// Numbas version: finer_feedback_settings
{"name": "John's copy of Finding bases for the null space and row space of a matrix", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"tags": ["bases", "basis", "column space", "dimension", "echelon form", "linear algebra", "linear independence", "linearly dependent", "matrices", "matrix type", "null space", "nullity", "rank", "row operations", "row reduced", "row space", "row-reduced"], "name": "John's copy of Finding bases for the null space and row space of a matrix", "variable_groups": [], "rulesets": {}, "ungrouped_variables": ["echelonform", "ech", "echform", "testmatrix", "record", "rref1", "record_ops_matrix", "record_ops_message", "rank", "maxint", "steps", "rdmat", "deter", "nullity", "echmatrix", "trtestmatrix", "nbasis_column", "coordcol", "freecols", "rows", "columns", "column_basis", "sum", "null_basis", "ec", "chosen", "student_cols_matrix", "not_chosen_col", "othercols", "c0", "c1", "colmark", "z"], "parts": [{"variableReplacements": [], "prompt": "<p>Rank of $A=\\;$[[0]]</p>\n<p>Nullity of $A=\\;$[[1]]</p>", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "gaps": [{"maxValue": "rank", "minValue": "rank", "correctAnswerFraction": false, "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": "0.5", "variableReplacements": [], "allowFractions": false, "showCorrectAnswer": true, "showPrecisionHint": false, "type": "numberentry"}, {"maxValue": "nullity", "minValue": "nullity", "correctAnswerFraction": false, "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": "0.5", "variableReplacements": [], "allowFractions": false, "showCorrectAnswer": true, "showPrecisionHint": false, "type": "numberentry"}]}, {"variableReplacements": [], "prompt": "<p>Determine a basis of $\\operatorname{row}(A)$ consisting of rows of $R$ by clicking on the appropriate choices of rows of $R$.</p>\n<p>[[0]]</p>", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "gaps": [{"maxAnswers": 0, "minAnswers": 0, "shuffleChoices": false, "maxMarks": "1", "variableReplacements": [], "warningType": "none", "matrix": ["0.3", "0.3", "0.4", 0, 0, 0], "displayType": "checkbox", "scripts": {}, "choices": ["<p>Row 1</p>", "<p>Row 2</p>", "<p>Row&nbsp;3</p>", "<p>Row 4</p>", "<p>Row&nbsp;5</p>", "<p>Row 6</p>"], "minMarks": 0, "marks": 0, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "displayColumns": 0, "type": "m_n_2", "distractors": ["", "", "", "", "", ""]}]}, {"maxAnswers": 0, "minAnswers": 0, "warningType": "none", "maxMarks": "1", "variableReplacements": [], "shuffleChoices": false, "variableReplacementStrategy": "originalfirst", "displayType": "checkbox", "prompt": "<p></p>\n<p>Click on the choices of columns of $A$ which form a basis for the column space.</p>\n\n<p>Note that there are more than one possible choice of columns and if the columns you choose form a basis then this will be marked as correct. If you Reveal the answer you will see one possible such choice.</p>\n<p style=\"color: red;\" id=\"explain\"></p>", "scripts": {"mark": {"script": "var trm=variables.trtestmatrix;\nvar echform=variables.echform;\nvar rank=variables.rank;\nvar trm=Numbas.util.copyarray(trm,true);\nthis.answered=true;\n\n\nvar tm=[];\nvar selected=[];\nvar not_selected=[];\nvar numTicks = 0;\nfor( var i=0;i<5;i++) {\n  if(this.ticks[i][0]) {\n    numTicks += 1;\n    tm.push(trm[i]);\n    selected.push(i+1);\n  }\n  else {not_selected.push(i+1);}\n}\n\nvar rank1 = question.functions.ref(tm);\nvar diff = false;\nvar explain = this.question.display.html.find('#explain');\nvar note = '';\nvar mess = '';\nfor(var i=0;i<3;i++) { \n  if(echform[i]!