{
  "index": "1978-A-2",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem A-2\nLet \\( a, b, p_{1}, p_{2}, \\ldots, p_{n} \\) be real numbers with \\( a \\neq b \\). Define \\( f(x)=\\left(p_{1}-x\\right)\\left(p_{2}-x\\right)\\left(p_{3}-x\\right) \\cdots\\left(p_{n}-x\\right) \\). Show that\n\\[\n\\operatorname{det}\\left(\\begin{array}{lllllll}\np_{1} & a & a & a & \\cdots & a & a \\\\\nb & p_{2} & a & a & \\cdots & a & a \\\\\nb & b & p_{3} & a & \\cdots & a & a \\\\\nb & b & b & p_{4} & \\cdots & a & a \\\\\n\\vdots & \\vdots & \\vdots & \\vdots & & \\vdots & \\vdots \\\\\nb & b & b & b & \\cdots & p_{n-1} & a \\\\\nb & b & b & b & \\cdots & b & p_{n}\n\\end{array}\\right)=\\frac{b f(a)-a f(b)}{b-a} .\n\\]",
  "solution": "A-2.\nLet \\( M_{t} \\) be the matrix obtained by subtracting \\( t \\) from each entry of the given matrix and let \\( G(t) \\) be the determinant of \\( M_{t} \\). By subtracting the entries of any row from the corresponding entries of each other row, one sees that \\( G(t) \\) is linear in \\( t \\). Then one notes that \\( G(a)=f(a) \\) and \\( G(b)=f(b) \\) using the fact that they are determinants of triangular matrices. Then linear interpolation shows that the desired determinant \\( G(0) \\) is\n\\[\n[b G(a)-a G(b)] /(b-a)=[b f(a)-a f(b)] /(b-a)\n\\]",
  "vars": [
    "x",
    "t",
    "f",
    "G",
    "M_t"
  ],
  "params": [
    "a",
    "b",
    "n",
    "p_1",
    "p_2",
    "p_3",
    "p_4",
    "p_n-1",
    "p_n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "t": "auxvar",
        "f": "polyfun",
        "G": "detfunc",
        "M_t": "shiftedmat",
        "a": "constanta",
        "b": "constantb",
        "n": "countnum",
        "p_1": "rootone",
        "p_2": "roottwo",
        "p_3": "rootthree",
        "p_4": "rootfour",
        "p_n-1": "rootpenult",
        "p_n": "rootlast"
      },
      "question": "Problem A-2\nLet \\( constanta, constantb, rootone, roottwo, \\ldots, rootlast \\) be real numbers with \\( constanta \\neq constantb \\). Define \\( polyfun(variable)=\\left(rootone-variable\\right)\\left(roottwo-variable\\right)\\left(rootthree-variable\\right) \\cdots\\left(rootlast-variable\\right) \\). Show that\n\\[\n\\operatorname{det}\\left(\\begin{array}{lllllll}\nrootone & constanta & constanta & constanta & \\cdots & constanta & constanta \\\\\nconstantb & roottwo & constanta & constanta & \\cdots & constanta & constanta \\\\\nconstantb & constantb & rootthree & constanta & \\cdots & constanta & constanta \\\\\nconstantb & constantb & constantb & rootfour & \\cdots & constanta & constanta \\\\\n\\vdots & \\vdots & \\vdots & \\vdots & & \\vdots & \\vdots \\\\\nconstantb & constantb & constantb & constantb & \\cdots & rootpenult & constanta \\\\\nconstantb & constantb & constantb & constantb & \\cdots & constantb & rootlast\n\\end{array}\\right)=\\frac{constantb\\, polyfun(constanta)-constanta\\, polyfun(constantb)}{constantb-constanta} .\n\\]",
      "solution": "A-2.\nLet \\( shiftedmat \\) be the matrix obtained by subtracting \\( auxvar \\) from each entry of the given matrix and let \\( detfunc(auxvar) \\) be the determinant of \\( shiftedmat \\). By subtracting the entries of any row from the corresponding entries of each other row, one sees that \\( detfunc(auxvar) \\) is linear in \\( auxvar \\). Then one notes that \\( detfunc(constanta)=polyfun(constanta) \\) and \\( detfunc(constantb)=polyfun(constantb) \\) using the fact that they are determinants of triangular matrices. Then linear interpolation shows that the desired determinant \\( detfunc(0) \\) is\n\\[\n[constantb\\, detfunc(constanta)-constanta\\, detfunc(constantb)] /(constantb-constanta)=[constantb\\, polyfun(constanta)-constanta\\, polyfun(constantb)] /(constantb-constanta)\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marbleseed",
