{
  "index": "1938-B-1",
  "type": "ALG",
  "tag": [
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "8. Take either (i) or (ii).\n(i) Let \\( A_{i k} \\) be the cofactor of \\( a_{i k} \\) in the determinant\n\\[\n\\boldsymbol{d}=\\left|\\begin{array}{llll}\na_{11} & a_{12} & a_{13} & a_{14} \\\\\na_{21} & a_{22} & a_{23} & a_{24} \\\\\na_{31} & a_{32} & a_{33} & a_{34} \\\\\na_{41} & a_{42} & a_{43} & a_{44}\n\\end{array}\\right|\n\\]\n\nLet \\( D \\) be the corresponding determinant with \\( a_{i k} \\) replaced by \\( A_{i k} \\). Prove \\( D= \\) \\( d^{3} \\).\n(page 86)\n(ii) Let \\( P(y)=A y^{2}+B y+C \\) be a quadratic polynomial in \\( y \\). If the roots of the quadratic equation \\( P(y)-y=0 \\) are \\( a \\) and \\( b(a \\neq b) \\), show that \\( a \\) and \\( b \\) are roots of the biquadratic equation \\( P[P(y)]-y=0 \\). Hence write down a quadratic equation which will give the other two roots, \\( c \\) and \\( d \\), of the biquadratic. Apply this result to solving the following biquadratic equation:\n\\[\n\\left(y^{2}-3 y+2\\right)^{2}-3\\left(y^{2}-3 y+2\\right)+2-y=0\n\\]",
  "solution": "Solution. Let \\( \\alpha \\) be the matrix of the given determinant with elements \\( a_{i k} \\) and let \\( \\beta \\) be the matrix of the cofactors \\( A_{i k} \\), and let \\( \\gamma \\) be the transpose of \\( \\beta \\). Then the product matrix \\( \\alpha \\gamma \\) is a diagonal matrix with all entries on the main diagonal equal to \\( d \\).\n\nThus \\( \\operatorname{det}(\\alpha \\gamma)=d^{4}=(\\operatorname{det} \\alpha)(\\operatorname{det} \\gamma)=(\\operatorname{det} \\alpha)(\\operatorname{det} \\beta)=d D \\). The equation\n\\[\nd D=d^{4} .\n\\]\nis an identity between polynomials in the 16 matrix entries regarded as independent indeterminates. Since there certainly exists a \\( 4 \\times 4 \\) matrix whose determinant is not zero, \\( d \\) is non-zero in the polynomial ring. Since the polynomial ring is an integral domain, the result\n\\[\nD=d^{3}\n\\]\nfollows from (1).\nRemark. The result can obviously be generalized to matrices of any dimension. The determinant of the matrix of cofactors of an \\( n \\times n \\) matrix is \\( d^{n-1} \\), where \\( d \\) is the determinant of the original matrix.\n8. (ii) Let \\( P(y)=A y^{2}+B y+C \\) be a quadratic polynomial in \\( y \\). If the roots of the quadratic equation \\( P(y)-y=0 \\) are \\( a \\) and \\( b(a \\neq b) \\), show that \\( a \\) and \\( b \\) are roots of the biquadratic equation \\( P[P(y)]-y=0 \\). Hence write down a quadratic equation which will give the other two roots, \\( c \\) and \\( d \\), of the biquadratic. Apply this result to solving the following biquadratic equation:\n\\[\n\\left(y^{2}-3 y+2\\right)^{2}-3\\left(y^{2}-3 y+2\\right)+2-y=0 .\n\\]\n\nSolution. Since \\( a \\) is a root of \\( P(y)-y=0 \\), we have \\( P(a)=a \\). Then \\( P(P(a))=P(a)=a \\), so \\( a \\) is a root of \\( P(P(y))-y=0 \\). Similarly, \\( b \\) is a root of this biquadratic.\nLet \\( Q(y)=P(P(y))-y \\). To find the other zeros of \\( Q \\), note that \\( P(y) \\) \\( -y=A y^{2}+(B-1) y+C=A(y-a)(y-b) \\), whence \\( A(a+b)= \\) \\( 1-B \\). Then\n\\[\n\\begin{aligned}\nQ(y)= & P(P(y))-P(y)+P(y)-y \\\\\n= & A\\{P(y)-a\\}\\{P(y)-b\\}+A(y-a)(y-b) \\\\\n= & A\\{A(y-a)(y-b)+y-a\\}\\{A(y-a)(y-b)+y-b\\} \\\\\n& \\quad+A(y-a)(y-b) \\\\\n= & A(y-a)(y-b) R(y),\n\\end{aligned}\n\\]\nwhere\n\\[\n\\begin{aligned}\nR(y) & =\\{A(y-b)+1\\}\\{A(y-a)+1\\}+1 \\\\\n& =A P(y)+A y-A(a+b)+2 \\\\\n& =A^{2} y^{2}+A(B+1) y+A C+B+1 .\n\\end{aligned}\n\\]\n\nThe roots \\( c \\) and \\( d \\) are the zeros of \\( R \\), so the required quadratic equation for \\( c \\) and \\( d \\) is\n\\[\nA^{2} y^{2}+A(B+1) y+A C+B+1=0 .\n\\]\n\nIn the special case given, \\( A=1, B=-3, C=2 \\), and \\( R(y)=y^{2}-2 y \\). The zeros of \\( P(y)-y \\) are \\( 2 \\pm \\sqrt{2} \\), so the zeros of \\( Q \\) are \\( 2 \\pm \\sqrt{2}, 0 \\), and 2 .",
  "vars": [
    "A_ik",
    "a_ik",
    "d",
    "D",
    "\\\\alpha",
    "\\\\beta",
    "A",
    "B",
    "C",
    "P",
    "y",
    "a",
    "b",
    "Q",
    "R",
    "c",
    "k",
    "n"
  ],
