{
  "index": "1951-A-1",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "1. Show that the determinant:\n\\[\n\\left|\\begin{array}{rrrr}\n0 & a & b & c \\\\\n-a & 0 & d & e \\\\\n-b & -d & 0 & f \\\\\n-c & -e & -f & 0\n\\end{array}\\right|\n\\]\nis non-negative, if its elements \\( a, b, c \\), etc., are real.",
  "solution": "Solution. Assume \\( f \\neq 0 \\). Divide the third row and the third column by \\( f \\), and write \\( \\bar{b}=b / f, \\bar{d}=d / f \\). Then, if \\( \\Delta \\) is the determinant,\n\\[\n\\frac{1}{f^{2}} \\Delta=\\left|\\begin{array}{rrrr}\n0 & a & \\bar{b} & c \\\\\n-a & 0 & \\bar{d} & e \\\\\n-\\bar{b} & -\\bar{d} & 0 & 1 \\\\\n-c & -e & -1 & 0\n\\end{array}\\right|\n\\]\n\nSubtract \\( c \\) times the third row from the first row and \\( e \\) times the third row from the second row.\n\\[\n\\frac{1}{f^{2}} \\Delta=\\left|\\begin{array}{cccc}\n\\bar{b} c & a+\\bar{d} c & \\bar{b} & 0 \\\\\n-a+\\bar{b} e & \\bar{d} e & \\bar{d} & 0 \\\\\n-\\bar{b} & -\\bar{d} & 0 & 1 \\\\\n-c & -e & -1 & 0\n\\end{array}\\right|\n\\]\n\nMake similar transformations of the columns.\n\\[\n\\begin{aligned}\n\\frac{1}{f^{2}} \\Delta & =\\left|\\begin{array}{cccc}\n0 & a+\\bar{d} c-\\bar{b} e & \\bar{b} & 0 \\\\\n-a+\\bar{b} e-\\bar{d} c & 0 & \\bar{d} & 0 \\\\\n-\\bar{b} & -\\bar{d} & 0 & 1 \\\\\n0 & 0 & -1 & 0\n\\end{array}\\right| \\\\\n& =\\left|\\begin{array}{cc}\n0 & a+c \\bar{d}-\\bar{b} e \\\\\n-a+\\bar{b} e-c \\bar{d} & 0\n\\end{array}\\right| \\\\\n& =(a+c \\bar{d}-\\bar{b} e)^{2} .\n\\end{aligned}\n\\]\n\nFinally, \\( \\Delta=(a f+c d-b e)^{2} \\), and therefore \\( \\Delta \\) is non-negative for any real choice of \\( a, b, c, d, e, f \\).\n\nNote that if \\( f=0 \\), the matrix has more zeros, and an elementary application of Laplace's expansion of a determinant gives \\( \\Delta=(c d-b e)^{2} \\). (Alternatively, this can also be seen by letting \\( f \\rightarrow 0 \\) in the above formula \\( \\Delta=(a f+c d-b e)^{2} \\).) More generally, since \\( \\Delta \\) is obviously a polynomial in the matrix entries, the assumption \\( f \\neq 0 \\) is really no loss of generality in evaluating \\( \\Delta \\). In the language of algebra, the computation was really made in the field \\( Q(a, b, c, d, e, f) \\) where \\( a, b, c, d, e \\), and \\( f \\) are indeterminates. In this field, \\( f \\neq 0 \\).\n\nRemark. This proof was chosen because it generalizes easily to prove the following fact by induction: The determinant of a skew-symmetric matrix of even order is the square of a polynomial in the matrix entries. (We show below that the determinant of a skew-symmetric matrix of odd order is zero.) In fact the determinant of the \\( 2 n \\times 2 n \\) skew-symmetric matrix \\( \\left(a_{i j}\\right) \\) is the square of\n\\[\n\\frac{1}{2 \\cdot 4 \\cdots 2 n} \\sum_{\\sigma}(-1)^{\\sigma} a_{\\sigma(1), \\sigma(2)} a_{\\sigma(3), \\sigma(4)} \\cdots a_{\\sigma(2 n-1), \\sigma(2 n)}\n\\]\nwhere \\( (-1)^{\\circ} \\) denotes the sign of the permutation \\( \\sigma \\) and the sum is taken over all members \\( \\sigma \\) of the symmetric group. This polynomial was called the Pfaffian by Cayley because Jacobi had used it in his work on Pfaff's problem. Because the \\( a \\) 's are skew-symmetric, the Pfaffian can be written as the sum of \\( 1 \\cdot 3 \\cdot 5 \\cdots(2 n-1) \\) products without denominators. See Thomas Muir, The Theory of. Determinants in the Historical Order of Development, Vol. 1, Dover, New York, pages 401-406.\n\nThe determinant of a skew-symmetric matrix \\( S \\) of odd order \\( n \\) is always zero. We have\n\\[\n(-1)^{n} \\operatorname{det} S=\\operatorname{det}(-S)=\\operatorname{det} S^{T}=\\operatorname{det} S ;\n\\]\nso, for \\( n \\) odd, \\( \\operatorname{det} S=0 \\).",