        "t": "granitebox",
        "f": "lanternfly",
        "G": "corridor",
        "M_t": "rainstorm",
        "a": "candlestick",
        "b": "arrowshield",
        "n": "hummingtop",
        "p_1": "sailcloth",
        "p_2": "treelizard",
        "p_3": "spiderling",
        "p_4": "canyonwind",
        "p_n-1": "cloudburst",
        "p_n": "moonflower"
      },
      "question": "Problem A-2\nLet \\( candlestick, arrowshield, sailcloth, treelizard, \\ldots, moonflower \\) be real numbers with \\( candlestick \\neq arrowshield \\). Define \\( lanternfly(marbleseed)=\\left(sailcloth-marbleseed\\right)\\left(treelizard-marbleseed\\right)\\left(spiderling-marbleseed\\right) \\cdots\\left(moonflower-marbleseed\\right) \\). Show that\n\\[\n\\operatorname{det}\\left(\\begin{array}{lllllll}\nsailcloth & candlestick & candlestick & candlestick & \\cdots & candlestick & candlestick \\\\\narrowshield & treelizard & candlestick & candlestick & \\cdots & candlestick & candlestick \\\\\narrowshield & arrowshield & spiderling & candlestick & \\cdots & candlestick & candlestick \\\\\narrowshield & arrowshield & arrowshield & canyonwind & \\cdots & candlestick & candlestick \\\\\n\\vdots & \\vdots & \\vdots & \\vdots & & \\vdots & \\vdots \\\\\narrowshield & arrowshield & arrowshield & arrowshield & \\cdots & p_{n-1} & candlestick \\\\\narrowshield & arrowshield & arrowshield & arrowshield & \\cdots & arrowshield & moonflower\n\\end{array}\\right)=\\frac{arrowshield \\, lanternfly(candlestick)-candlestick \\, lanternfly(arrowshield)}{arrowshield-candlestick} .\n\\]",
      "solution": "A-2.\nLet \\( rainstorm \\) be the matrix obtained by subtracting \\( granitebox \\) from each entry of the given matrix and let \\( corridor(granitebox) \\) be the determinant of \\( rainstorm \\). By subtracting the entries of any row from the corresponding entries of each other row, one sees that \\( corridor(granitebox) \\) is linear in \\( granitebox \\). Then one notes that \\( corridor(candlestick)=lanternfly(candlestick) \\) and \\( corridor(arrowshield)=lanternfly(arrowshield) \\) using the fact that they are determinants of triangular matrices. Then linear interpolation shows that the desired determinant \\( corridor(0) \\) is\n\\[\n[arrowshield \\, corridor(candlestick)-candlestick \\, corridor(arrowshield)] /(arrowshield-candlestick)=[arrowshield \\, lanternfly(candlestick)-candlestick \\, lanternfly(arrowshield)] /(arrowshield-candlestick)\n\\]\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "t": "timeless",
        "f": "unchanging",
        "G": "triviality",
        "M_t": "singlescalar",
        "a": "limitless",
        "b": "confined",
        "n": "infinite",
        "p_1": "expansive",
        "p_2": "boundless",
        "p_3": "measureless",
        "p_4": "unending",
        "p_n-1": "ceaseless",
        "p_n": "eternity"
      },