  "params": [],
  "sci_consts": [
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "A_ik": "cofacentry",
        "a_ik": "matrixentry",
        "d": "detorig",
        "D": "detcofac",
        "\\alpha": "alphamatrix",
        "\\beta": "betamatrix",
        "A": "quadlead",
        "B": "quadlinear",
        "C": "quadconst",
        "P": "quadpoly",
        "y": "variable",
        "a": "rootone",
        "b": "roottwo",
        "Q": "biquadpoly",
        "R": "auxipoly",
        "c": "rootthree",
        "k": "indexkvar",
        "n": "dimension"
      },
      "question": "8. Take either (i) or (ii).\n(i) Let \\( cofacentry \\) be the cofactor of \\( matrixentry \\) in the determinant\n\\[\n\\boldsymbol{detorig}=\\left|\\begin{array}{llll}\na_{11} & a_{12} & a_{13} & a_{14} \\\\\na_{21} & a_{22} & a_{23} & a_{24} \\\\\na_{31} & a_{32} & a_{33} & a_{34} \\\\\na_{41} & a_{42} & a_{43} & a_{44}\n\\end{array}\\right|\n\\]\n\nLet \\( detcofac \\) be the corresponding determinant with \\( matrixentry \\) replaced by \\( cofacentry \\). Prove \\( detcofac = detorig^{3} \\).\n(page 86)\n(ii) Let \\( quadpoly(variable)=quadlead \\, variable^{2}+quadlinear \\, variable+quadconst \\) be a quadratic polynomial in \\( variable \\). If the roots of the quadratic equation \\( quadpoly(variable)-variable=0 \\) are \\( rootone \\) and \\( roottwo(rootone \\neq roottwo) \\), show that \\( rootone \\) and \\( roottwo \\) are roots of the biquadratic equation \\( quadpoly[quadpoly(variable)]-variable=0 \\). Hence write down a quadratic equation which will give the other two roots, \\( rootthree \\) and \\( detorig \\), of the biquadratic. Apply this result to solving the following biquadratic equation:\n\\[\n\\left(variable^{2}-3 variable+2\\right)^{2}-3\\left(variable^{2}-3 variable+2\\right)+2-variable=0\n\\]\n",
      "solution": "Solution. Let \\( alphamatrix \\) be the matrix of the given determinant with elements \\( matrixentry \\) and let \\( betamatrix \\) be the matrix of the cofactors \\( cofacentry \\), and let \\( \\gamma \\) be the transpose of \\( betamatrix \\). Then the product matrix \\( alphamatrix \\gamma \\) is a diagonal matrix with all entries on the main diagonal equal to \\( detorig \\).\n\nThus \\( \\operatorname{det}(alphamatrix \\gamma)=detorig^{4}=(\\operatorname{det} alphamatrix)(\\operatorname{det} \\gamma)=(\\operatorname{det} alphamatrix)(\\operatorname{det} betamatrix)=detorig \\, detcofac \\). The equation\n\\[\n detorig \\, detcofac = detorig^{4} .\n\\]\nis an identity between polynomials in the 16 matrix entries regarded as independent indeterminates. Since there certainly exists a \\( 4 \\times 4 \\) matrix whose determinant is not zero, \\( detorig \\) is non-zero in the polynomial ring. Since the polynomial ring is an integral domain, the result\n\\[\n detcofac = detorig^{3}\n\\]\nfollows from (1).\n\nRemark. The result can obviously be generalized to matrices of any dimension. The determinant of the matrix of cofactors of an \\( dimension \\times dimension \\) matrix is \\( detorig^{dimension-1} \\), where \\( detorig \\) is the determinant of the original matrix.\n\n8. (ii) Let \\( quadpoly(variable)=quadlead \\, variable^{2}+quadlinear \\, variable+quadconst \\) be a quadratic polynomial in \\( variable \\). If the roots of the quadratic equation \\( quadpoly(variable)-variable=0 \\) are \\( rootone \\) and \\( roottwo(rootone \\neq roottwo) \\), show that \\( rootone \\) and \\( roottwo \\) are roots of the biquadratic equation \\( quadpoly[quadpoly(variable)]-variable=0 \\). Hence write down a quadratic equation which will give the other two roots, \\( rootthree \\) and \\( detorig \\), of the biquadratic. Apply this result to solving the following biquadratic equation:\n\\[\n\\left(variable^{2}-3 variable+2\\right)^{2}-3\\left(variable^{2}-3 variable+2\\right)+2-variable=0 .\n\\]\n\nSolution. Since \\( rootone \\) is a root of \\( quadpoly(variable)-variable=0 \\), we have \\( quadpoly(rootone)=rootone \\). Then \\( quadpoly(quadpoly(rootone))=quadpoly(rootone)=rootone \\), so \\( rootone \\) is a root of \\( quadpoly(quadpoly(variable))-variable=0 \\). Similarly, \\( roottwo \\) is a root of this biquadratic.