  "vars": [
    "a",
    "b",
    "c",
    "d",
    "e",
    "f",
    "\\\\Delta",
    "n",
    "\\\\sigma",
    "S",
    "i",
    "j",
    "a_ij"
  ],
  "params": [
    "Q"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "varalpha",
        "b": "varbeta",
        "c": "vargamma",
        "d": "vardelta",
        "e": "varepsilon",
        "f": "varphi",
        "\\Delta": "determinant",
        "n": "dimension",
        "\\sigma": "permutation",
        "S": "skewsymmetric",
        "i": "rowindex",
        "j": "colindex",
        "a_ij": "entrygeneric",
        "Q": "rationalfield"
      },
      "question": "1. Show that the determinant:\n\\[\n\\left|\\begin{array}{rrrr}\n0 & varalpha & varbeta & vargamma \\\\\n-varalpha & 0 & vardelta & varepsilon \\\\\n-varbeta & -vardelta & 0 & varphi \\\\\n-vargamma & -varepsilon & -varphi & 0\n\\end{array}\\right|\n\\]\nis non-negative, if its elements \\( varalpha, varbeta, vargamma \\), etc., are real.",
      "solution": "Solution. Assume \\( varphi \\neq 0 \\). Divide the third row and the third column by \\( varphi \\), and write \\( \\bar{varbeta}=varbeta / varphi, \\bar{vardelta}=vardelta / varphi \\). Then, if \\( determinant \\) is the determinant,\n\\[\n\\frac{1}{varphi^{2}}\\, determinant=\\left|\\begin{array}{rrrr}\n0 & varalpha & \\bar{varbeta} & vargamma \\\\\n-varalpha & 0 & \\bar{vardelta} & varepsilon \\\\\n-\\bar{varbeta} & -\\bar{vardelta} & 0 & 1 \\\\\n-vargamma & -varepsilon & -1 & 0\n\\end{array}\\right|\n\\]\n\nSubtract \\( vargamma \\) times the third row from the first row and \\( varepsilon \\) times the third row from the second row.\n\\[\n\\frac{1}{varphi^{2}}\\, determinant=\\left|\\begin{array}{cccc}\n\\bar{varbeta}\\, vargamma & varalpha+\\bar{vardelta}\\, vargamma & \\bar{varbeta} & 0 \\\\\n-varalpha+\\bar{varbeta}\\, varepsilon & \\bar{vardelta}\\, varepsilon & \\bar{vardelta} & 0 \\\\\n-\\bar{varbeta} & -\\bar{vardelta} & 0 & 1 \\\\\n-vargamma & -varepsilon & -1 & 0\n\\end{array}\\right|\n\\]\n\nMake similar transformations of the columns.\n\\[\n\\begin{aligned}\n\\frac{1}{varphi^{2}}\\, determinant & =\\left|\\begin{array}{cccc}\n0 & varalpha+vargamma \\bar{vardelta}-\\bar{varbeta}\\, varepsilon & \\bar{varbeta} & 0 \\\\\n-varalpha+\\bar{varbeta}\\, varepsilon-\\bar{vardelta}\\, vargamma & 0 & \\bar{vardelta} & 0 \\\\\n-\\bar{varbeta} & -\\bar{vardelta} & 0 & 1 \\\\\n0 & 0 & -1 & 0\n\\end{array}\\right| \\\\\n& =\\left|\\begin{array}{cc}\n0 & varalpha+vargamma \\bar{vardelta}-\\bar{varbeta}\\, varepsilon \\\\\n-varalpha+\\bar{varbeta}\\, varepsilon-\\bar{vardelta}\\, vargamma & 0\n\\end{array}\\right| \\\\\n& =(varalpha+vargamma \\bar{vardelta}-\\bar{varbeta}\\, varepsilon)^{2} .\n\\end{aligned}\n\\]\n\nFinally, \\( determinant=(varalpha varphi+vargamma vardelta-varbeta varepsilon)^{2} \\), and therefore \\( determinant \\) is non-negative for any real choice of \\( varalpha, varbeta, vargamma, vardelta, varepsilon, varphi \\).\n\nNote that if \\( varphi=0 \\), the matrix has more zeros, and an elementary application of Laplace's expansion of a determinant gives \\( determinant=(vargamma vardelta-varbeta varepsilon)^{2} \\). (Alternatively, this can also be seen by letting \\( varphi \\rightarrow 0 \\) in the above formula \\( determinant=(varalpha varphi+vargamma vardelta-varbeta varepsilon)^{2} \\).) More generally, since \\( determinant \\) is obviously a polynomial in the matrix entries, the assumption \\( varphi \\neq 0 \\) is really no loss of generality in evaluating \\( determinant \\). In the language of algebra, the computation was really made in the field \\( rationalfield(varalpha, varbeta, vargamma, vardelta, varepsilon, varphi) \\) where \\( varalpha, varbeta, vargamma, vardelta, varepsilon \\), and \\( varphi \\) are indeterminates. In this field, \\( varphi \\neq 0 \\).