      "question": "Problem A-2\nLet \\( limitless, confined, expansive, boundless, \\ldots, eternity \\) be real numbers with \\( limitless \\neq confined \\). Define \\( unchanging(fixedvalue)=\\left(expansive-fixedvalue\\right)\\left(boundless-fixedvalue\\right)\\left(measureless-fixedvalue\\right) \\cdots\\left(eternity-fixedvalue\\right) \\). Show that\n\\[\n\\operatorname{det}\\left(\\begin{array}{lllllll}\nexpansive & limitless & limitless & limitless & \\cdots & limitless & limitless \\\\\nconfined & boundless & limitless & limitless & \\cdots & limitless & limitless \\\\\nconfined & confined & measureless & limitless & \\cdots & limitless & limitless \\\\\nconfined & confined & confined & unending & \\cdots & limitless & limitless \\\\\n\\vdots & \\vdots & \\vdots & \\vdots & & \\vdots & \\vdots \\\\\nconfined & confined & confined & confined & \\cdots & ceaseless & limitless \\\\\nconfined & confined & confined & confined & \\cdots & confined & eternity\n\\end{array}\\right)=\\frac{confined\\, unchanging(limitless)-limitless\\, unchanging(confined)}{confined-limitless} .\n\\]",
      "solution": "A-2.\nLet \\( singlescalar \\) be the matrix obtained by subtracting \\( timeless \\) from each entry of the given matrix and let \\( triviality(timeless) \\) be the determinant of \\( singlescalar \\). By subtracting the entries of any row from the corresponding entries of each other row, one sees that \\( triviality(timeless) \\) is linear in \\( timeless \\). Then one notes that \\( triviality(limitless)=unchanging(limitless) \\) and \\( triviality(confined)=unchanging(confined) \\) using the fact that they are determinants of triangular matrices. Then linear interpolation shows that the desired determinant \\( triviality(0) \\) is\n\\[\n[confined\\, triviality(limitless)-limitless\\, triviality(confined)] /(confined-limitless)=[confined\\, unchanging(limitless)-limitless\\, unchanging(confined)] /(confined-limitless)\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "vaqsitlum",
        "t": "zyoqkner",
        "f": "hdusipwq",
        "G": "iklabsyv",
        "M_t": "qmaorbel",
        "a": "wexafnup",
        "b": "cidromal",
        "n": "ybregtis",
        "p_1": "ufmaclep",
        "p_2": "zirdonex",
        "p_3": "qbafliro",
        "p_4": "mgrailto",
        "p_n-1": "yodgreps",
        "p_n": "hanjoklu"
      },
      "question": "Problem A-2\nLet \\( wexafnup, cidromal, ufmaclep, zirdonex, \\ldots, hanjoklu \\) be real numbers with \\( wexafnup \\neq cidromal \\). Define \\( hdusipwq(vaqsitlum)=\\left(ufmaclep-vaqsitlum\\right)\\left(zirdonex-vaqsitlum\\right)\\left(qbafliro-vaqsitlum\\right) \\cdots\\left(hanjoklu-vaqsitlum\\right) \\). Show that\n\\[\n\\operatorname{det}\\left(\\begin{array}{lllllll}\nufmaclep & wexafnup & wexafnup & wexafnup & \\cdots & wexafnup & wexafnup \\\\\ncidromal & zirdonex & wexafnup & wexafnup & \\cdots & wexafnup & wexafnup \\\\\ncidromal & cidromal & qbafliro & wexafnup & \\cdots & wexafnup & wexafnup \\\\\ncidromal & cidromal & cidromal & mgrailto & \\cdots & wexafnup & wexafnup \\\\\n\\vdots & \\vdots & \\vdots & \\vdots & & \\vdots & \\vdots \\\\\ncidromal & cidromal & cidromal & cidromal & \\cdots & yodgreps & wexafnup \\\\\ncidromal & cidromal & cidromal & cidromal & \\cdots & cidromal & hanjoklu\n\\end{array}\\right)=\\frac{cidromal \\, hdusipwq(wexafnup)-wexafnup \\, hdusipwq(cidromal)}{cidromal-wexafnup} .\n\\]",