\n\nLet \\( biquadpoly(variable)=quadpoly(quadpoly(variable))-variable \\). To find the other zeros of \\( biquadpoly \\), note that\n\\[\n quadpoly(variable)-variable = quadlead \\, variable^{2} + (quadlinear-1) \\, variable + quadconst = quadlead (variable-rootone)(variable-roottwo),\n\\]\nwhence \\( quadlead(rootone+roottwo)=1-quadlinear \\). Then\n\\[\n\\begin{aligned}\nbiquadpoly(variable)= & quadpoly(quadpoly(variable)) - quadpoly(variable) + quadpoly(variable) - variable \\\\\n= & quadlead \\{ quadpoly(variable)-rootone \\} \\{ quadpoly(variable)-roottwo \\} + quadlead (variable-rootone)(variable-roottwo) \\\\\n= & quadlead \\{ quadlead (variable-rootone)(variable-roottwo) + variable - rootone \\} \\{ quadlead (variable-rootone)(variable-roottwo) + variable - roottwo \\} \\\\\n& \\quad + quadlead (variable-rootone)(variable-roottwo) \\\\\n= & quadlead (variable-rootone)(variable-roottwo) \\, auxipoly(variable),\n\\end{aligned}\n\\]\nwhere\n\\[\n\\begin{aligned}\nauxipoly(variable) & = \\{ quadlead(variable-roottwo)+1 \\} \\{ quadlead(variable-rootone)+1 \\} + 1 \\\\\n& = quadlead \\, quadpoly(variable) + quadlead \\, variable - quadlead(rootone+roottwo) + 2 \\\\\n& = quadlead^{2} \\, variable^{2} + quadlead(quadlinear+1) \\, variable + quadlead \\, quadconst + quadlinear + 1 .\n\\end{aligned}\n\\]\n\nThe roots \\( rootthree \\) and \\( detorig \\) are the zeros of \\( auxipoly \\), so the required quadratic equation for \\( rootthree \\) and \\( detorig \\) is\n\\[\n quadlead^{2} \\, variable^{2} + quadlead(quadlinear+1) \\, variable + quadlead \\, quadconst + quadlinear + 1 = 0 .\n\\]\n\nIn the special case given, \\( quadlead = 1, quadlinear = -3, quadconst = 2 \\), and \\( auxipoly(variable) = variable^{2} - 2 \\, variable \\). The zeros of \\( quadpoly(variable)-variable \\) are \\( 2 \\pm \\sqrt{2} \\), so the zeros of \\( biquadpoly \\) are \\( 2 \\pm \\sqrt{2}, 0 \\), and 2 .\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "A_ik": "sunflower",
        "a_ik": "backboard",
        "d": "buttercup",
        "D": "salamander",
        "\\alpha": "arrowroot",
        "\\beta": "lighthouse",
        "A": "strawhat",
        "B": "moonlight",
        "C": "periscope",
        "P": "tangerine",
        "y": "waterfall",
        "a": "dragonfly",
        "b": "pineapple",
        "Q": "whirlwind",
        "R": "journeyer",
        "c": "snowflake",
        "k": "afterglow",
        "n": "harbinger"
      },
      "question": "8. Take either (i) or (ii).\n(i) Let \\( sunflower_{i afterglow} \\) be the cofactor of \\( backboard_{i afterglow} \\) in the determinant\n\\[\n\\boldsymbol{buttercup}=\\left|\\begin{array}{llll}\nbackboard_{11} & backboard_{12} & backboard_{13} & backboard_{14} \\\\\nbackboard_{21} & backboard_{22} & backboard_{23} & backboard_{24} \\\\\nbackboard_{31} & backboard_{32} & backboard_{33} & backboard_{34} \\\\\nbackboard_{41} & backboard_{42} & backboard_{43} & backboard_{44}\n\\end{array}\\right|\n\\]\n\nLet \\( salamander \\) be the corresponding determinant with \\( backboard_{i afterglow} \\) replaced by \\( sunflower_{i afterglow} \\). Prove \\( salamander= \\) \\( buttercup^{3} \\).\n(page 86)\n(ii) Let \\( tangerine(waterfall)=strawhat waterfall^{2}+moonlight waterfall+periscope \\) be a quadratic polynomial in \\( waterfall \\). If the roots of the quadratic equation \\( tangerine(waterfall)-waterfall=0 \\) are \\( dragonfly \\) and \\( pineapple(dragonfly \\neq pineapple) \\), show that \\( dragonfly \\) and \\( pineapple \\) are roots of the biquadratic equation \\( tangerine[tangerine(waterfall)]-waterfall=0 \\). Hence write down a quadratic equation which will give the other two roots, \\( snowflake \\) and \\( buttercup \\), of the biquadratic. Apply this result to solving the following biquadratic equation:\n\\[\n\\left(waterfall^{2}-3 waterfall+2\\right)^{2}-3\\left(waterfall^{2}-3 waterfall+2\\right)+2-waterfall=0\n\\]",