\n\nRemark. This proof was chosen because it generalizes easily to prove the following fact by induction: The determinant of a skew-symmetric matrix of even order is the square of a polynomial in the matrix entries. (We show below that the determinant of a skew-symmetric matrix of odd order is zero.) In fact the determinant of the \\( 2 \\dimension \\times 2 \\dimension \\) skew-symmetric matrix \\( \\left(entrygeneric_{rowindex colindex}\\right) \\) is the square of\n\\[\n\\frac{1}{2 \\cdot 4 \\cdots 2 \\dimension} \\sum_{\\permutation}(-1)^{\\permutation}\\, entrygeneric_{\\permutation(1), \\permutation(2)}\\, entrygeneric_{\\permutation(3), \\permutation(4)} \\cdots entrygeneric_{\\permutation(2 \\dimension-1), \\permutation(2 \\dimension)}\n\\]\nwhere \\( (-1)^{\\permutation} \\) denotes the sign of the permutation \\( \\permutation \\) and the sum is taken over all members \\( \\permutation \\) of the symmetric group. This polynomial was called the Pfaffian by Cayley because Jacobi had used it in his work on Pfaff's problem. Because the entrygeneric's are skew-symmetric, the Pfaffian can be written as the sum of \\( 1 \\cdot 3 \\cdot 5 \\cdots(2 \\dimension-1) \\) products without denominators. See Thomas Muir, The Theory of Determinants in the Historical Order of Development, Vol. 1, Dover, New York, pages 401-406.\n\nThe determinant of a skew-symmetric matrix \\( skewsymmetric \\) of odd order \\( \\dimension \\) is always zero. We have\n\\[\n(-1)^{\\dimension} \\operatorname{det} skewsymmetric=\\operatorname{det}(-skewsymmetric)=\\operatorname{det} skewsymmetric^{T}=\\operatorname{det} skewsymmetric ;\n\\]\nso, for \\( \\dimension \\) odd, \\( \\operatorname{det} skewsymmetric=0 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "armchair",
        "b": "blackbird",
        "c": "cinnamon",
        "d": "daydream",
        "e": "evergreen",
        "f": "flagstaff",
        "\\\\Delta": "tapestry",
        "n": "nightfall",
        "\\\\sigma": "silhouette",
        "S": "partridge",
        "i": "porcelain",
        "j": "junction",
        "a_ij": "buttercup",
        "Q": "riverbank"
      },
      "question": "1. Show that the determinant:\n\\[\n\\left|\\begin{array}{rrrr}\n0 & armchair & blackbird & cinnamon \\\\\n-armchair & 0 & daydream & evergreen \\\\\n-blackbird & -daydream & 0 & flagstaff \\\\\n-cinnamon & -evergreen & -flagstaff & 0\n\\end{array}\\right|\n\\]\nis non-negative, if its elements \\( armchair, blackbird, cinnamon \\), etc., are real.",
      "solution": "Solution. Assume \\( flagstaff \\neq 0 \\). Divide the third row and the third column by \\( flagstaff \\), and write \\( \\bar{blackbird}=blackbird / flagstaff, \\bar{daydream}=daydream / flagstaff \\). Then, if \\( tapestry \\) is the determinant,\n\\[\n\\frac{1}{flagstaff^{2}} tapestry=\\left|\\begin{array}{rrrr}\n0 & armchair & \\bar{blackbird} & cinnamon \\\\\n-armchair & 0 & \\bar{daydream} & evergreen \\\\\n-\\bar{blackbird} & -\\bar{daydream} & 0 & 1 \\\\\n-cinnamon & -evergreen & -1 & 0\n\\end{array}\\right|\n\\]\n\nSubtract \\( cinnamon \\) times the third row from the first row and \\( evergreen \\) times the third row from the second row.