      "solution": "A-2.\nLet \\( qmaorbel \\) be the matrix obtained by subtracting \\( zyoqkner \\) from each entry of the given matrix and let \\( iklabsyv(zyoqkner) \\) be the determinant of \\( qmaorbel \\). By subtracting the entries of any row from the corresponding entries of each other row, one sees that \\( iklabsyv(zyoqkner) \\) is linear in \\( zyoqkner \\). Then one notes that \\( iklabsyv(wexafnup)=hdusipwq(wexafnup) \\) and \\( iklabsyv(cidromal)=hdusipwq(cidromal) \\) using the fact that they are determinants of triangular matrices. Then linear interpolation shows that the desired determinant \\( iklabsyv(0) \\) is\n\\[\n[cidromal \\, iklabsyv(wexafnup)-wexafnup \\, iklabsyv(cidromal)] /(cidromal-wexafnup)=[cidromal \\, hdusipwq(wexafnup)-wexafnup \\, hdusipwq(cidromal)] /(cidromal-wexafnup)\n\\]"
    },
    "kernel_variant": {
      "question": "Let $n\\ge 2$ be an integer.  Fix real numbers  \n\n\\[\np_{1},p_{2},\\ldots ,p_{n}\\qquad (\\text{the diagonal data})\n\\]\n\nand two $n$-vectors  \n\n\\[\nu=(u_{1},u_{2},\\ldots ,u_{n})^{\\mathsf T},\\qquad \nv=(v_{1},v_{2},\\ldots ,v_{n})^{\\mathsf T}\\qquad \n(\\text{the row/column parameters})\n\\]\n\nsuch that  \n\\[\np_{i}\\neq u_{i}+v_{i}\\qquad\\text{for every }i=1,\\ldots ,n. \\tag{$\\dagger$}\n\\]\n\nDefine the $n\\times n$ matrix $M=(m_{ij})$ by  \n\n\\[\nm_{ii}=p_{i},\\qquad m_{ij}=u_{i}+v_{j}\\;\\;(i\\neq j). \\tag{$\\star$}\n\\]\n\nPut  \n\n\\[\nd_{i}=p_{i}-u_{i}-v_{i}\\quad(\\text{so }d_{i}\\neq 0\\text{ by }(\\dagger)), \n\\qquad D=\\operatorname{diag}(d_{1},\\ldots ,d_{n}).\n\\]\n\n(a) Prove that  \n\n\\[\n\\det M=\\Bigl(\\prod_{i=1}^{n} d_{i}\\Bigr)\\,\\bigl[(1+A)(1+B)-C\\,E\\bigr], \\tag{1}\n\\]\n\nwhere  \n\n\\[\nA=\\sum_{i=1}^{n}\\frac{u_{i}}{d_{i}},\\qquad \nB=\\sum_{i=1}^{n}\\frac{v_{i}}{d_{i}},\\qquad \nC=\\sum_{i=1}^{n}\\frac{1}{d_{i}},\\qquad \nE=\\sum_{i=1}^{n}\\frac{u_{i}v_{i}}{d_{i}}. \\tag{2}\n\\]\n\n(b) Show that the determinant collapses to the rank-$1$ form  \n\n\\[\n\\det M=\\Bigl(\\prod_{i=1}^{n} d_{i}\\Bigr)\\,(1-E) \\tag{3}\n\\]\n\nif and only if the four rational sums $(A,B,C,E)$ satisfy the single algebraic relation  \n\n\\[\nA+B+AB=(C-1)E. \\tag{4}\n\\]\n\nGive two non-trivial, explicitly parametrised families of data $(p,u,v)$ that meet condition (4).\n\nProvide complete justification for every assertion.",
      "solution": "Throughout write  \n\n\\[\nd_{i}=p_{i}-u_{i}-v_{i},\\qquad D=\\operatorname{diag}(d_{1},\\ldots ,d_{n}).\n\\]\n\n--------------------------------------------------------------------\nPart (a).  Writing $M$ as a rank-two perturbation of $D$\n--------------------------------------------------------------------\nIntroduce the vectors and covectors  \n\n\\[\nx:=u,\\qquad y:=\\mathbf 1,\\qquad \n\\alpha^{\\mathsf T}:=\\mathbf 1^{\\mathsf T},\\qquad \n\\beta^{\\mathsf T}:=v^{\\mathsf T},\n\\]\n\nwhere $\\mathbf 1$ denotes the column vector $(1,1,\\ldots ,1)^{\\mathsf T}$.  \nFor $i\\neq j$ we have  \n\n\\[\n(D+x\\alpha^{\\mathsf T}+y\\beta^{\\mathsf T})_{ij}=0+u_{i}\\!\\cdot\\!1+1\\!\\cdot\\! v_{j}=u_{i}+v_{j},\n\\]\n\nwhile for $i=j$  \n\n\\[\n(D+x\\alpha^{\\mathsf T}+y\\beta^{\\mathsf T})_{ii}=d_{i}+u_{i}\\!\\cdot\\! 1+1\\!\\cdot\\! v_{i}=p_{i}.\n\\]\n\nHence  \n\\[\nM=D+x\\alpha^{\\mathsf T}+y\\beta^{\\mathsf T}. \\tag{5}\n\\]\nThus $M$ is a rank-$2$ update of the invertible diagonal matrix $D$.\n\n--------------------------------------------------------------------\nWoodbury’s determinant lemma\n--------------------------------------------------------------------\nFor any invertible matrix $N$ and any $n\\times k$ matrices $U,V$ one has  \n\n\\[\n\\det(N+UV^{\\mathsf T})=\\det N\\cdot\\det(I_{k}+V^{\\mathsf T}N^{-1}U). \\tag{6}\n\\]\n\nHere $k=2$ and  \n\n\\[\nU=[\\,x\\; y\\,],\\qquad V=[\\,\\alpha\\; \\beta\\,].\n\\]\n\nConsequently  \n\n\\[\n\\det M=\\det D\\;\\det\\!