=selected[i]) {\n    diff=true;\n  }\n}\nif(diff) {\n  mess='You have chosen correctly, but the easiest choice to make is those columns which have pivots, in this case columns '+echform[0]+', '+echform[1]+', '+echform[2]+'. ';\n}\nif(numTicks<rank) {\n  this.setCredit(0,'You have not put in enough columns, the rank is greater than '+numTicks+'.');\n} else if(numTicks>rank) {\n  this.setCredit(0, 'You have included too many columns, the rank is less than '+numTicks+'.');\n} else if(rank1==rank) {\n  if(diff) {\n    mess='You have chosen correctly, but the easiest choice to make is those columns which have pivots, in this case columns '+echform[0]+', '+echform[1]+', '+echform[2]+'. ';\n  } else { \n    mess='';\n  }\n  this.setCredit(1,mess);\n  if(diff){ \n    note='Click on the Show feedback button for more information on your choices.';\n  } else {\n    note='';\n  }\n  explain.text(note);\n} else {\n  this.setCredit(0,'These columns are not linearly independent.');\n}\n\n\nthis.validation.numTicks = numTicks;", "order": "instead"}}, "matrix": "colmark", "minMarks": 0, "marks": 0, "showCorrectAnswer": true, "displayColumns": 0, "type": "m_n_2", "choices": ["<p>Column 1</p>", "<p>Column 2</p>", "<p>Column 3</p>", "<p>Column 4</p>", "<p>Column 5</p>"]}, {"variableReplacements": [], "prompt": "<p>Express the following columns in terms of the columns $c_\\var{echform[0]}$,&nbsp;$c_\\var{echform[1]}$,&nbsp;$c_\\var{echform[2]}$</p>\n<p>$c_{\\var{othercols[0]}}$=[[0]]$c_\\var{echform[0]}$+[[1]]$c_{\\var{echform[1]}}$+[[2]]$c_{\\var{echform[2]}}$</p>\n<p>$c_{\\var{othercols[1]}}$=[[3]]$c_\\var{echform[0]}$+[[4]]$c_{\\var{echform[1]}}$+[[5]]$c_{\\var{echform[2]}}$</p>", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "gaps": [{"maxValue": "c0[0]", "minValue": "c0[0]", "correctAnswerFraction": false, "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": "0.5", "variableReplacements": [], "allowFractions": false, "showCorrectAnswer": true, "showPrecisionHint": false, "type": "numberentry"}, {"maxValue": "c0[1]", "minValue": "c0[1]", "correctAnswerFraction": false, "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": "0.5", "variableReplacements": [], "allowFractions": false, "showCorrectAnswer": true, "showPrecisionHint": false, "type": "numberentry"}, {"maxValue": "c0[2]", "minValue": "c0[2]", "correctAnswerFraction": false, "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": "0.5", "variableReplacements": [], "allowFractions": false, "showCorrectAnswer": true, "showPrecisionHint": false, "type": "numberentry"}, {"maxValue": "c1[0]", "minValue": "c1[0]", "correctAnswerFraction": false, "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": "0.5", "variableReplacements": [], "allowFractions": false, "showCorrectAnswer": true, "showPrecisionHint": false, "type": "numberentry"}, {"maxValue": "c1[1]", "minValue": "c1[1]", "correctAnswerFraction": false, "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": "0.5", "variableReplacements": [], "allowFractions": false, "showCorrectAnswer": true, "showPrecisionHint": false, "type": "numberentry"}, {"maxValue": "c1[2]", "minValue": "c1[2]", "correctAnswerFraction": false, "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": "0.5", "variableReplacements": [], "allowFractions": false, "showCorrectAnswer": true, "showPrecisionHint": false, "type": "numberentry"}]}, {"allowResize": true, "correctAnswer": "transpose(matrix(null_basis))", "numColumns": "1", "variableReplacements": [], "allowFractions": false, "correctAnswerFractions": false, "variableReplacementStrategy": "originalfirst", "prompt": "<p>Determine a basis for $\\operatorname{Null}(A)$.