x).     (2)\n\n--------------------------------------------------------------------\n5. Evaluate Sₘ(x) at x = a.  \nPutting x = a in (1) gives  \n\n Sₘ(a)=⎛ ma   mb   mc  \n      mb a+(m−1)c  ma  \n      mc  ma a+(m−1)b ⎞.\n\nFor a 3×3 matrix M=(m_{ij}) the identity  \n\n det M = m₁₁m₂₂m₃₃+2m₁₂m₂₃m₃₁−m₁₁m₂₃²−m₂₂m₃₁²−m₃₃m₁₂²  \n\nyields, after routine algebra,\n\n det Sₘ(a)=  \n ma (a+(m−1)b)(a+(m−1)c)  \n  − m³a³ + 2m³abc  \n  − m²b²(a+(m−1)b)  \n  − m²c²(a+(m−1)c).                             (3)\n\n--------------------------------------------------------------------\n6. Assemble the limit.  \nFrom (2) and continuity of det Sₘ,  \n\n Lₘ = (a−b)^{m−1}(a−c)^{m−1} det Sₘ(a).\n\nSubstituting (3):\n\nLₘ = (a−b)^{m−1}(a−c)^{m−1} · [  \n ma (a+(m−1)b)(a+(m−1)c)  \n − m³a³ + 2m³abc  \n − m²b²(a+(m−1)b)  \n − m²c²(a+(m−1)c)  \n].                                       (4)\n\n--------------------------------------------------------------------\n7. Degree check.  \nThe prefactor (a−b)^{m−1}(a−c)^{m−1} has total degree 2(m−1);  \ndet Sₘ(a) has total degree 3.  \nHence Lₘ is a polynomial of total degree  \n\n 2(m−1)+3 = 2m+1.\n\n--------------------------------------------------------------------\nAnswer.  \n\n Lₘ = (a−b)^{m−1}(a−c)^{m−1} [  \n  ma (a+(m−1)b)(a+(m−1)c)  \n  − m³a³ + 2m³abc  \n  − m²b²(a+(m−1)b)  \n  − m²c²(a+(m−1)c)  \n ].  \n\nThis polynomial in a, b, c has total degree 2m+1, and all coefficients are integral polynomials in m.\n\n——————————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.673910",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension and modular structure – the matrix size grows to \\(3m\\times3m\\) and its entries are governed by congruence classes modulo 3, not merely parity.  \n2. Multiple off–diagonal parameters – three distinct constants \\(a,b,c\\) interact rather than two, producing three different families of eigenvalues.  \n3. Two–level spectral decomposition – one must first diagonalise each \\(m\\times m\\) block via the \\(I_{m}/J_{m}\\) decomposition and then solve a non-trivial \\(3\\times 3\\) determinant on the complementary subspace.  \n4. Delicate limit – only one of the three linear factors vanishes at \\(x=a\\), so the correct order \\((x-a)^{m-1}\\) must be identified; any other power yields either zero or infinity.  \n5. Substantial algebra – the final determinant \\(\\det S_{m}(a)\\) involves quintic expressions in \\(m\\) and cubic expressions in \\(a,b,c\\), considerably more intricate than the quadratic outcome of the original problem.\n\nThese additions force contestants to combine block-matrix algebra, representation-theoretic insight (for the cyclic group of order 3), and heavy symbolic manipulation; simple pattern matching or rank-two observations no longer suffice."
      }
    },
    "original_kernel_variant": {
      "question": "For a positive integer m ≥ 1 and three distinct real numbers a, b, c, define the 3m × 3m matrix  \n  Qₘ(x) = (q_{ij})_{1≤i,j≤3m}, where  \n\n  q_{ij} = ⎧  \n    x,       if i = j,  \n    a,       if i≠j and i+j≡0 (mod 3),  \n    b,       if i≠j and i+j≡1 (mod 3),  \n    c.       if i≠j and i+j≡2 (mod 3).  \n\nEvaluate the limit  \n\n  Lₘ = lim_{x→a} det Qₘ(x) ∕ (x − a)^{\\,m-1},  \n\nand give Lₘ explicitly as a polynomial in a, b, c and m.\n\n——————————————————————————————————————————",
      "solution": "Notation.  \nIₘ = identity_{m×m},   Jₘ = all-ones_{m×m},          \n1 = (1,…,1)ᵀ ∈ ℝᵐ.\n\n--------------------------------------------------------------------\n1. 3 × 3 block form.  \nWrite every index as i = 3s+r (s = 0,…,m–1, r = 0,1,2).  \nAfter cyclicly regrouping rows and columns (0-block first, then 1-block, then 2-block) the matrix becomes a 3 × 3 block matrix  \n\n                     Qₘ(x)=⎛    D₀   B₀₁ B₀₂  \n                                    B₁₀ D₁ B₁₂  \n                                    B₂₀ B₂₁ D₂ ⎞,  \n\nwith  \n\n D₀ = (x−a)Iₘ + aJₘ,             B₀₁ = B₁₀ᵀ = bJₘ,  \n D₁ = (x−c)Iₘ + cJₘ,             B₀₂ = B₂₀ᵀ = cJₘ,  \n D₂ = (x−b)Iₘ + bJₘ,             B₁₂ = B₂₁ᵀ = aJₘ.