// Numbas version: exam_results_page_options {"name": "Gareth's copy of Matrix entry part demo", "type": "exam", "duration": 0, "metadata": {"notes": "", "description": "

A few questions which show off the new matrix entry part.

", "licence": "Creative Commons Attribution 4.0 International"}, "allQuestions": true, "shuffleQuestions": false, "questions": [], "percentPass": 0, "timing": {"allowPause": true, "timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "pickQuestions": 0, "navigation": {"onleave": {"action": "none", "message": ""}, "reverse": true, "allowregen": true, "preventleave": true, "browse": true, "showfrontpage": false, "showresultspage": "never"}, "feedback": {"showtotalmark": true, "advicethreshold": 0, "showanswerstate": true, "showactualmark": true, "allowrevealanswer": true}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": [{"name": "Multiply two matrices", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "functions": {}, "ungrouped_variables": ["matrices"], "tags": ["matrix", "matrix entry"], "preamble": {"css": "", "js": ""}, "advice": "", "rulesets": {}, "parts": [{"prompt": "

What is $\\simplify{{matrices[0]}*{matrices[1]}}$?

[[0]]

", "marks": 0, "gaps": [{"numColumns": 1, "type": "matrix", "allowFractions": false, "correctAnswerFractions": false, "markPerCell": false, "numRows": 1, "showCorrectAnswer": true, "correctAnswer": "matrices[0]*matrices[1]", "scripts": {}, "marks": 1, "tolerance": 0, "allowResize": true}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}], "statement": "", "variable_groups": [], "progress": "in-progress", "variablesTest": {"maxRuns": 100, "condition": ""}, "type": "question", "variables": {"matrices": {"definition": "repeat(matrix(repeat(repeat(random(0..6)*random(1,1,1,-1),2),2)),2)", "templateType": "anything", "group": "Ungrouped variables", "name": "matrices", "description": "

A couple of random matrices to multiply together

"}}, "metadata": {"notes": "", "description": "

A really simple question which asks the student to multiply two matrices.

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Determinant and inverse of 2x2 matrices", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "functions": {"inverse": {"definition": "matrix([\n [m[1][1],-m[0][1]],\n [-m[1][0],m[0][0]]\n])/det(m)", "type": "matrix", "language": "jme", "parameters": [["m", "matrix"]]}}, "ungrouped_variables": [], "tags": ["determinant of a matrix", "inverse", "inverse matrix", "matrices", "matrix", "matrix inverse", "matrix multiplication", "multiplication of matrices", "tested1"], "preamble": {"css": "", "js": ""}, "advice": "

Determinant of a $2 \\times 2$ matrix

The determinant of a matrix $\\mathrm{M} = \\begin{pmatrix} a&b \\\\ c&d \\end{pmatrix}$ is given by

\\[ \\det\\left(\\mathrm{M}\\right) = ad-bc \\]

If we have two $n \\times n$ matrices $M$ and $N$, then

\\[ \\det\\left(\\mathrm{MN}\\right) = \\det\\left(\\mathrm{M}\\right)\\det\\left(\\mathrm{N}\\right) \\]

And it follows that if we have a third matrix $P$,

\\[ \\det\\left(\\mathrm{MNP}\\right) = \\det\\left(\\mathrm{M}\\right)\\det\\left(\\mathrm{N}\\right)\\det\\left(\\mathrm{P}\\right) \\]

a)

Thus for our example we have:

\\begin{align}
\\det\\left(\\mathrm{A}\\right) &= \\simplify[]{{a11}*{a22}-{a12}*{a21} = {det(a)}} \\\\
\\det\\left(\\mathrm{B}\\right) &= \\simplify[]{{b11}*{b22}-{b12}*{b21} = {det(b)}} \\\\
\\det\\left(\\mathrm{C}\\right) &= \\simplify[]{{c11}*{c22}-{c12}*{c21} = {det(c)}}
\\end{align}

\\begin{align}
\\det\\left( \\mathrm{ABC} \\right) &= \\det(\\mathrm{A}) \\det(\\mathrm{B}) \\det(\\mathrm{C}) \\\\
&= \\simplify[]{{det(a)}*{det(b)}*{det(c)}} \\\\
&= \\var{det(a*b*c)}
\\end{align}

Inverse of a $2 \\times 2$ matrix

Suppose $\\mathrm{M} = \\begin{pmatrix} a&b \\\\ c&d \\end{pmatrix}$ is a $2 \\times 2$ matrix and $\\det\\left(\\mathrm{M}\\right) = \\Delta \\neq 0$.