</p>\n<p>Enter the basis elements as the columns of a matrix:</p>\n<p>Note that any such basis will be marked as correct. If you Reveal the answer you will see one possible such basis.</p>\n<p></p>", "scripts": {"mark": {"script": "var nullity=variables.nullity;\nvar columns=variables.columns;\nvar rows=variables.rows;\nvar testmatrix=variables.testmatrix;\nvar ans=this.studentAnswer;\nvar ans_rows=this.studentAnswerRows;\nthis.answered=true;\n\nif(ans_rows!=columns){\n  this.setCredit(0,'These vectors are not of the correct dimension.')\n  return;\n}\nvar ans_cols=this.studentAnswerColumns;\nif(ans_cols!=nullity){\n  this.setCredit(0,'Not the correct number of vectors.')\n  return;\n}\nvar dim_null_space_ans=question.functions.ref(ans);\nif(dim_null_space_ans!=nullity){\n  this.setCredit(0, 'These vectors are not linearly independent.');\n  return;\n}\nfunction check_in_kernel(m1,m2) {\n  var d = 0;\n  for(var i=0;i<rows;i++) {\n    for(var k=0;k<nullity;k++){\n    var s=0;\n      for (var j=0;j<columns;j++)\n  {s+=m1[i][j]*m2[j][k];}\n  d+=Math.abs(s);\n    }             \n  }\n  return d;\n}\nvar check=check_in_kernel(testmatrix,ans);\nif(check!=0){\n  this.setCredit(0,'Although these vectors are linearly independent, one or more is not in the null space.')\n}\nelse this.setCredit(1);", "order": "instead"}}, "markPerCell": false, "marks": "4", "showCorrectAnswer": true, "tolerance": 0, "type": "matrix", "numRows": "5"}], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "notes": "<p>Fix rank to be 3.</p>", "description": "<p>Given a&nbsp;matrix in row reduced form use this to find bases for the null, column and row spaces of the matrix.</p>"}, "preamble": {"js": "question.functions={};\nquestion.functions.ref=function(m){\n  \n  m = util.copyarray(m,true);\n  \n  function finish() {\n\n    return rank;\n  }\n\n\n\n  var lead = 0;\n  var rank = 0;\n  var echelon=[];\n  var rows = m.length;\n  for(var i=0;i<rows;i++){echelon.push(0);}\n  var columns=m[0].length;\n  for (var r = 0; r < rows; r++) {\n    if (columns <= lead) {\n      return finish();\n    }\n    var i = r;\n    while (m[i][lead] == 0) {\n      i++;\n      if (rows == i) {\n\n        i = r;\n        lead++;\n        if (columns == lead) {\n          return finish();\n        }\n      }\n    }\n\n    var tmp = m[i];\n    m[i] = m[r];\n    m[r] = tmp;\n    echelon[r]=lead+1;\n    rank+=1;\n    //var val = m[r][lead];\n    //var inval;\n    //nval=1/val;\n\n    // divide this row by the leading value\n    /*\n        for (var j = 0; j < columns; j++) {\n            m[r][j] = m[r][j]*inval;\n        }\n      */\n\n    // for each row, find a linear combination of m[r] and m[i] so that m[i][lead] becomes 0\n    for (var i = 0; i < rows; i++) {\n      if (i == r) continue;\n      var v_i = m[i][lead];\n      if(v_i!