      "solution": "Solution. Let \\( arrowroot \\) be the matrix of the given determinant with elements \\( backboard_{i afterglow} \\) and let \\( lighthouse \\) be the matrix of the cofactors \\( sunflower_{i afterglow} \\), and let \\( \\gamma \\) be the transpose of \\( lighthouse \\). Then the product matrix \\( arrowroot \\gamma \\) is a diagonal matrix with all entries on the main diagonal equal to \\( buttercup \\).\n\nThus \\( \\operatorname{det}(arrowroot \\gamma)=buttercup^{4}=(\\operatorname{det} arrowroot)(\\operatorname{det} \\gamma)=(\\operatorname{det} arrowroot)(\\operatorname{det} lighthouse)=buttercup salamander \\). The equation\n\\[\nbuttercup salamander=buttercup^{4} .\n\\]\nis an identity between polynomials in the 16 matrix entries regarded as independent indeterminates. Since there certainly exists a \\( 4 \\times 4 \\) matrix whose determinant is not zero, \\( buttercup \\) is non-zero in the polynomial ring. Since the polynomial ring is an integral domain, the result\n\\[\nsalamander=buttercup^{3}\n\\]\nfollows from (1).\nRemark. The result can obviously be generalized to matrices of any dimension. The determinant of the matrix of cofactors of an \\( harbinger \\times harbinger \\) matrix is \\( buttercup^{harbinger-1} \\), where \\( buttercup \\) is the determinant of the original matrix.\n8. (ii) Let \\( tangerine(waterfall)=strawhat waterfall^{2}+moonlight waterfall+periscope \\) be a quadratic polynomial in \\( waterfall \\). If the roots of the quadratic equation \\( tangerine(waterfall)-waterfall=0 \\) are \\( dragonfly \\) and \\( pineapple(dragonfly \\neq pineapple) \\), show that \\( dragonfly \\) and \\( pineapple \\) are roots of the biquadratic equation \\( tangerine[tangerine(waterfall)]-waterfall=0 \\). Hence write down a quadratic equation which will give the other two roots, \\( snowflake \\) and \\( buttercup \\), of the biquadratic. Apply this result to solving the following biquadratic equation:\n\\[\n\\left(waterfall^{2}-3 waterfall+2\\right)^{2}-3\\left(waterfall^{2}-3 waterfall+2\\right)+2-waterfall=0 .\n\\]\n\nSolution. Since \\( dragonfly \\) is a root of \\( tangerine(waterfall)-waterfall=0 \\), we have \\( tangerine(dragonfly)=dragonfly \\). Then \\( tangerine(tangerine(dragonfly))=tangerine(dragonfly)=dragonfly \\), so \\( dragonfly \\) is a root of \\( tangerine(tangerine(waterfall))-waterfall=0 \\). Similarly, \\( pineapple \\) is a root of this biquadratic.\nLet \\( whirlwind(waterfall)=tangerine(tangerine(waterfall))-waterfall \\). To find the other zeros of \\( whirlwind \\), note that \\( tangerine(waterfall) \\)\n \\(-waterfall=strawhat waterfall^{2}+(moonlight-1) waterfall+periscope=strawhat(waterfall-dragonfly)(waterfall-pineapple) \\), whence \\( strawhat(dragonfly+pineapple)= \\) \\( 1-moonlight \\). Then\n\\[\n\\begin{aligned}\nwhirlwind(waterfall)= & tangerine(tangerine(waterfall))-tangerine(waterfall)+tangerine(waterfall)-waterfall \\\\\n= & strawhat\\{tangerine(waterfall)-dragonfly\\}\\{tangerine(waterfall)-pineapple\\}+strawhat(waterfall-dragonfly)(waterfall-pineapple) \\\\\n= & strawhat\\{strawhat(waterfall-dragonfly)(waterfall-pineapple)+waterfall-dragonfly\\}\\{strawhat(waterfall-dragonfly)(waterfall-pineapple)+waterfall-pineapple\\} \\\\\n& \\quad+strawhat(waterfall-dragonfly)(waterfall-pineapple) \\\\\n= & strawhat(waterfall-dragonfly)(waterfall-pineapple) journeyer(waterfall),\n\\end{aligned}\n\\]\nwhere\n\\[\n\\begin{aligned}\njourneyer(waterfall) & =\\{strawhat(waterfall-pineapple)+1\\}\\{strawhat(waterfall-dragonfly)+1\\}+1 \\\\\n& =strawhat tangerine(waterfall)+strawhat waterfall-strawhat(dragonfly+pineapple)+2 \\\\\n& =strawhat^{2} waterfall^{2}+strawhat(moonlight+1) waterfall+strawhat periscope+moonlight+1 .\n\\end{aligned}\n\\]\n\nThe roots \\( snowflake \\) and \\( buttercup \\) are the zeros of \\( journeyer \\), so the required quadratic equation for \\( snowflake \\) and \\( buttercup \\) is\n\\[\nstrawhat^{2} waterfall^{2}+strawhat(moonlight+1) waterfall+strawhat periscope+moonlight+1=0 .\n\\]\n\nIn the special case given, \\( strawhat=1, moonlight=-3, periscope=2 \\), and \\( journeyer(waterfall)=waterfall^{2}-2 waterfall \\). The zeros of \\( tangerine(waterfall)-waterfall \\) are \\( 2 \\pm \\sqrt{2} \\), so the zeros of \\( whirlwind \\) are \\( 2 \\pm \\sqrt{2}, 0 \\), and 2 ."