\n\n--------------------------------------------------------------------\n2. Orthogonal decomposition with respect to Jₘ.  \nLet  \n\n W = {v∈ℝᵐ | 1ᵀv = 0}  (dim W = m−1),  \n\nso Jₘ acts as 0 on W and as multiplication by m on the 1-dimensional complement ⟨1⟩.\n\n--------------------------------------------------------------------\n3. Eigenvalues coming from W.  \nFor each r ∈ {0,1,2} and v∈W insert v in block r and zeros elsewhere.  \nBecause Jₘv = 0 such vectors are eigenvectors of Qₘ(x) with eigenvalue  \n\n λ_r(x) = ⎧x−a if r=0; x−c if r=1; x−b if r=2⎫.\n\nTherefore  \n\n (x−a)^{m−1}(x−b)^{m−1}(x−c)^{m−1}  \n\nis a factor of det Qₘ(x).\n\n--------------------------------------------------------------------\n4. Compression to a 3-dimensional subspace.  \nDefine  \n\n 1_r (r=0,1,2) = vector that equals 1 in block r and 0 elsewhere.\n\nThe vectors 1_r ∕√m are orthonormal and span the orthogonal complement of the 3(m−1)-dimensional space from Step 3.  \nRestricting Qₘ(x) to this 3-space yields the 3×3 matrix  \n\n Sₘ(x)=⎛ x−a+ma   mb   mc  \n      mb x−c+mc  ma  \n      mc  ma x−b+mb ⎞.             (1)\n\nConsequently  \n\n det Qₘ(x) = (x−a)^{m−1}(x−b)^{m−1}(x−c)^{m−1} det Sₘ(x).     (2)\n\n--------------------------------------------------------------------\n5. Evaluate Sₘ(x) at x = a.  \nPutting x = a in (1) gives  \n\n Sₘ(a)=⎛ ma   mb   mc  \n      mb a+(m−1)c  ma  \n      mc  ma a+(m−1)b ⎞.\n\nFor a 3×3 matrix M=(m_{ij}) the identity  \n\n det M = m₁₁m₂₂m₃₃+2m₁₂m₂₃m₃₁−m₁₁m₂₃²−m₂₂m₃₁²−m₃₃m₁₂²  \n\nyields, after routine algebra,\n\n det Sₘ(a)=  \n ma (a+(m−1)b)(a+(m−1)c)  \n  − m³a³ + 2m³abc  \n  − m²b²(a+(m−1)b)  \n  − m²c²(a+(m−1)c).                             (3)\n\n--------------------------------------------------------------------\n6. Assemble the limit.  \nFrom (2) and continuity of det Sₘ,  \n\n Lₘ = (a−b)^{m−1}(a−c)^{m−1} det Sₘ(a).\n\nSubstituting (3):\n\nLₘ = (a−b)^{m−1}(a−c)^{m−1} · [  \n ma (a+(m−1)b)(a+(m−1)c)  \n − m³a³ + 2m³abc  \n − m²b²(a+(m−1)b)  \n − m²c²(a+(m−1)c)  \n].                                       (4)\n\n--------------------------------------------------------------------\n7. Degree check.  \nThe prefactor (a−b)^{m−1}(a−c)^{m−1} has total degree 2(m−1);  \ndet Sₘ(a) has total degree 3.  \nHence Lₘ is a polynomial of total degree  \n\n 2(m−1)+3 = 2m+1.\n\n--------------------------------------------------------------------\nAnswer.  \n\n Lₘ = (a−b)^{m−1}(a−c)^{m−1} [  \n  ma (a+(m−1)b)(a+(m−1)c)  \n  − m³a³ + 2m³abc  \n  − m²b²(a+(m−1)b)  \n  − m²c²(a+(m−1)c)  \n ].  \n\nThis polynomial in a, b, c has total degree 2m+1, and all coefficients are integral polynomials in m.\n\n——————————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.528873",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension and modular structure – the matrix size grows to \\(3m\\times3m\\) and its entries are governed by congruence classes modulo 3, not merely parity.  \n2. Multiple off–diagonal parameters – three distinct constants \\(a,b,c\\) interact rather than two, producing three different families of eigenvalues.  \n3. Two–level spectral decomposition – one must first diagonalise each \\(m\\times m\\) block via the \\(I_{m}/J_{m}\\) decomposition and then solve a non-trivial \\(3\\times 3\\) determinant on the complementary subspace.  \n4. Delicate limit – only one of the three linear factors vanishes at \\(x=a\\), so the correct order \\((x-a)^{m-1}\\) must be identified; any other power yields either zero or infinity.  \n5. Substantial algebra – the final determinant \\(\\det S_{m}(a)\\) involves quintic expressions in \\(m\\) and cubic expressions in \\(a,b,c\\), considerably more intricate than the quadratic outcome of the original problem.\n\nThese additions force contestants to combine block-matrix algebra, representation-theoretic insight (for the cyclic group of order 3), and heavy symbolic manipulation; simple pattern matching or rank-two observations no longer suffice."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}