=0) {\n        var v_r = m[r][lead];\n        var lcm = Numbas.math.lcm(Math.abs(v_i),Math.abs(v_r));\n        var a = lcm/Math.abs(v_i);\n        var b = -Numbas.math.sign(v_i*v_r)*lcm/Math.abs(v_r);\n        for (var j = 0; j < columns; j++) {\n          m[i][j] = a*m[i][j] + b*m[r][j];\n        }\n      }\n    }\n    lead++;\n  }\n  return finish();\n}\n\n", "css": ""}, "variablesTest": {"condition": "deter<>0\nand\nmax(map(testing(record[0][x]),x,0..length(record[0])-1))=0", "maxRuns": "200"}, "question_groups": [{"name": "", "pickQuestions": 0, "pickingStrategy": "all-ordered", "questions": []}], "functions": {"rref": {"parameters": [["a", "list"], ["p", "number"]], "definition": "\nfunction findinv(a,q){\n  var s=0;\n  for(var j=1;j<q;j++){\n    if(modulo(a*j,q)==1){s=j;}\n  }\n  return s;\n}\nfunction copy_matrix(m){\n  return Numbas.util.copyarray(m,true);\n}\nfunction modulo(n,r){\n  if(r==0){return n;} else {return (r+n%r)%r;}\n}\n    var m=copy_matrix(a);\n    var lead = 0;\n    var echelon=[];\n    var deter=1;\n    var rows = m.length;\n    for(var i=0;i<rows;i++){echelon.push(0);}\n    var rank=rows;\n    var steps = 0;\n    var record_ops_matrix=[a];\n    var record_ops_message=['Starting with the matrix.'];\n    var columns=m[0].length;\n    for (var r = 0; r < rows; r++) {\n        if (columns <= lead) {\n          \n          record_ops_matrix.push(copy_matrix(m));\n          record_ops_message.push('Finished!');\n          return [record_ops_matrix, record_ops_message,steps,deter];\n        }\n        var i = r;\n        while (modulo(m[i][lead],p) == 0) {\n            i++;\n            if (rows == i) {\n                deter=0;\n                i = r;\n                rank-=1;\n                lead++;\n                if (columns == lead) {\n                    steps+=1;\n                    record_ops_matrix.push(copy_matrix(m));\n                    record_ops_message.push('Finished!');\n               return [record_ops_matrix, record_ops_message,steps,deter];\n                }\n            }\n        }\n \n        var tmp = m[i];\n        m[i] = m[r];\n        m[r] = tmp;\n        echelon[r]=lead+1;\n        var val = modulo(m[r][lead],p);\n        var inval;\n      if(p==0){inval=1/val;} else {inval=findinv(val,p);}\n      //if we have switched the ith and rth row i.e. a transposition, we have to multiply by -1\n      deter=(i==r?1:-1)*val*deter;  \n      for (var j = 0; j < columns; j++) {\n            m[r][j] = modulo(m[r][j]*inval,p);\n        }\n                    steps+=1;\n      var sw='';\n      var tw='';\n      if(val!=1){tw='We have divided throughout row '+(r+1)+' by $\\\\simplify[fractionNumbers]{{'+val+'}}$ so that the pivot is 1.';}\n      if(i!=r)\n      {sw='We switch rows '+(r+1)+' and '+(i+1)+'.';\n      }\n      if(val!=1||i!=r){record_ops_matrix.push(copy_matrix(m));\n                 record_ops_message.push(sw+tw);}\n      \n        for (var i = 0; i < rows; i++) {\n            if (i == r) continue;\n            val = modulo(m[i][lead],p);\n            for (var j = 0; j < columns; j++) {\n                m[i][j] = modulo(m[i][j]-val * m[r][j],p);\n            }\n        }\n                    steps+=1;\n\n                    record_ops_matrix.push(copy_matrix(m));\n                    record_ops_message.push('We have reduced every other entry in column '+ (lead+1)+' to zero by adding suitable multiples of row '+(r+1)+ ' to the other rows.');\n      \n        lead++;\n    }\n          steps+=1;          \n          record_ops_matrix.push(copy_matrix(m));\n          record_ops_message.push('Finished!');\n          return [record_ops_matrix, record_ops_message,steps,deter];\n", "type": "list", "language": "javascript"}, "testing": {"parameters": [["m", "list"]], "definition": "\nvar check=0;\nhere:\nfor(var i=0;i<m.length;i++){\n  for(var j=0;j<m[0].length;j++)\n  {if(Math.round(m[i][j])!