    },
    "descriptive_long_misleading": {
      "map": {
        "A_ik": "noncofactor",
        "a_ik": "majorelement",
        "d": "antideterminant",
        "D": "indeterminant",
        "\\alpha": "omegaarray",
        "\\beta": "alphaarray",
        "A": "zerocoeff",
        "B": "microcoeff",
        "C": "varyingpart",
        "P": "transcendfunc",
        "y": "constantval",
        "a": "leafvalue",
        "b": "stemvalue",
        "Q": "basepoly",
        "R": "simpleroot",
        "c": "branchval",
        "k": "unindexed",
        "n": "unbounded"
      },
      "question": "Problem:\n<<<\n8. Take either (i) or (ii).\n(i) Let \\( noncofactor_{i unindexed} \\) be the cofactor of \\( majorelement_{i unindexed} \\) in the determinant\n\\[\n\\boldsymbol{antideterminant}=\\left|\\begin{array}{llll}\nmajorelement_{11} & majorelement_{12} & majorelement_{13} & majorelement_{14} \\\\\nmajorelement_{21} & majorelement_{22} & majorelement_{23} & majorelement_{24} \\\\\nmajorelement_{31} & majorelement_{32} & majorelement_{33} & majorelement_{34} \\\\\nmajorelement_{41} & majorelement_{42} & majorelement_{43} & majorelement_{44}\n\\end{array}\\right|\n\\]\n\nLet \\( indeterminant \\) be the corresponding determinant with \\( majorelement_{i unindexed} \\) replaced by \\( noncofactor_{i unindexed} \\). Prove \\( indeterminant= \\) \\( antideterminant^{3} \\).\n(page 86)\n(ii) Let \\( transcendfunc(constantval)=zerocoeff \\, constantval^{2}+microcoeff \\, constantval+varyingpart \\) be a quadratic polynomial in \\( constantval \\). If the roots of the quadratic equation \\( transcendfunc(constantval)-constantval=0 \\) are \\( leafvalue \\) and \\( stemvalue(leafvalue \\neq stemvalue) \\), show that \\( leafvalue \\) and \\( stemvalue \\) are roots of the biquadratic equation \\( transcendfunc[transcendfunc(constantval)]-constantval=0 \\). Hence write down a quadratic equation which will give the other two roots, \\( branchval \\) and \\( antideterminant \\), of the biquadratic. Apply this result to solving the following biquadratic equation:\n\\[\n\\left(constantval^{2}-3 \\, constantval+2\\right)^{2}-3\\left(constantval^{2}-3 \\, constantval+2\\right)+2-constantval=0\n\\]\n>>>\n",
      "solution": "Solution:\n<<<\nSolution. Let \\( omegaarray \\) be the matrix of the given determinant with elements \\( majorelement_{i unindexed} \\) and let \\( alphaarray \\) be the matrix of the cofactors \\( noncofactor_{i unindexed} \\), and let \\( \\gamma \\) be the transpose of \\( alphaarray \\). Then the product matrix \\( omegaarray \\gamma \\) is a diagonal matrix with all entries on the main diagonal equal to \\( antideterminant \\).\n\nThus \\( \\operatorname{det}(omegaarray \\gamma)=antideterminant^{4}=(\\operatorname{det} \\, omegaarray)(\\operatorname{det} \\, \\gamma)=(\\operatorname{det} \\, omegaarray)(\\operatorname{det} \\, alphaarray)=antideterminant \\, indeterminant \\). The equation\n\\[\nantideterminant \\, indeterminant=antideterminant^{4} .\n\\]\nis an identity between polynomials in the 16 matrix entries regarded as independent indeterminates. Since there certainly exists a \\( 4 \\times 4 \\) matrix whose determinant is not zero, \\( antideterminant \\) is non-zero in the polynomial ring. Since the polynomial ring is an integral domain, the result\n\\[\nindeterminant=antideterminant^{3}\n\\]\nfollows from (1).\nRemark. The result can obviously be generalized to matrices of any dimension. The determinant of the matrix of cofactors of an \\( unbounded \\times unbounded \\) matrix is \\( antideterminant^{unbounded-1} \\), where \\( antideterminant \\) is the determinant of the original matrix.