\n\\[\n\\frac{1}{flagstaff^{2}} tapestry=\\left|\\begin{array}{cccc}\n\\bar{blackbird} cinnamon & armchair+\\bar{daydream} cinnamon & \\bar{blackbird} & 0 \\\\\n-armchair+\\bar{blackbird} evergreen & \\bar{daydream} evergreen & \\bar{daydream} & 0 \\\\\n-\\bar{blackbird} & -\\bar{daydream} & 0 & 1 \\\\\n-cinnamon & -evergreen & -1 & 0\n\\end{array}\\right|\n\\]\n\nMake similar transformations of the columns.\n\\[\n\\begin{aligned}\n\\frac{1}{flagstaff^{2}} tapestry & =\\left|\\begin{array}{cccc}\n0 & armchair+cinnamon \\bar{daydream}-\\bar{blackbird} evergreen & \\bar{blackbird} & 0 \\\\\n-armchair+\\bar{blackbird} evergreen-\\bar{daydream} cinnamon & 0 & \\bar{daydream} & 0 \\\\\n-\\bar{blackbird} & -\\bar{daydream} & 0 & 1 \\\\\n0 & 0 & -1 & 0\n\\end{array}\\right| \\\\\n& =\\left|\\begin{array}{cc}\n0 & armchair+cinnamon \\bar{daydream}-\\bar{blackbird} evergreen \\\\\n-armchair+\\bar{blackbird} evergreen-cinnamon \\bar{daydream} & 0\n\\end{array}\\right| \\\\\n& =(armchair+cinnamon \\bar{daydream}-\\bar{blackbird} evergreen)^{2} .\n\\end{aligned}\n\\]\n\nFinally, \\( tapestry=(armchair flagstaff+cinnamon daydream-blackbird evergreen)^{2} \\), and therefore \\( tapestry \\) is non-negative for any real choice of \\( armchair, blackbird, cinnamon, daydream, evergreen, flagstaff \\).\n\nNote that if \\( flagstaff=0 \\), the matrix has more zeros, and an elementary application of Laplace's expansion of a determinant gives \\( tapestry=(cinnamon daydream-blackbird evergreen)^{2} \\). (Alternatively, this can also be seen by letting \\( flagstaff \\rightarrow 0 \\) in the above formula \\( tapestry=(armchair flagstaff+cinnamon daydream-blackbird evergreen)^{2} \\).) More generally, since \\( tapestry \\) is obviously a polynomial in the matrix entries, the assumption \\( flagstaff \\neq 0 \\) is really no loss of generality in evaluating \\( tapestry \\). In the language of algebra, the computation was really made in the field \\( riverbank(armchair, blackbird, cinnamon, daydream, evergreen, flagstaff) \\) where \\( armchair, blackbird, cinnamon, daydream, evergreen \\), and \\( flagstaff \\) are indeterminates. In this field, \\( flagstaff \\neq 0 \\).\n\nRemark. This proof was chosen because it generalizes easily to prove the following fact by induction: The determinant of a skew-symmetric matrix of even order is the square of a polynomial in the matrix entries. (We show below that the determinant of a skew-symmetric matrix of odd order is zero.) In fact the determinant of the \\( 2 nightfall \\times 2 nightfall \\) skew-symmetric matrix \\( \\left(armchair_{porcelain junction}\\right) \\) is the square of\n\\[\n\\frac{1}{2 \\cdot 4 \\cdots 2 nightfall} \\sum_{silhouette}(-1)^{silhouette} armchair_{silhouette(1), silhouette(2)} armchair_{silhouette(3), silhouette(4)} \\cdots armchair_{silhouette(2 nightfall-1), silhouette(2 nightfall)}\n\\]\nwhere \\( (-1)^{\\circ} \\) denotes the sign of the permutation \\( silhouette \\) and the sum is taken over all members \\( silhouette \\) of the symmetric group. This polynomial was called the Pfaffian by Cayley because Jacobi had used it in his work on Pfaff's problem. Because the \\( armchair \\) 's are skew-symmetric, the Pfaffian can be written as the sum of \\( 1 \\cdot 3 \\cdot 5 \\cdots(2 nightfall-1) \\) products without denominators. See Thomas Muir, The Theory of. Determinants in the Historical Order of Development, Vol. 1, Dover, New York, pages 401-406.\n\nThe determinant of a skew-symmetric matrix \\( partridge \\) of odd order \\( nightfall \\) is always zero. We have\n\\[\n(-1)^{nightfall} \\operatorname{det} partridge=\\operatorname{det}(-partridge)=\\operatorname{det} partridge^{T}=\\operatorname{det} partridge ;\n\\]\nso, for \\( nightfall \\) odd, \\( \\operatorname{det} partridge=0 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "lastletter",