\\bigl(I_{2}+V^{\\mathsf T}D^{-1}U\\bigr). \\tag{7}\n\\]\n\n--------------------------------------------------------------------\nEvaluating the $2\\times 2$ auxiliary determinant\n--------------------------------------------------------------------\nBecause $D$ is diagonal, $D^{-1}=\\operatorname{diag}(1/d_{1},\\ldots ,1/d_{n})$.  \nA direct multiplication gives  \n\n\\[\nV^{\\mathsf T}D^{-1}U=\n\\begin{pmatrix}\n\\alpha^{\\mathsf T}D^{-1}x & \\alpha^{\\mathsf T}D^{-1}y\\\\[2pt]\n\\beta^{\\mathsf T}D^{-1}x  & \\beta^{\\mathsf T}D^{-1}y\n\\end{pmatrix}\n=\n\\begin{pmatrix}\nA & C\\\\[2pt]\nE & B\n\\end{pmatrix}, \\tag{8}\n\\]\nwith $A,B,C,E$ as in (2). Hence  \n\n\\[\n\\det\\!\\bigl(I_{2}+V^{\\mathsf T}D^{-1}U\\bigr)\n=\\det\n\\begin{pmatrix}\n1+A & C\\\\[2pt]\nE & 1+B\n\\end{pmatrix}\n=(1+A)(1+B)-C\\,E. \\tag{9}\n\\]\n\n--------------------------------------------------------------------\nPutting the pieces together\n--------------------------------------------------------------------\nSince $\\det D=\\prod_{i=1}^{n}d_{i}$, equations (7) and (9) give precisely formula (1), completing part (a). \\hfill $\\square$\n\n\n\n--------------------------------------------------------------------\nPart (b).  When does the determinant reduce to rank $1$?\n--------------------------------------------------------------------\n\nNecessity.  Imposing the rank-$1$ collapse (3) on the general formula (1) requires  \n\n\\[\n(1+A)(1+B)-C\\,E=1-E. \\tag{10}\n\\]\n\nExpanding the left side yields  \n\n\\[\n1+A+B+AB-CE=1-E. \\tag{11}\n\\]\n\nCancelling the leading $1$’s shows that (10) is equivalent to  \n\n\\[\nA+B+AB=(C-1)E, \\tag{12}\n\\]\ni.e. to condition (4).  Thus (4) is necessary.\n\nSufficiency.  Conversely, if (4) holds then the identity (11) is valid, so (10) follows and with it (3).  Hence (4) is also sufficient. \\hfill $\\square$\n\n\n\n--------------------------------------------------------------------\nTwo non-trivial parametrised families satisfying (4)\n--------------------------------------------------------------------\nWe rewrite (4) once more as  \n\\[\n(A+1)(B+1)=1+CE. \\tag{13}\n\\]\n\nBoth examples below meet this equation for all admissible data in the family.\n\nFamily I – sign-anti-symmetric parameters  \nFix positive numbers $p_{1},\\ldots ,p_{n}$ with $\\sum_{i=1}^{n}1/p_{i}=1$.  \nLet $w=(w_{1},\\ldots ,w_{n})^{\\mathsf T}$ be any vector satisfying  \n$\\sum_{i=1}^{n}w_{i}/p_{i}=0$, and set  \n\n\\[\nu=w,\\qquad v=-w.\n\\]\n\nThen $d_{i}=p_{i}$, so  \n\\[\nC=\\sum_{i=1}^{n}\\frac{1}{p_{i}}=1,\\qquad \nA=\\sum_{i=1}^{n}\\frac{w_{i}}{p_{i}}=0,\\qquad \nB=-A=0,\\qquad \nE=\\sum_{i=1}^{n}\\frac{w_{i}(-w_{i})}{p_{i}}=-\\sum_{i=1}^{n}\\frac{w_{i}^{2}}{p_{i}}.\n\\]\nHence $A+B+AB=0$ and $(C-1)E=0$, verifying (4) for every choice of $w$.  \nThe determinant therefore collapses to (3).\n\nFamily II – constant row/column parameters  \nFix arbitrary diagonal data $p_{1},\\ldots ,p_{n}$ and choose constants $a,b\\in\\mathbb R$ satisfying  \n\n\\[\nab+a+b=0\\qquad\\Longleftrightarrow\\qquad (1+a)(1+b)=1. \\tag{14}\n\\]\n\n(Thus neither constant equals $-1$.)  Define  \n\n\\[\nu=(a,a,\\ldots ,a)^{\\mathsf T},\\qquad v=(b,b,\\ldots ,b)^{\\mathsf T}.\n\\]\n\nThen $d_{i}=p_{i}-(a+b)\\neq 0$ by assumption, and  \n\n\\[\nC=\\sum_{i=1}^{n}\\frac{1}{d_{i}},\\qquad \nA=aC,\\qquad \nB=bC,\\qquad \nE=abC.\n\\]\n\nSubstituting into (4) gives  \n\n\\[\nA+B+AB=aC+bC+abC^{2},\\qquad (C-1)E=(C-1)abC.\n\\]\n\nEquality holds because $a+b+ab=0$ by (14).  Thus every admissible choice of $(p_{i})$ with $p_{i}\\neq a+b$ satisfies (4).  \nFor instance $a=1,\\;b=-\\tfrac12$ or $a=-\\tfrac13,\\;b=\\tfrac12$ produce valid data.\n\nBoth families are genuinely multi-parameter and illustrate the delicate balancing embodied in equation (4). \\hfill $\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.634872",