\n\n8. (ii) Let \\( transcendfunc(constantval)=zerocoeff \\, constantval^{2}+microcoeff \\, constantval+varyingpart \\) be a quadratic polynomial in \\( constantval \\). If the roots of the quadratic equation \\( transcendfunc(constantval)-constantval=0 \\) are \\( leafvalue \\) and \\( stemvalue(leafvalue \\neq stemvalue) \\), show that \\( leafvalue \\) and \\( stemvalue \\) are roots of the biquadratic equation \\( transcendfunc[transcendfunc(constantval)]-constantval=0 \\). Hence write down a quadratic equation which will give the other two roots, \\( branchval \\) and \\( antideterminant \\), of the biquadratic. Apply this result to solving the following biquadratic equation:\n\\[\n\\left(constantval^{2}-3 \\, constantval+2\\right)^{2}-3\\left(constantval^{2}-3 \\, constantval+2\\right)+2-constantval=0 .\n\\]\n\nSolution. Since \\( leafvalue \\) is a root of \\( transcendfunc(constantval)-constantval=0 \\), we have \\( transcendfunc(leafvalue)=leafvalue \\). Then \\( transcendfunc(transcendfunc(leafvalue))=transcendfunc(leafvalue)=leafvalue \\), so \\( leafvalue \\) is a root of \\( transcendfunc(transcendfunc(constantval))-constantval=0 \\). Similarly, \\( stemvalue \\) is a root of this biquadratic.\nLet \\( basepoly(constantval)=transcendfunc(transcendfunc(constantval))-constantval \\). To find the other zeros of \\( basepoly \\), note that \\( transcendfunc(constantval)-constantval=zerocoeff \\, constantval^{2}+(microcoeff-1) \\, constantval+varyingpart=zerocoeff(constantval-leafvalue)(constantval-stemvalue) \\), whence \\( zerocoeff(leafvalue+stemvalue)=1-microcoeff \\). Then\n\\[\n\\begin{aligned}\nbasepoly(constantval)= & transcendfunc(transcendfunc(constantval))-transcendfunc(constantval)+transcendfunc(constantval)-constantval \\\\\n= & zerocoeff\\{transcendfunc(constantval)-leafvalue\\}\\{transcendfunc(constantval)-stemvalue\\}+zerocoeff(constantval-leafvalue)(constantval-stemvalue) .\n\\end{aligned}\n\\]\nContinuing,\n\\[\n\\begin{aligned}\n= & zerocoeff\\{zerocoeff(constantval-leafvalue)(constantval-stemvalue)+constantval-leafvalue\\} \\\\\n& \\qquad \\times\\{zerocoeff(constantval-leafvalue)(constantval-stemvalue)+constantval-stemvalue\\} \\\\\n& +zerocoeff(constantval-leafvalue)(constantval-stemvalue) \\\\\n= & zerocoeff(constantval-leafvalue)(constantval-stemvalue)\\, simpleroot(constantval),\n\\end{aligned}\n\\]\nwhere\n\\[\n\\begin{aligned}\nsimpleroot(constantval) & =\\{zerocoeff(constantval-stemvalue)+1\\}\\{zerocoeff(constantval-leafvalue)+1\\}+1 \\\\\n& =zerocoeff \\, transcendfunc(constantval)+zerocoeff \\, constantval-zerocoeff(leafvalue+stemvalue)+2 \\\\\n& =zerocoeff^{2} \\, constantval^{2}+zerocoeff(microcoeff+1) \\, constantval+zerocoeff \\, varyingpart+microcoeff+1 .\n\\end{aligned}\n\\]\n\nThe roots \\( branchval \\) and \\( antideterminant \\) are the zeros of \\( simpleroot \\), so the required quadratic equation for \\( branchval \\) and \\( antideterminant \\) is\n\\[\nzerocoeff^{2} \\, constantval^{2}+zerocoeff(microcoeff+1) \\, constantval+zerocoeff \\, varyingpart+microcoeff+1=0 .\n\\]\n\nIn the special case given, \\( zerocoeff=1, \\, microcoeff=-3, \\, varyingpart=2 \\), and \\( simpleroot(constantval)=constantval^{2}-2 \\, constantval \\). The zeros of \\( transcendfunc(constantval)-constantval \\) are \\( 2 \\pm \\sqrt{2} \\), so the zeros of \\( basepoly \\) are \\( 2 \\pm \\sqrt{2}, 0 \\), and 2 .\n>>>\n"
    },
    "garbled_string": {
      "map": {
        "A_ik": "qzxwvtnp",
        "a_ik": "hjgrksla",
        "d": "vmbqcloe",
        "D": "turyznas",
        "\\alpha": "kwjfiezr",
        "\\beta": "plxhrmdu",
        "A": "xsgltpdo",
        "B": "nqbfrzke",
        "C": "wyndocva",
        "P": "cehsturz",
        "y": "mglvhdas",
        "a": "frtqkebi",
        "b": "skydmevo",
        "Q": "afvhrnci",
        "R": "zjplqtdx",
        "c": "ouhnimvr",
        "k": "gmlapxni",
        "n": "lqmetrsa"
      },