=m[i][j]) {check= 1;break here;}\n  }\n}\nreturn check;\n      ", "type": "number", "language": "javascript"}, "rref1": {"parameters": [["m", "list"]], "definition": "function finish() {\n  for(var i=0;i<rows;i++) {\n    var leader = null;\n    for(var j=0;j<columns;j++) {\n      if(leader!==null) {\n        m[i][j] /= leader;\n      } else if(m[i][j]!=0) {\n        leader = m[i][j];\n        m[i][j] = 1;\n      }\n    }\n  }\n  return [m,echelon,rank];\n}\n\n//This question used the function rref(m) defined in Functions & Rulesets to find the row reduced echelon form of a matrix m.\n\n//rref(m)[0] is the row reduced matrix, rref(m)[1] the echelon form.\n\n    var lead = 0;\n    var rank=0;\n    var echelon=[];\n    var rows = m.length;\n    for(var i=0;i<rows;i++){echelon.push(0);}\n    var columns=m[0].length;\n    for (var r = 0; r < rows; r++) {\n        if (columns <= lead) {\n            return finish();\n        }\n        var i = r;\n        while (m[i][lead] == 0) {\n            i++;\n            if (rows == i) {\n              \n                i = r;\n                lead++;\n                if (columns == lead) {\n                    return finish();\n                }\n            }\n        }\n \n        var tmp = m[i];\n        m[i] = m[r];\n        m[r] = tmp;\n        echelon[r]=lead+1;\n        rank+=1;\n        //var val = m[r][lead];\n        //var inval;\n        //nval=1/val;\n\n        // divide this row by the leading value\n      /*\n        for (var j = 0; j < columns; j++) {\n            m[r][j] = m[r][j]*inval;\n        }\n      */\n \n      // for each row, find a linear combination of m[r] and m[i] so that m[i][lead] becomes 0\n       for (var i = 0; i < rows; i++) {\n         if (i == r) continue;\n         var v_i = m[i][lead];\n         if(v_i!=0) {\n           var v_r = m[r][lead];\n           var lcm = Numbas.math.lcm(Math.abs(v_i),Math.abs(v_r));\n           var a = lcm/Math.abs(v_i);\n           var b = -Numbas.math.sign(v_i*v_r)*lcm/Math.abs(v_r);\n           for (var j = 0; j < columns; j++) {\n             m[i][j] = a*m[i][j] + b*m[r][j];\n           }\n         }\n        }\n        lead++;\n    }\n    return finish();\n", "type": "list", "language": "javascript"}, "fill": {"parameters": [["a", "number"], ["b", "number"], ["c", "number"]], "definition": "if(a>=b,0,c)", "type": "number", "language": "jme"}, "echelon_form_matrix": {"parameters": [["n", "number"], ["r", "number"], ["m", "number"], ["max", "number"]], "definition": "\nfunction getRandomInt(min, max) {\n  return Math.floor(Math.random() * (max - min + 1)) + min;\n};\n//generate a random echelon form\nfunction random_echelon_form(columns,rank){\nvar l=[];\nvar ech_form=[], ech_form1=[],b=[];\nfor(var i=1;i<columns;i++) {l.push(i);}\nech_form1=Numbas.math.shuffle(l);\nech_form1.splice(0,0,0);\nech_form=ech_form1.slice(0,rank);\nb=ech_form.sort(function(x,y){return x-y;});\nreturn b;\n}\n\n\n\nvar c=random_echelon_form(m,r);\n\nvar ind;\nvar temp;\nvar ech_matrix=[];\nfor(var i=0;i<r;i++){\n  ind=c[i];\n  temp=[];\n  for(var j=0;j<m;j++){\n    j>=ind?temp.push(getRandomInt(1,max)):temp.push(0);\n  }\n  for(var k=0;k<r;k++){temp.splice(c[k],1,0);}\n  