        "was_fixed": false,
        "difficulty_analysis": "1. Rank-Two Perturbation:  \n   The original problem involved a single auxiliary parameter and could be handled\n   by elementary row operations leading to a linear polynomial.  The enhanced\n   variant treats a diagonal matrix modified simultaneously by two independent\n   rank-one updates, forcing the use of the full Woodbury (rank-k) determinant\n   formula.  This is a substantial theoretical leap.\n\n2. More Variables and Parameters:  \n   Instead of two scalars (a,b) the new matrix depends on 3n independent\n   parameters (pᵢ, uᵢ, vᵢ), creating a vastly richer configuration space and far\n   more complicated algebraic expressions.\n\n3. Interaction of Concepts:  \n   The solution blends several advanced topics – low-rank updates, block\n   determinants, and careful bookkeeping of symmetric sums – none of which appear\n   in the original statement.\n\n4. Non-trivial Specialisation:  \n   Part (b) asks for an additional simplification under linear constraints on the\n   data, compelling the contestant to manipulate the general formula further and\n   detect hidden cancellations; this layer of subtlety is entirely absent from\n   the source problem.\n\n5. Absence of Pattern Matching:  \n   Elementary pattern recognition (e.g. “subtract a from each entry”) is no\n   longer adequate.  One must recognise the rank-two structure, recall an\n   advanced determinant identity, and carry out a sequence of non-routine\n   computations.\n\nConsequently the enhanced variant is markedly more demanding both technically\nand conceptually than the original kernel problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n\\ge 2$ be an integer.  Fix real numbers  \n\n\\[\np_{1},p_{2},\\ldots ,p_{n}\\qquad (\\text{the diagonal data})\n\\]\n\nand two $n$-vectors  \n\n\\[\nu=(u_{1},u_{2},\\ldots ,u_{n})^{\\mathsf T},\\qquad \nv=(v_{1},v_{2},\\ldots ,v_{n})^{\\mathsf T}\\qquad \n(\\text{the row/column parameters})\n\\]\n\nsuch that  \n\\[\np_{i}\\neq u_{i}+v_{i}\\qquad\\text{for every }i=1,\\ldots ,n. \\tag{$\\dagger$}\n\\]\n\nDefine the $n\\times n$ matrix $M=(m_{ij})$ by  \n\n\\[\nm_{ii}=p_{i},\\qquad m_{ij}=u_{i}+v_{j}\\;\\;(i\\neq j). \\tag{$\\star$}\n\\]\n\nPut  \n\n\\[\nd_{i}=p_{i}-u_{i}-v_{i}\\quad(\\text{so }d_{i}\\neq 0\\text{ by }(\\dagger)), \n\\qquad D=\\operatorname{diag}(d_{1},\\ldots ,d_{n}).\n\\]\n\n(a) Prove that  \n\n\\[\n\\det M=\\Bigl(\\prod_{i=1}^{n} d_{i}\\Bigr)\\,\\bigl[(1+A)(1+B)-C\\,E\\bigr], \\tag{1}\n\\]\n\nwhere  \n\n\\[\nA=\\sum_{i=1}^{n}\\frac{u_{i}}{d_{i}},\\qquad \nB=\\sum_{i=1}^{n}\\frac{v_{i}}{d_{i}},\\qquad \nC=\\sum_{i=1}^{n}\\frac{1}{d_{i}},\\qquad \nE=\\sum_{i=1}^{n}\\frac{u_{i}v_{i}}{d_{i}}. \\tag{2}\n\\]\n\n(b) Show that the determinant collapses to the rank-$1$ form  \n\n\\[\n\\det M=\\Bigl(\\prod_{i=1}^{n} d_{i}\\Bigr)\\,(1-E) \\tag{3}\n\\]\n\nif and only if the four rational sums $(A,B,C,E)$ satisfy the single algebraic relation  \n\n\\[\nA+B+AB=(C-1)E. \\tag{4}\n\\]\n\nGive two non-trivial, explicitly parametrised families of data $(p,u,v)$ that meet condition (4).\n\nProvide complete justification for every assertion.",