      "question": "8. Take either (i) or (ii).\n(i) Let \\( qzxwvtnp \\) be the cofactor of \\( hjgrksla \\) in the determinant\n\\[\n\\boldsymbol{vmbqcloe}=\\left|\\begin{array}{llll}\nfrtqkebi_{11} & frtqkebi_{12} & frtqkebi_{13} & frtqkebi_{14} \\\\\nfrtqkebi_{21} & frtqkebi_{22} & frtqkebi_{23} & frtqkebi_{24} \\\\\nfrtqkebi_{31} & frtqkebi_{32} & frtqkebi_{33} & frtqkebi_{34} \\\\\nfrtqkebi_{41} & frtqkebi_{42} & frtqkebi_{43} & frtqkebi_{44}\n\\end{array}\\right|\n\\]\n\nLet \\( turyznas \\) be the corresponding determinant with \\( hjgrksla \\) replaced by \\( qzxwvtnp \\). Prove \\( turyznas= \\) \\( vmbqcloe^{3} \\).\n(page 86)\n(ii) Let \\( cehsturz(mglvhdas)=xsgltpdo\\,mglvhdas^{2}+nqbfrzke\\,mglvhdas+wyndocva \\) be a quadratic polynomial in \\( mglvhdas \\). If the roots of the quadratic equation \\( cehsturz(mglvhdas)-mglvhdas=0 \\) are \\( frtqkebi \\) and \\( skydmevo(frtqkebi \\neq skydmevo) \\), show that \\( frtqkebi \\) and \\( skydmevo \\) are roots of the biquadratic equation \\( cehsturz[cehsturz(mglvhdas)]-mglvhdas=0 \\). Hence write down a quadratic equation which will give the other two roots, \\( ouhnimvr \\) and \\( vmbqcloe \\), of the biquadratic. Apply this result to solving the following biquadratic equation:\n\\[\n\\left(mglvhdas^{2}-3 mglvhdas+2\\right)^{2}-3\\left(mglvhdas^{2}-3 mglvhdas+2\\right)+2-mglvhdas=0\n\\]",
      "solution": "Solution. Let \\( kwjfiezr \\) be the matrix of the given determinant with elements \\( hjgrksla \\) and let \\( plxhrmdu \\) be the matrix of the cofactors \\( qzxwvtnp \\), and let \\( \\gamma \\) be the transpose of \\( plxhrmdu \\). Then the product matrix \\( kwjfiezr \\gamma \\) is a diagonal matrix with all entries on the main diagonal equal to \\( vmbqcloe \\).\n\nThus \\( \\operatorname{det}(kwjfiezr \\gamma)=vmbqcloe^{4}=(\\operatorname{det} kwjfiezr)(\\operatorname{det} \\gamma)=(\\operatorname{det} kwjfiezr)(\\operatorname{det} plxhrmdu)=vmbqcloe\\,turyznas \\). The equation\n\\[\nvmbqcloe\\,turyznas=vmbqcloe^{4} .\n\\]\nis an identity between polynomials in the 16 matrix entries regarded as independent indeterminates. Since there certainly exists a \\( 4 \\times 4 \\) matrix whose determinant is not zero, \\( vmbqcloe \\) is non-zero in the polynomial ring. Since the polynomial ring is an integral domain, the result\n\\[\nturyznas=vmbqcloe^{3}\n\\]\nfollows from (1).\nRemark. The result can obviously be generalized to matrices of any dimension. The determinant of the matrix of cofactors of an \\( lqmetrsa \\times lqmetrsa \\) matrix is \\( vmbqcloe^{lqmetrsa-1} \\), where \\( vmbqcloe \\) is the determinant of the original matrix.\n\n8. (ii) Let \\( cehsturz(mglvhdas)=xsgltpdo\\,mglvhdas^{2}+nqbfrzke\\,mglvhdas+wyndocva \\) be a quadratic polynomial in \\( mglvhdas \\). If the roots of the quadratic equation \\( cehsturz(mglvhdas)-mglvhdas=0 \\) are \\( frtqkebi \\) and \\( skydmevo(frtqkebi \\neq skydmevo) \\), show that \\( frtqkebi \\) and \\( skydmevo \\) are roots of the biquadratic equation \\( cehsturz[cehsturz(mglvhdas)]-mglvhdas=0 \\). Hence write down a quadratic equation which will give the other two roots, \\( ouhnimvr \\) and \\( vmbqcloe \\), of the biquadratic. Apply this result to solving the following biquadratic equation:\n\\[\n\\left(mglvhdas^{2}-3 mglvhdas+2\\right)^{2}-3\\left(mglvhdas^{2}-3 mglvhdas+2\\right)+2-mglvhdas=0 .\n\\]\n\nSolution. Since \\( frtqkebi \\) is a root of \\( cehsturz(mglvhdas)-mglvhdas=0 \\), we have \\( cehsturz(frtqkebi)=frtqkebi \\). Then \\( cehsturz(cehsturz(frtqkebi))=cehsturz(frtqkebi)=frtqkebi \\), so \\( frtqkebi \\) is a root of \\( cehsturz(cehsturz(mglvhdas))-mglvhdas=0 \\). Similarly, \\( skydmevo \\) is a root of this biquadratic.