Then $\\mathrm{M}$ is invertible and

\\[ \\mathrm{M}^{-1} = \\frac{1}{\\Delta} \\begin{pmatrix} d & -b\\\\ -c& a \\end{pmatrix}=\\begin{pmatrix} \\frac{d}{\\Delta} & -\\frac{b}{\\Delta}\\\\ -\\frac{c}{\\Delta}& \\frac{a}{\\Delta} \\end{pmatrix}\\]

Applying this to these examples we obtain:

b)

\\[ \\simplify[fractionnumbers]{matrix:A^(-1)={inverse(a)}} \\]

c)

\\[ \\simplify[fractionnumbers]{matrix:B^(-1)={inverse(b)}} \\]

d)

\\[ \\simplify[fractionnumbers]{matrix:C^(-1)={inverse(c)}} \\]

", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers"]}, "parts": [{"prompt": "

Let

\\[\\mathrm{A} = \\var{a},\\;\\; \\mathrm{B} = \\var{b},\\;\\; \\mathrm{C} = \\var{c}\\]

Calculate the determinants of these matrices:

$\\det\\left(\\mathrm{A}\\right) = $ [[0]]

$\\det\\left(\\mathrm{B}\\right) = $ [[1]]

$\\det\\left(\\mathrm{C}\\right) = $ [[2]]

$\\det\\left(\\mathrm{ABC}\\right) = $ [[3]]

", "marks": 0, "gaps": [{"allowFractions": false, "marks": 0.5, "maxValue": "det(a)", "minValue": "det(a)", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "marks": 0.5, "maxValue": "det(b)", "minValue": "det(b)", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "marks": 0.5, "maxValue": "det(c)", "minValue": "det(c)", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "marks": 0.5, "maxValue": "det(a*b*c)", "minValue": "det(a*b*c)", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "

Find the inverses of the matrices given above. Input all matrix entries as fractions or integers and not as decimals.

$\\mathrm{A}^{-1} = $ [[0]]

", "marks": 0, "gaps": [{"numColumns": "2", "type": "matrix", "allowFractions": true, "correctAnswerFractions": true, "markPerCell": false, "numRows": "2", "showCorrectAnswer": true, "correctAnswer": "inverse(a)", "scripts": {}, "marks": 1, "tolerance": "0.01", "allowResize": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "

$\\mathrm{B}^{-1} = $ [[0]]

", "marks": 0, "gaps": [{"numColumns": "2", "type": "matrix", "allowFractions": true, "correctAnswerFractions": true, "markPerCell": false, "numRows": "2", "showCorrectAnswer": true, "correctAnswer": "inverse(b)", "scripts": {}, "marks": 1, "tolerance": "0.01", "allowResize": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "

$\\mathrm{C}^{-1} = $ [[0]]

", "marks": 0, "gaps": [{"numColumns": "2", "type": "matrix", "allowFractions": true, "correctAnswerFractions": true, "markPerCell": false, "numRows": "2", "showCorrectAnswer": true, "correctAnswer": "inverse(c)", "scripts": {}, "marks": 1, "tolerance": "0.01", "allowResize": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}], "statement": "

Do the following matrix problems.