        "b": "antibravo",
        "c": "anticharlie",
        "d": "oppositedelta",
        "e": "inverseecho",
        "f": "antifoxtrot",
        "\\\\Delta": "voidvalue",
        "n": "infiniteindex",
        "\\\\sigma": "unorderedperm",
        "S": "symmetricmat",
        "i": "globalindex",
        "j": "localindex",
        "a_ij": "oppositeentry",
        "Q": "nonrationals"
      },
      "question": "1. Show that the determinant:\n\\[\n\\left|\\begin{array}{rrrr}\n0 & lastletter & antibravo & anticharlie \\\\\n-lastletter & 0 & oppositedelta & inverseecho \\\\\n-antibravo & -oppositedelta & 0 & antifoxtrot \\\\\n-anticharlie & -inverseecho & -antifoxtrot & 0\n\\end{array}\\right|\n\\]\nis non-negative, if its elements \\( lastletter, antibravo, anticharlie \\), etc., are real.",
      "solution": "Solution. Assume \\( antifoxtrot \\neq 0 \\). Divide the third row and the third column by \\( antifoxtrot \\), and write \\( \\bar{antibravo}=antibravo / antifoxtrot,\\ \\bar{oppositedelta}=oppositedelta / antifoxtrot \\). Then, if \\( voidvalue \\) is the determinant,\n\\[\n\\frac{1}{antifoxtrot^{2}} voidvalue=\\left|\\begin{array}{rrrr}\n0 & lastletter & \\bar{antibravo} & anticharlie \\\\\n-lastletter & 0 & \\bar{oppositedelta} & inverseecho \\\\\n-\\bar{antibravo} & -\\bar{oppositedelta} & 0 & 1 \\\\\n-anticharlie & -inverseecho & -1 & 0\n\\end{array}\\right|\n\\]\n\nSubtract \\( anticharlie \\) times the third row from the first row and \\( inverseecho \\) times the third row from the second row.\n\\[\n\\frac{1}{antifoxtrot^{2}} voidvalue=\\left|\\begin{array}{cccc}\n\\bar{antibravo} anticharlie & lastletter+\\bar{oppositedelta} anticharlie & \\bar{antibravo} & 0 \\\\\n-lastletter+\\bar{antibravo} inverseecho & \\bar{oppositedelta} inverseecho & \\bar{oppositedelta} & 0 \\\\\n-\\bar{antibravo} & -\\bar{oppositedelta} & 0 & 1 \\\\\n-anticharlie & -inverseecho & -1 & 0\n\\end{array}\\right|\n\\]\n\nMake similar transformations of the columns.\n\\[\n\\begin{aligned}\n\\frac{1}{antifoxtrot^{2}} voidvalue & =\\left|\\begin{array}{cccc}\n0 & lastletter+anticharlie \\bar{oppositedelta}-\\bar{antibravo} inverseecho & \\bar{antibravo} & 0 \\\\\n-lastletter+\\bar{antibravo} inverseecho-\\bar{oppositedelta} anticharlie & 0 & \\bar{oppositedelta} & 0 \\\\\n-\\bar{antibravo} & -\\bar{oppositedelta} & 0 & 1 \\\\\n0 & 0 & -1 & 0\n\\end{array}\\right| \\\\\n& =\\left|\\begin{array}{cc}\n0 & lastletter+anticharlie \\bar{oppositedelta}-\\bar{antibravo} inverseecho \\\\\n-lastletter+\\bar{antibravo} inverseecho-anticharlie \\bar{oppositedelta} & 0\n\\end{array}\\right| \\\\\n& =(lastletter+anticharlie \\bar{oppositedelta}-\\bar{antibravo} inverseecho)^{2} .\n\\end{aligned}\n\\]\n\nFinally, \\( voidvalue=(lastletter\\, antifoxtrot+anticharlie\\, oppositedelta-antibravo\\, inverseecho)^{2} \\), and therefore \\( voidvalue \\) is non-negative for any real choice of \\( lastletter, antibravo, anticharlie, oppositedelta, inverseecho, antifoxtrot \\).\n\nNote that if \\( antifoxtrot=0 \\), the matrix has more zeros, and an elementary application of Laplace's expansion of a determinant gives \\( voidvalue=(anticharlie\\, oppositedelta-antibravo\\, inverseecho)^{2} \\). (Alternatively, this can also be seen by letting \\( antifoxtrot \\rightarrow 0 \\) in the above formula \\( voidvalue=(lastletter\\, antifoxtrot+anticharlie\\, oppositedelta-antibravo\\, inverseecho)^{2} \\).) More generally, since \\( voidvalue \\) is obviously a polynomial in the matrix entries, the assumption \\( antifoxtrot \\neq 0 \\) is really no loss of generality in evaluating \\( voidvalue \\). In the language of algebra, the computation was really made in the field \\( nonrationals(lastletter, antibravo, anticharlie, oppositedelta, inverseecho, antifoxtrot) \\) where \\( lastletter, antibravo, anticharlie, oppositedelta, inverseecho \\), and \\( antifoxtrot \\) are indeterminates. In this field, \\( antifoxtrot \\neq 0 \\).