      "solution": "Throughout write  \n\n\\[\nd_{i}=p_{i}-u_{i}-v_{i},\\qquad D=\\operatorname{diag}(d_{1},\\ldots ,d_{n}).\n\\]\n\n--------------------------------------------------------------------\nPart (a).  Writing $M$ as a rank-two perturbation of $D$\n--------------------------------------------------------------------\nIntroduce the vectors and covectors  \n\n\\[\nx:=u,\\qquad y:=\\mathbf 1,\\qquad \n\\alpha^{\\mathsf T}:=\\mathbf 1^{\\mathsf T},\\qquad \n\\beta^{\\mathsf T}:=v^{\\mathsf T},\n\\]\n\nwhere $\\mathbf 1$ denotes the column vector $(1,1,\\ldots ,1)^{\\mathsf T}$.  \nFor $i\\neq j$ we have  \n\n\\[\n(D+x\\alpha^{\\mathsf T}+y\\beta^{\\mathsf T})_{ij}=0+u_{i}\\!\\cdot\\!1+1\\!\\cdot\\! v_{j}=u_{i}+v_{j},\n\\]\n\nwhile for $i=j$  \n\n\\[\n(D+x\\alpha^{\\mathsf T}+y\\beta^{\\mathsf T})_{ii}=d_{i}+u_{i}\\!\\cdot\\! 1+1\\!\\cdot\\! v_{i}=p_{i}.\n\\]\n\nHence  \n\\[\nM=D+x\\alpha^{\\mathsf T}+y\\beta^{\\mathsf T}. \\tag{5}\n\\]\nThus $M$ is a rank-$2$ update of the invertible diagonal matrix $D$.\n\n--------------------------------------------------------------------\nWoodbury’s determinant lemma\n--------------------------------------------------------------------\nFor any invertible matrix $N$ and any $n\\times k$ matrices $U,V$ one has  \n\n\\[\n\\det(N+UV^{\\mathsf T})=\\det N\\cdot\\det(I_{k}+V^{\\mathsf T}N^{-1}U). \\tag{6}\n\\]\n\nHere $k=2$ and  \n\n\\[\nU=[\\,x\\; y\\,],\\qquad V=[\\,\\alpha\\; \\beta\\,].\n\\]\n\nConsequently  \n\n\\[\n\\det M=\\det D\\;\\det\\!\\bigl(I_{2}+V^{\\mathsf T}D^{-1}U\\bigr). \\tag{7}\n\\]\n\n--------------------------------------------------------------------\nEvaluating the $2\\times 2$ auxiliary determinant\n--------------------------------------------------------------------\nBecause $D$ is diagonal, $D^{-1}=\\operatorname{diag}(1/d_{1},\\ldots ,1/d_{n})$.  \nA direct multiplication gives  \n\n\\[\nV^{\\mathsf T}D^{-1}U=\n\\begin{pmatrix}\n\\alpha^{\\mathsf T}D^{-1}x & \\alpha^{\\mathsf T}D^{-1}y\\\\[2pt]\n\\beta^{\\mathsf T}D^{-1}x  & \\beta^{\\mathsf T}D^{-1}y\n\\end{pmatrix}\n=\n\\begin{pmatrix}\nA & C\\\\[2pt]\nE & B\n\\end{pmatrix}, \\tag{8}\n\\]\nwith $A,B,C,E$ as in (2). Hence  \n\n\\[\n\\det\\!\\bigl(I_{2}+V^{\\mathsf T}D^{-1}U\\bigr)\n=\\det\n\\begin{pmatrix}\n1+A & C\\\\[2pt]\nE & 1+B\n\\end{pmatrix}\n=(1+A)(1+B)-C\\,E. \\tag{9}\n\\]\n\n--------------------------------------------------------------------\nPutting the pieces together\n--------------------------------------------------------------------\nSince $\\det D=\\prod_{i=1}^{n}d_{i}$, equations (7) and (9) give precisely formula (1), completing part (a). \\hfill $\\square$\n\n\n\n--------------------------------------------------------------------\nPart (b).  When does the determinant reduce to rank $1$?\n--------------------------------------------------------------------\n\nNecessity.  Imposing the rank-$1$ collapse (3) on the general formula (1) requires  \n\n\\[\n(1+A)(1+B)-C\\,E=1-E. \\tag{10}\n\\]\n\nExpanding the left side yields  \n\n\\[\n1+A+B+AB-CE=1-E. \\tag{11}\n\\]\n\nCancelling the leading $1$’s shows that (10) is equivalent to  \n\n\\[\nA+B+AB=(C-1)E, \\tag{12}\n\\]\ni.e. to condition (4).  Thus (4) is necessary.\n\nSufficiency.  Conversely, if (4) holds then the identity (11) is valid, so (10) follows and with it (3).  Hence (4) is also sufficient. \\hfill $\\square$\n\n\n\n--------------------------------------------------------------------\nTwo non-trivial parametrised families satisfying (4)\n--------------------------------------------------------------------\nWe rewrite (4) once more as  \n\\[\n(A+1)(B+1)=1+CE. \\tag{13}\n\\]\n\nBoth examples below meet this equation for all admissible data in the family.