", "variable_groups": [{"variables": ["a11", "a12", "a21", "a22", "a"], "name": "Matrix A"}, {"variables": ["b11", "b12", "b21", "b22", "b"], "name": "Matrix B"}, {"variables": ["c11", "c12", "c21", "c22", "c"], "name": "Matrix C"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "type": "question", "variables": {"c22": {"definition": "random(1..9 except c21*c12/c11)", "templateType": "anything", "group": "Matrix C", "name": "c22", "description": ""}, "a21": {"definition": "random(-6..6 except 0) ", "templateType": "anything", "group": "Matrix A", "name": "a21", "description": ""}, "a22": {"definition": "random(1..9 except a21*a12/a11)", "templateType": "anything", "group": "Matrix A", "name": "a22", "description": ""}, "c21": {"definition": "random(2..5)", "templateType": "anything", "group": "Matrix C", "name": "c21", "description": ""}, "a11": {"definition": "random(-9..9 except 0)", "templateType": "anything", "group": "Matrix A", "name": "a11", "description": ""}, "a12": {"definition": "random(-5..5)", "templateType": "anything", "group": "Matrix A", "name": "a12", "description": ""}, "c": {"definition": "matrix([\n [c11,c12],\n [c21,c22]\n])", "templateType": "anything", "group": "Matrix C", "name": "c", "description": ""}, "b21": {"definition": "random(-6..6 except 0)", "templateType": "anything", "group": "Matrix B", "name": "b21", "description": ""}, "b22": {"definition": "random(-9..9 except [0,b21*b12/b11])", "templateType": "anything", "group": "Matrix B", "name": "b22", "description": ""}, "b": {"definition": "matrix([\n [b11,b12],\n [b21,b22]\n])", "templateType": "anything", "group": "Matrix B", "name": "b", "description": ""}, "b12": {"definition": "random(-5..5)", "templateType": "anything", "group": "Matrix B", "name": "b12", "description": ""}, "a": {"definition": "matrix([\n [a11,a12],\n [a21,a22]\n])", "templateType": "anything", "group": "Matrix A", "name": "a", "description": ""}, "b11": {"definition": "random(1..9 except a11)\n//if(a11=tr2,tr2+1,tr2)", "templateType": "anything", "group": "Matrix B", "name": "b11", "description": ""}, "c12": {"definition": "a12+b12", "templateType": "anything", "group": "Matrix C", "name": "c12", "description": ""}, "c11": {"definition": "random(1,2,4)", "templateType": "anything", "group": "Matrix C", "name": "c11", "description": ""}}, "metadata": {"notes": "

10/07/2012:

Added tags.

Question appears to be working correctly.

Corrected a typo in the Advice section.

24/12/2012:

Checked calculations, OK. Added tested1 tag.

", "description": "

Find the determinant and inverse of three $2 \\times 2$ invertible matrices.

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Inverse Matrix", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "functions": {}, "ungrouped_variables": ["ra", "rb", "rc", "f", "g"], "tags": ["augmented matrices", "augmented matrix", "inverse matrix", "invertible matrices", "inverting a matrix", "linear algebra", "mas104220122013CBA4_3", "matrices", "matrix inverse", "matrix inversion"], "preamble": {"css": "", "js": ""}, "advice": "

a) Introduce zeros in the first column

The first entry in the second row is $\\var{a[1][0]}$, and we have a $1$ in the first entry of the first row, so multiply the first row by $\\var{abs(zero_2_1)}$ and {if(zero_2_1<0,'subtract from','add to')} row 2 to get

\\[ \\simplify[rowvector,all]{{step0[1]}+{zero_2_1}*{step0[0]} = {step1[1]}} \\]

The first entry in the third row is $\\var{a[2][0]}$, so multiply the first row by $\\var{abs(zero_3_1)}$ and {if(zero_3_1<0,'subtract from','add to')} row 3 to get

\\[ \\simplify[rowvector,all]{{step0[2]}+{zero_3_1}*{step0[0]} = {step1[2]}} \\]

The matrix at this stage is

\\[ \\var{step1} \\]

b) Introduce a zero in the second column of the third row

The second entry in the third row is $\\var{step1[2][1]}$, and we have a $1$ in the second entry of the second row, so multiply the second row by $\\var{abs(zero_3_2)}$ and {if(zero_3_2<0,'subtract from','add to')} row 3 to get

\\[ \\simplify[rowvector,all]{{step1[2]}+{zero_3_2}*{step1[1]} = {step1[2]+zero_3_2*step1[1]}} \\]

{if(one_3_3<>1,one_3_3_message,'')}

The matrix at this stage is

\\[ \\var{step2} \\]

c) Introduce zeros in the third column

The third entry in the first row is $\\var{step2[0][2]}$, and we have a $1$ in the third entry of the third row, so multiply the third row by $\\var{abs(zero_1_3)}$ and {if(zero_1_3<0,'subtract from','add to')} row 1 to get

\\[ \\simplify[rowvector,all]{{step2[0]}+{zero_1_3}*{step2[2]} = {step2[0]+zero_1_3*step2[2]}} \\]