\n\nRemark. This proof was chosen because it generalizes easily to prove the following fact by induction: The determinant of a skew-symmetric matrix of even order is the square of a polynomial in the matrix entries. (We show below that the determinant of a skew-symmetric matrix of odd order is zero.) In fact the determinant of the \\( 2 infiniteindex \\times 2 infiniteindex \\) skew-symmetric matrix \\( \\left(lastletter_{globalindex \\, localindex}\\right) \\) is the square of\n\\[\n\\frac{1}{2 \\cdot 4 \\cdots 2 infiniteindex} \\sum_{unorderedperm}(-1)^{unorderedperm} lastletter_{unorderedperm(1), unorderedperm(2)} lastletter_{unorderedperm(3), unorderedperm(4)} \\cdots lastletter_{unorderedperm(2 infiniteindex-1), unorderedperm(2 infiniteindex)}\n\\]\nwhere \\( (-1)^{\\circ} \\) denotes the sign of the permutation \\( unorderedperm \\) and the sum is taken over all members \\( unorderedperm \\) of the symmetric group. This polynomial was called the Pfaffian by Cayley because Jacobi had used it in his work on Pfaff's problem. Because the \\( lastletter \\)'s are skew-symmetric, the Pfaffian can be written as the sum of \\( 1 \\cdot 3 \\cdot 5 \\cdots(2 infiniteindex-1) \\) products without denominators. See Thomas Muir, The Theory of. Determinants in the Historical Order of Development, Vol. 1, Dover, New York, pages 401-406.\n\nThe determinant of a skew-symmetric matrix \\( symmetricmat \\) of odd order \\( infiniteindex \\) is always zero. We have\n\\[\n(-1)^{infiniteindex} \\operatorname{det} symmetricmat=\\operatorname{det}(-symmetricmat)=\\operatorname{det} symmetricmat^{T}=\\operatorname{det} symmetricmat ;\n\\]\nso, for \\( infiniteindex \\) odd, \\( \\operatorname{det} symmetricmat=0 \\)."
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "mfldpqoze",
        "d": "xrjvuknh",
        "e": "pgtswyoc",
        "f": "blcznqer",
        "\\Delta": "snvkdjfe",
        "n": "vhmrkzoi",
        "\\sigma": "ulqxvbry",
        "S": "tgjlsfkm",
        "i": "rywqmpoz",
        "j": "lpsndwxa",
        "a_ij": "vlcspton",
        "Q": "zbnvrhks"
      },
      "question": "1. Show that the determinant:\n\\[\n\\left|\\begin{array}{rrrr}\n0 & qzxwvtnp & hjgrksla & mfldpqoze \\\\\n-qzxwvtnp & 0 & xrjvuknh & pgtswyoc \\\\\n-hjgrksla & -xrjvuknh & 0 & blcznqer \\\\\n-mfldpqoze & -pgtswyoc & -blcznqer & 0\n\\end{array}\\right|\n\\]\nis non-negative, if its elements \\( qzxwvtnp, hjgrksla, mfldpqoze \\), etc., are real.",
      "solution": "Solution. Assume \\( blcznqer \\neq 0 \\). Divide the third row and the third column by \\( blcznqer \\), and write \\( \\bar{hjgrksla}=hjgrksla / blcznqer, \\bar{xrjvuknh}=xrjvuknh / blcznqer \\). Then, if \\( \\snvkdjfe \\) is the determinant,\n\\[\n\\frac{1}{blcznqer^{2}} \\snvkdjfe=\\left|\\begin{array}{rrrr}\n0 & qzxwvtnp & \\bar{hjgrksla} & mfldpqoze \\\\\n-qzxwvtnp & 0 & \\bar{xrjvuknh} & pgtswyoc \\\\\n-\\bar{hjgrksla} & -\\bar{xrjvuknh} & 0 & 1 \\\\\n-mfldpqoze & -pgtswyoc & -1 & 0\n\\end{array}\\right|\n\\]\n\nSubtract \\( mfldpqoze \\) times the third row from the first row and \\( pgtswyoc \\) times the third row from the second row.