temp.splice(ind,1,1);\n\n  ech_matrix.push(temp);\n}\nfor(var i1=r;i1<n;i1++){\n var tempz=[];\n\n for(j=0;j<m;j++){tempz.push(0);} \n ech_matrix.push(tempz);\n}\n\n\nreturn [ech_matrix,c];\n\n\n\n\n\n", "type": "list", "language": "javascript"}, "colmar": {"parameters": [["z", "list"], ["ef", "list"]], "definition": "var ar=z;\nfor (var i=0;i<3;i++) {\n  ar[ef[i]-1]=1;\n}\nreturn ar;", "type": "list", "language": "javascript"}, "rdmat": {"parameters": [["n", "number"]], "definition": "function getRandomInt(min, max) {\nreturn Math.floor(Math.random() * (max - min + 1)) + min;\n};\nvar m=[];\nvar temp;\nfor(var i=0;i<n;i++){\n  temp=[];\n  for(var j=0;j<n;j++)\n  {temp.push(getRandomInt(-3,3));\n  }\n  m.push(temp);\n}\nm[0][0]=1;\nreturn m;", "type": "list", "language": "javascript"}, "solution": {"parameters": [["matrix_list", "list"], ["message_list", "list"]], "definition": "function st(o){\n  return '['+o.toString()+']';\n}\nfunction display_format(mat){\nvar s= st(mat.map(st));\n  return '$\\\\simplify[fractionNumbers]{{matrix('+s+')}}$';\n}\nvar table = $('<table>');\nvar n=matrix_list.length;\nvar m='';\nvar mess='';\nfor(var i=0;i<n;i++){\nm=display_format(matrix_list[i]);\nmess= message_list[i];\ntable.append('<tr><td style=\"text-align:left;\">'+mess+'</td></tr>');\ntable.append('<tr><td style=\"text-align:left;\"></td></tr>');\ntable.append('<tr><td style=\"text-align:left;\">'+m+'</td></tr>');  \n}\nreturn table;\n", "type": "html", "language": "javascript"}}, "showQuestionGroupNames": false, "statement": "<p>The row-reduced echelon form $R$ of the following matrix $A$</p>\n<p>$A=\\var{testmatrix}$ is given by $R=\\var{echmatrix}$.</p>\n<p><span id=\"type\"></span></p>", "variables": {"echmatrix": {"name": "echmatrix", "definition": "matrix(ech)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "coordcol": {"name": "coordcol", "definition": "map(id(5)[nbasis_column[x]-1],x,0..length(nbasis_column)-1)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "c0": {"name": "c0", "definition": "map(fill(echform[x]-1,othercols[0]-1,echmatrix[x][othercols[0]-1]),x,0..2)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "echform": {"name": "echform", "definition": "map(x+1,x,echelonform[1])", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "echelonform": {"name": "echelonform", "definition": "echelon_form_matrix(rows,rank,columns,maxint)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "columns": {"name": "columns", "definition": "5", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "record_ops_matrix": {"name": "record_ops_matrix", "definition": "record[0]", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "not_chosen_col": {"name": "not_chosen_col", "definition": "transpose(matrix(map(transpose(echmatrix)[n],n,filter(!chosen[j],j,0..4)))[0])", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "student_cols_matrix": {"name": "student_cols_matrix", "definition": "transpose(matrix(map(transpose(echmatrix)[n],n,filter(chosen[j],j,0..4))))", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "deter": {"name": "deter", "definition": "rref(rdmat,0)[3]", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "nullity": {"name": "nullity", "definition": "5-rank", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "rref1": {"name": "rref1", "definition": "rref1(list(testmatrix))[0]", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "column_basis": {"