\n\nFamily I – sign-anti-symmetric parameters  \nFix positive numbers $p_{1},\\ldots ,p_{n}$ with $\\sum_{i=1}^{n}1/p_{i}=1$.  \nLet $w=(w_{1},\\ldots ,w_{n})^{\\mathsf T}$ be any vector satisfying  \n$\\sum_{i=1}^{n}w_{i}/p_{i}=0$, and set  \n\n\\[\nu=w,\\qquad v=-w.\n\\]\n\nThen $d_{i}=p_{i}$, so  \n\\[\nC=\\sum_{i=1}^{n}\\frac{1}{p_{i}}=1,\\qquad \nA=\\sum_{i=1}^{n}\\frac{w_{i}}{p_{i}}=0,\\qquad \nB=-A=0,\\qquad \nE=\\sum_{i=1}^{n}\\frac{w_{i}(-w_{i})}{p_{i}}=-\\sum_{i=1}^{n}\\frac{w_{i}^{2}}{p_{i}}.\n\\]\nHence $A+B+AB=0$ and $(C-1)E=0$, verifying (4) for every choice of $w$.  \nThe determinant therefore collapses to (3).\n\nFamily II – constant row/column parameters  \nFix arbitrary diagonal data $p_{1},\\ldots ,p_{n}$ and choose constants $a,b\\in\\mathbb R$ satisfying  \n\n\\[\nab+a+b=0\\qquad\\Longleftrightarrow\\qquad (1+a)(1+b)=1. \\tag{14}\n\\]\n\n(Thus neither constant equals $-1$.)  Define  \n\n\\[\nu=(a,a,\\ldots ,a)^{\\mathsf T},\\qquad v=(b,b,\\ldots ,b)^{\\mathsf T}.\n\\]\n\nThen $d_{i}=p_{i}-(a+b)\\neq 0$ by assumption, and  \n\n\\[\nC=\\sum_{i=1}^{n}\\frac{1}{d_{i}},\\qquad \nA=aC,\\qquad \nB=bC,\\qquad \nE=abC.\n\\]\n\nSubstituting into (4) gives  \n\n\\[\nA+B+AB=aC+bC+abC^{2},\\qquad (C-1)E=(C-1)abC.\n\\]\n\nEquality holds because $a+b+ab=0$ by (14).  Thus every admissible choice of $(p_{i})$ with $p_{i}\\neq a+b$ satisfies (4).  \nFor instance $a=1,\\;b=-\\tfrac12$ or $a=-\\tfrac13,\\;b=\\tfrac12$ produce valid data.\n\nBoth families are genuinely multi-parameter and illustrate the delicate balancing embodied in equation (4). \\hfill $\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.505344",
        "was_fixed": false,
        "difficulty_analysis": "1. Rank-Two Perturbation:  \n   The original problem involved a single auxiliary parameter and could be handled\n   by elementary row operations leading to a linear polynomial.  The enhanced\n   variant treats a diagonal matrix modified simultaneously by two independent\n   rank-one updates, forcing the use of the full Woodbury (rank-k) determinant\n   formula.  This is a substantial theoretical leap.\n\n2. More Variables and Parameters:  \n   Instead of two scalars (a,b) the new matrix depends on 3n independent\n   parameters (pᵢ, uᵢ, vᵢ), creating a vastly richer configuration space and far\n   more complicated algebraic expressions.\n\n3. Interaction of Concepts:  \n   The solution blends several advanced topics – low-rank updates, block\n   determinants, and careful bookkeeping of symmetric sums – none of which appear\n   in the original statement.\n\n4. Non-trivial Specialisation:  \n   Part (b) asks for an additional simplification under linear constraints on the\n   data, compelling the contestant to manipulate the general formula further and\n   detect hidden cancellations; this layer of subtlety is entirely absent from\n   the source problem.\n\n5. Absence of Pattern Matching:  \n   Elementary pattern recognition (e.g. “subtract a from each entry”) is no\n   longer adequate.  One must recognise the rank-two structure, recall an\n   advanced determinant identity, and carry out a sequence of non-routine\n   computations.\n\nConsequently the enhanced variant is markedly more demanding both technically\nand conceptually than the original kernel problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}