\n\\[\n\\frac{1}{blcznqer^{2}} \\snvkdjfe=\\left|\\begin{array}{cccc}\n\\bar{hjgrksla} mfldpqoze & qzxwvtnp+\\bar{xrjvuknh} mfldpqoze & \\bar{hjgrksla} & 0 \\\\\n-qzxwvtnp+\\bar{hjgrksla} pgtswyoc & \\bar{xrjvuknh} pgtswyoc & \\bar{xrjvuknh} & 0 \\\\\n-\\bar{hjgrksla} & -\\bar{xrjvuknh} & 0 & 1 \\\\\n-mfldpqoze & -pgtswyoc & -1 & 0\n\\end{array}\\right|\n\\]\n\nMake similar transformations of the columns.\n\\[\n\\begin{aligned}\n\\frac{1}{blcznqer^{2}} \\snvkdjfe & =\\left|\\begin{array}{cccc}\n0 & qzxwvtnp+mfldpqoze \\bar{xrjvuknh}-\\bar{hjgrksla} pgtswyoc & \\bar{hjgrksla} & 0 \\\\\n-qzxwvtnp+\\bar{hjgrksla} pgtswyoc-\\bar{xrjvuknh} mfldpqoze & 0 & \\bar{xrjvuknh} & 0 \\\\\n-\\bar{hjgrksla} & -\\bar{xrjvuknh} & 0 & 1 \\\\\n0 & 0 & -1 & 0\n\\end{array}\\right| \\\\\n& =\\left|\\begin{array}{cc}\n0 & qzxwvtnp+mfldpqoze \\bar{xrjvuknh}-\\bar{hjgrksla} pgtswyoc \\\\\n-qzxwvtnp+\\bar{hjgrksla} pgtswyoc-mfldpqoze \\bar{xrjvuknh} & 0\n\\end{array}\\right| \\\\\n& =(qzxwvtnp+mfldpqoze \\bar{xrjvuknh}-\\bar{hjgrksla} pgtswyoc)^{2} .\n\\end{aligned}\n\\]\n\nFinally, \\( \\snvkdjfe=(qzxwvtnp blcznqer+mfldpqoze xrjvuknh-hjgrksla pgtswyoc)^{2} \\), and therefore \\( \\snvkdjfe \\) is non-negative for any real choice of \\( qzxwvtnp, hjgrksla, mfldpqoze, xrjvuknh, pgtswyoc, blcznqer \\).\n\nNote that if \\( blcznqer=0 \\), the matrix has more zeros, and an elementary application of Laplace's expansion of a determinant gives \\( \\snvkdjfe=(mfldpqoze xrjvuknh-hjgrksla pgtswyoc)^{2} \\). (Alternatively, this can also be seen by letting \\( blcznqer \\rightarrow 0 \\) in the above formula \\( \\snvkdjfe=(qzxwvtnp blcznqer+mfldpqoze xrjvuknh-hjgrksla pgtswyoc)^{2} \\).) More generally, since \\( \\snvkdjfe \\) is obviously a polynomial in the matrix entries, the assumption \\( blcznqer \\neq 0 \\) is really no loss of generality in evaluating \\( \\snvkdjfe \\). In the language of algebra, the computation was really made in the field \\( zbnvrhks(qzxwvtnp, hjgrksla, mfldpqoze, xrjvuknh, pgtswyoc, blcznqer) \\) where \\( qzxwvtnp, hjgrksla, mfldpqoze, xrjvuknh, pgtswyoc \\), and \\( blcznqer \\) are indeterminates. In this field, \\( blcznqer \\neq 0 \\).\n\nRemark. This proof was chosen because it generalizes easily to prove the following fact by induction: The determinant of a skew-symmetric matrix of even order is the square of a polynomial in the matrix entries. (We show below that the determinant of a skew-symmetric matrix of odd order is zero.) In fact the determinant of the \\( 2 vhmrkzoi \\times 2 vhmrkzoi \\) skew-symmetric matrix \\( \\left(vlcspton_{rywqmpoz lpsndwxa}\\right) \\) is the square of\n\\[\n\\frac{1}{2 \\cdot 4 \\cdots 2 vhmrkzoi} \\sum_{ulqxvbry}(-1)^{ulqxvbry} qzxwvtnp_{ulqxvbry(1), ulqxvbry(2)} qzxwvtnp_{ulqxvbry(3), ulqxvbry(4)} \\cdots qzxwvtnp_{ulqxvbry(2 vhmrkzoi-1), ulqxvbry(2 vhmrkzoi)}\n\\]\nwhere \\( (-1)^{\\circ} \\) denotes the sign of the permutation \\( ulqxvbry \\) and the sum is taken over all members \\( ulqxvbry \\) of the symmetric group. This polynomial was called the Pfaffian by Cayley because Jacobi had used it in his work on Pfaff's problem. Because the \\( qzxwvtnp \\) 's are skew-symmetric, the Pfaffian can be written as the sum of \\( 1 \\cdot 3 \\cdot 5 \\cdots(2 vhmrkzoi-1) \\) products without denominators. See Thomas Muir, The Theory of. Determinants in the Historical Order of Development, Vol. 1, Dover, New York, pages 401-406.\n\nThe determinant of a skew-symmetric matrix \\( tgjlsfkm \\) of odd order \\( vhmrkzoi \\) is always zero. We have\n\\[\n(-1)^{vhmrkzoi} \\operatorname{det} tgjlsfkm=\\operatorname{det}(-tgjlsfkm)=\\operatorname{det} tgjlsfkm^{T}=\\operatorname{det} tgjlsfkm ;\n\\]\nso, for \\( vhmrkzoi \\) odd, \\( \\operatorname{det} tgjlsfkm=0 \\)."