name": "column_basis", "definition": "set(map(trtestmatrix[x],x,echelonform[1]))", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "chosen": {"name": "chosen", "definition": "[true,true,false,true,false]", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "z": {"name": "z", "definition": "[0,0,0,0,0]", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "ec": {"name": "ec", "definition": "echmatrix", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "record_ops_message": {"name": "record_ops_message", "definition": "record[1]", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "freecols": {"name": "freecols", "definition": "map(transpose(echmatrix)[nbasis_column[x]-1][0..columns-1],x,0..length(nbasis_column)-1)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "testmatrix": {"name": "testmatrix", "definition": "matrix(rdmat)*matrix(ech)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "sum": {"name": "sum", "definition": "sum(echform)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "nbasis_column": {"name": "nbasis_column", "definition": "list(set(1,2,3,4,5)-set(echform))", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "othercols": {"name": "othercols", "definition": "list(set(1,2,3,4,5)-set(echform))", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "maxint": {"name": "maxint", "definition": "4", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "rdmat": {"name": "rdmat", "definition": "rdmat(rows)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "null_basis": {"name": "null_basis", "definition": "switch(sum=6,[[ec[0][2]*ec[2][3]-ec[0][3],ec[1][2]*ec[2][3]-ec[1][3],-ec[2][3],1,0],\n    [ec[0][2]*ec[2][4]-ec[0][4],ec[1][2]*ec[2][4]-ec[1][4],-ec[2][4],0,1]],\n  sum=7,[[-ec[0][2],-ec[1][2],1,0,0],[-ec[0][4],-ec[1][4],0,-ec[2][4],1]],\n  (sum=8) and(5 in echform),[[-ec[0][2],-ec[1][2],1,0,0],[-ec[0][3],-ec[1][3],0,1,0]],\n  sum=8,[[-ec[0][1],1,0,0,0],[-ec[0][4],0,-ec[1][4],-ec[2][4],1]],\n  sum=9,[[-ec[0][1],1,0,0,0],[-ec[0][3],0,-ec[1][3],1,0]],\n  [[-ec[0][1],1,0,0,0],[-ec[0][2],0,1,0,0]])", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "ech": {"name": "ech", "definition": "echelonform[0]", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "trtestmatrix": {"name": "trtestmatrix", "definition": "transpose(testmatrix)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "c1": {"name": "c1", "definition": "map(fill(echform[x]-1,othercols[1]-1,echmatrix[x][othercols[1]-1]),x,0..2)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "rows": {"name": "rows", "definition": "6", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "rank": {"name": "rank", "definition": "3\n", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "record": {"name": "record", "definition": "rref(list(testmatrix),0)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "colmark": {"name": "colmark", "definition": "colmar(z,echform)", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "steps": {"name": "steps", "definition": "rref(list(testmatrix),0)[2]", "group": "Ungrouped variables", "templateType": "anything", "description": ""}}, "type": "question", "advice": "<p><strong>a)</strong> The following shows how $A$ is reduced to row-echelon form.</p>\n<p></p>\n<p>{solution(record_ops_matrix,record_ops_message)}</p>\n<p>The rank of A is the number of columns of R with pivots, i.e. 3.</p>\n<p>The nullity of A is the number of columns of R without pivots, i.e. 2.