\nLet \\( afvhrnci(mglvhdas)=cehsturz(cehsturz(mglvhdas))-mglvhdas \\). To find the other zeros of \\( afvhrnci \\), note that \\( cehsturz(mglvhdas)-mglvhdas=xsgltpdo\\,mglvhdas^{2}+(nqbfrzke-1)mglvhdas+wyndocva=xsgltpdo(mglvhdas-frtqkebi)(mglvhdas-skydmevo) \\), whence \\( xsgltpdo(frtqkebi+skydmevo)=1-nqbfrzke \\). Then\n\\[\n\\begin{aligned}\nafvhrnci(mglvhdas)=&\\;cehsturz(cehsturz(mglvhdas))-cehsturz(mglvhdas)+cehsturz(mglvhdas)-mglvhdas\\\\\n=&\\;xsgltpdo\\{cehsturz(mglvhdas)-frtqkebi\\}\\{cehsturz(mglvhdas)-skydmevo\\}+xsgltpdo(mglvhdas-frtqkebi)(mglvhdas-skydmevo)\\\\\n=&\\;xsgltpdo\\{xsgltpdo(mglvhdas-frtqkebi)(mglvhdas-skydmevo)+mglvhdas-frtqkebi\\}\\{xsgltpdo(mglvhdas-frtqkebi)(mglvhdas-skydmevo)+mglvhdas-skydmevo\\}\\\\\n&\\;+xsgltpdo(mglvhdas-frtqkebi)(mglvhdas-skydmevo)\\\\\n=&\\;xsgltpdo(mglvhdas-frtqkebi)(mglvhdas-skydmevo)\\,zjplqtdx(mglvhdas),\n\\end{aligned}\n\\]\nwhere\n\\[\n\\begin{aligned}\nzjplqtdx(mglvhdas)=&\\;\\{xsgltpdo(mglvhdas-skydmevo)+1\\}\\{xsgltpdo(mglvhdas-frtqkebi)+1\\}+1\\\\\n=&\\;xsgltpdo\\,cehsturz(mglvhdas)+xsgltpdo\\,mglvhdas-xsgltpdo(frtqkebi+skydmevo)+2\\\\\n=&\\;xsgltpdo^{2}mglvhdas^{2}+xsgltpdo(nqbfrzke+1)mglvhdas+xsgltpdo\\,wyndocva+nqbfrzke+1 .\n\\end{aligned}\n\\]\n\nThe roots \\( ouhnimvr \\) and \\( vmbqcloe \\) are the zeros of \\( zjplqtdx \\), so the required quadratic equation for \\( ouhnimvr \\) and \\( vmbqcloe \\) is\n\\[\nxsgltpdo^{2}mglvhdas^{2}+xsgltpdo(nqbfrzke+1)mglvhdas+xsgltpdo\\,wyndocva+nqbfrzke+1=0 .\n\\]\n\nIn the special case given, \\( xsgltpdo=1, nqbfrzke=-3, wyndocva=2 \\), and \\( zjplqtdx(mglvhdas)=mglvhdas^{2}-2mglvhdas \\). The zeros of \\( cehsturz(mglvhdas)-mglvhdas \\) are \\( 2 \\pm \\sqrt{2} \\), so the zeros of \\( afvhrnci \\) are \\( 2 \\pm \\sqrt{2}, 0 \\), and 2 ."
    },
    "kernel_variant": {
      "question": "Let  \n\n  d = det α ,  α = (a_{ik})_{1≤i,k≤5},  a_{ik} independent indeterminates.  \n\nFor every pair (i,k) let A_{ik} be the cofactor of a_{ik} in d and put  \n\n  β = (A_{ik})_{1≤i,k≤5}.  \n\nFinally write γ = β^T = adj α and δ = adj γ (i.e. the adjugate taken twice).\n\nEstablish, with full justification, the following three statements.\n\n(i) det β = d⁴.  \n\n(ii) δ = d³ α.  \n\n(iii) det δ = d¹⁵ and, if d ≠ 0, the matrices α and β have the same rank and the same right–kernel: for every column-vector v one has αv = 0 ⇔ βv = 0.\n\n———",
      "solution": "Throughout “det” means determinant and I₅ is the 5 × 5 identity.  \nBecause all 25 indeterminates are algebraically independent, the polynomial ring in which we work is an integral domain; hence factors may be cancelled once we know they are non–zero for at least one numerical specialization.\n\nStep 1.  Proof of (i): determinant of the first adjugate.  \nObserve that γ = β^T is the classical adjugate of α, so α γ = γ α = d I₅.  \nTaking determinants gives  \n det(α γ) = det(d I₅) = d⁵.  \nBut det(α γ) = det α · det γ = d · det β, whence d · det β = d⁵.  \nBecause d can be non–zero (take α = I₅), cancellation yields det β = d⁴, proving (i).\n\nStep 2.  A lemma on “the adjugate of the adjugate”.  \nFor an invertible n × n matrix M one has  \n adj(adj M) = (det M)^{n–2} M.             (★)  \nIndeed adj M = (det M) M⁻¹, so  \n adj(adj M) = det(adj M) · (adj M)⁻¹  \n      = (det M)^{n–1} · (det M)^{-1} M  \n      = (det M)^{n–2} M.  \nBecause both sides of (★) are polynomial in the entries of M, the identity extends to all M (even singular) by the integral-domain argument used in Step 1.\n\nStep 3.  Proof of (ii): substitute n = 5 and M = α in (★).  \nWith n = 5 we get  \n adj(adj α) = d³ α.  \nBut adj α = γ, so adj γ = δ = d³ α, completing (ii).\n\nStep 4.  Proof of (iii): determinant and kernel comparison.  \nFirst, det δ = det(d³ α) = d^{3·5} = d¹⁵, as claimed.  \n\nNext, assume d ≠ 0 (so α is invertible).  \nBecause γ = adj α satisfies α γ = d I₅, we have γ = d α⁻¹.  \nIn particular β = γ^T = d (α⁻¹)^T.  \nHence β = d α^{-T}.  Multiplying by α gives α β = d I₅^{T} = d I₅, so α β = d I₅ = β α.  \n\nNow let v be any column vector.  \nIf αv = 0, multiply by β: d v = β α v = 0 ⇒ v = 0.  \nConversely, if βv = 0, multiply by α: d v = α β v = 0 ⇒ v = 0.  \nThus, for d ≠ 0, both kernels are trivial and hence identical; the corresponding ranks are equal as well.  \n\nThe three requested assertions are therefore established. ∎\n\n———",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.072393",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}