    },
    "kernel_variant": {
      "question": "Let \\(a,b,c,d,e,f\\in\\mathbb R\\).  Prove that the determinant\n\\[\n\\Delta=\n\\det\\!\n\\begin{pmatrix}\n 0 & a & d & b\\\\\n -a& 0 & c & e\\\\\n -d& -c& 0 & f\\\\\n -b& -e& -f& 0\n\\end{pmatrix}\n\\]\n is always non–negative and find a closed-form expression for it.",
      "solution": "1. (Choice of a pivot and scaling.)  Assume first that e≠0.  Because the (2,4)–entry equals e, pick it as the pivot.  Divide the second row and the second column by e.  Writing\n\\[\\bar a=\\tfrac{a}{e},\\qquad \\bar c=\\tfrac{c}{e}\\]\nand observing that the determinant picks up the factor e^{−2}, we get\n\\[\n\\frac{\\Delta}{e^{2}}=\\det\n\\begin{pmatrix}\n 0 &\\;\\bar a& d & b\\\\\n -\\bar a & 0 & \\bar c & 1\\\\\n -d & -\\bar c & 0 & f\\\\\n -b & -1 & -f & 0\n\\end{pmatrix}\n=:M_1.\n\\]\n\n2. (Clearing selected entries while preserving the determinant.)  Subtract b times the second row from the first row and f times the second row from the third row.  Then perform the corresponding column operations (subtract the same multiples of the second column from the first and third columns).  These simultaneous row–column operations leave the determinant unchanged, and we reach\n\\[\nM_2=\n\\begin{pmatrix}\n 0 & \\bar a & d-b\\bar c-f\\bar a & 0\\\\\n -\\bar a & 0 & \\bar c & 1\\\\\n -d+b\\bar c+f\\bar a & -\\bar c & 0 & 0\\\\\n 0 & -1 & 0 & 0\n\\end{pmatrix}.\n\\]\n\n3. (Laplace expansion.)  All entries in the last row (and last column) vanish except the (4,2)–entry.  A Laplace expansion along the fourth row shows\n\\[\n\\det M_2=(d-b\\bar c-f\\bar a)^2.\n\\]\n\n4. (Undoing the scaling.)  Hence\n\\[\n\\frac{\\Delta}{e^{2}}=(d-b\\bar c-f\\bar a)^{2}\n\\quad\\Longrightarrow\\quad\n\\Delta=(ed-bc-fa)^{2}=(af-de+bc)^{2}\\ge0.\n\\]\n\n5. (The zero–pivot case.)  If e=0, one finds by direct expansion (or by continuity) that \\(\\Delta=(fa+bc)^2\\), in agreement with the limit of the formula above.  Thus for all real a,b,c,d,e,f,\n\\[\n\\Delta=(af-de+bc)^2\\ge0,\n\\]\nas required. ■",
      "_meta": {
        "core_steps": [
          "Pick a non-zero off-diagonal entry and scale its row & column by that value, accounting for the determinant’s ±factor.",
          "Use determinant-preserving operations (add multiples of the pivoted row/column) to clear selected entries and create a block form with many zeros.",
          "Observe that the determinant now factors into the determinant of a 2×2 skew-symmetric block times ±1.",
          "Compute that 2×2 determinant; it is a perfect square of a linear combination of the original entries.",
          "Undo the initial scaling; the full determinant is still a square, hence ≥ 0, and the zero-pivot case follows by continuity/Laplace expansion."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Choice of the pivot (the off-diagonal entry singled out for scaling).  Any non-zero entry would work.",
            "original": "f"
          },
          "slot2": {
            "description": "Labelling/placement of the six independent entries; simultaneous permutations of corresponding rows & columns merely permute the final linear combination.",
            "original": "matrix ordering (a,b,c,d,e,f as in the statement)"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}