</p>\n<p><strong>b)</strong> Given the row-echelon form we see that the first three rows of $R$ form a basis for the row space of $A$.</p>\n<p><strong>c)</strong> One basis for the column-space of $A$ is given by the original columns which in the row-reduced echelon form have a pivot.</p>\n<p>For example, $\\{c_i:i \\in \\var{set(echform)}\\}$</p>\n<p>$\\var{column_basis}$</p>\n<p>There are other bases which you could have chosen and would also have been marked as correct.</p>\n<p><strong>d)</strong> If you look at the columns $c_{\\var{othercols[0]}}$ and&nbsp;$c_{\\var{othercols[1]}}$ we can simply read off these columns as combinations of the other columns by looking at the numbers above the pivot element in each of the columns&nbsp;$c_{\\var{othercols[0]}}$ and&nbsp;$c_{\\var{othercols[1]}}$.</p>\n<p><strong>e)&nbsp;</strong>Basis for the null-space.</p>\n<p>The dimension of the null-space is $2$ and so we should have $2$ vectors in the basis.</p>\n<p>The null space is given by $\\operatorname{Null}(A)=\\operatorname{Null}(R)= $ &nbsp;\\[\\left\\{(x_1,x_2,x_3,x_4,x_5):\\simplify{{ec}*vector(x_1,x_2,x_3,x_4,x_5)=vector(0,0,0,0,0,0)}\\right\\}\\]</p>\n<p>This gives&nbsp;three equations:</p>\n<p>\\begin{align}\\simplify{{ec[0][0]}x_1+{ec[0][1]}x_2+{ec[0][2]}x_3+{ec[0][3]}x_4+{ec[0][4]}x_5}&amp;=0\\\\\\simplify{{ec[1][0]}x_1+{ec[1][1]}x_2+{ec[1][2]}x_3+{ec[1][3]}x_4+{ec[1][4]}x_5}&amp;=0\\\\\\simplify{{ec[2][0]}x_1+{ec[2][1]}x_2+{ec[2][2]}x_3+{ec[2][3]}x_4+{ec[2][4]}x_5}&amp;=0 \\end{align}</p>\n<p>The free columns (without a pivot element) are $c_{\\var{nbasis_column[0]}},c_{\\var{nbasis_column[1]}}$</p>\n<p>We obtain $2$ vectors which are linearly independent by:</p>\n<p>1) Setting: $x_{\\var{nbasis_column[0]}}=1,x_{\\var{nbasis_column[1]}}=0$ gives $\\var{rowvector(null_basis[0])}$ as a basis element for the null-space.</p>\n<p>2)&nbsp;Setting: $x_{\\var{nbasis_column[0]}}=0,x_{\\var{nbasis_column[1]}}=1$ gives $\\var{rowvector(null_basis[1])}$&nbsp;as another basis element for the null-space.</p>\n<p>Hence&nbsp;$\\{\\var{rowvector(null_basis[0])},\\var{rowvector(null_basis[1])}\\}$</p>\n<p>is a basis for the null-space.</p>\n<p>Note that this is just one possible basis, and if you input another correct basis for the null-space then it will be marked as correct.</p>\n<p><strong>Alternatively.</strong></p>\n<p>Note that on using the equations above, the variables&nbsp;$x_{\\var{echform[0]}},\\;x_{\\var{echform[1]}}$ and &nbsp;$x_{\\var{echform[2]}}$ can be expressed in terms of the variables&nbsp; $x_{\\var{nbasis_column[0]}},x_{\\var{nbasis_column[1]}}$.</p>\n<p>If we do so we find that:</p>\n<p>$&nbsp;(x_1,x_2,x_3,x_4,x_5) \\in \\operatorname{Null}(A) \\iff$ \\[(x_1,x_2,x_3,x_4,x_5)=&nbsp;x_{\\var{nbasis_column[0]}}\\var{rowvector(null_basis[0])}+x_{\\var{nbasis_column[1]}}\\var{rowvector(null_basis[1])}.\\]</p>\n<p>As $&nbsp;x_{\\var{nbasis_column[0]}},\\;x_{\\var{nbasis_column[1]}}$ can take any values we see that the vectors</p>\n<p>$\\{\\var{rowvector(null_basis[0])},\\var{rowvector(null_basis[1])}\\}$ form a basis as they span &nbsp;$\\operatorname{Null}(A)$ and there are $2$ vectors which agrees with the nullity.</p>\n<p></p>\n<p></p>", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "John Steele", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2218/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "John Steele", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2218/"}]}