The third entry in the second row is $\\var{step2[1][2]}$, so multiply the third row by $\\var{abs(zero_2_3)}$ and {if(zero_2_3<0,'subtract from','add to')} row 2 to get

\\[ \\simplify[rowvector,all]{{step2[1]}+{zero_2_3}*{step2[2]} = {step2[1]+zero_2_3*step2[2]}} \\]

The matrix at this stage is

\\[ \\var{step3} \\]

d) Introduce a zero in the second column of the first row

The second entry in the first row is $\\var{step3[0][1]}$, and we have a $1$ in the second entry of the second row, so multiply the second row by $\\var{abs(zero_1_2)}$ and {if(zero_1_2<0,'subtract from','add to')} row 1 to get

\\[ \\simplify[rowvector,all]{{step3[0]}+{zero_1_2}*{step3[1]} = {step3[0]+zero_1_2*step3[1]}} \\]

The matrix at this stage is

\\[ \\var{step4} \\]

Note that we have a $3 \\times 3$ identity matrix in the left three columns, and $\\mathrm{A}^{-1}$ is written in the three columns on the right. That is,

\\[ \\mathrm{A}^{-1} = \\var{inverse} \\]

", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "

Introduce zeros in the first column below the first entry by adding suitable multiples of the first row to rows 2 and 3.

If necessary, multiply the second row by a suitable constant so that the second entry in the second row is $1$.

$\\left(\\begin{array}{ccc|ccc} \\var{step0[0][0]} & \\var{step0[0][1]} & \\var{step0[0][2]} & \\var{step0[0][3]} & \\var{step0[0][4]} & \\var{step0[0][5]} \\\\ 0 & 1 & ? & ? & ? & ? \\\\ 0 & ? & ? & ? & ? & ? \\end{array}\\right) = $ [[0]]

", "marks": 0, "gaps": [{"numColumns": "6", "type": "matrix", "allowFractions": false, "correctAnswerFractions": false, "markPerCell": false, "numRows": "3", "showCorrectAnswer": true, "correctAnswer": "step1", "scripts": {}, "marks": 1, "tolerance": 0, "allowResize": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "

Now, using this matrix, introduce a zero in the second column of the third row.

If necessary, multiply the third row by a suitable constant so that the third entry in the third row is $1$.

", "marks": 0, "gaps": [{"numColumns": "6", "type": "matrix", "allowFractions": false, "correctAnswerFractions": false, "markPerCell": false, "numRows": "3", "showCorrectAnswer": true, "correctAnswer": "step2", "scripts": {}, "marks": 1, "tolerance": 0, "allowResize": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "

Next, introduce zeros into the third column of the second and first rows, by adding suitable multiples of the third row to those rows.

$\\left(\\begin{array}{ccc|ccc} ? & ? & 0 & ? & ? & ? \\\\ 0 & 1 & 0 & ? & ? & ? \\\\ 0 & 0 & 1 & ? & ? & ? \\end{array}\\right) = $ [[0]]

", "marks": 0, "gaps": [{"numColumns": "6", "type": "matrix", "allowFractions": false, "correctAnswerFractions": false, "markPerCell": false, "numRows": "3", "showCorrectAnswer": true, "correctAnswer": "step3", "scripts": {}, "marks": 1, "tolerance": 0, "allowResize": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "

One final operation is needed: add a multiple of the second row to the first row to introduce a zero in the second column of the first row.

$\\left(\\begin{array}{ccc|ccc} 1 & 0 & 0 & ? & ? & ? \\\\ 0 & 1 & 0 & ? & ? & ? \\\\ 0 & 0 & 1 & ? & ? & ? \\end{array}\\right) = $ [[0]]

", "marks": 0, "gaps": [{"numColumns": "6", "type": "matrix", "allowFractions": false, "correctAnswerFractions": false, "markPerCell": false, "numRows": "3", "showCorrectAnswer": true, "correctAnswer": "step4", "scripts": {}, "marks": 1, "tolerance": 0, "allowResize": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}], "statement": "

Find the inverse of the following matrix:
\\[ \\simplify{matrix:A = {a}} \\]

Form the $3 \\times 6$ augmented matrix $\\mathrm{B}$ by placing $I_3$ to the right of $\\mathrm{A}$ as below:

\\[\\mathrm{B} = \\left(\\begin{array}{rrr|ccc} \\var{a[0][0]} & \\var{a[0][1]} & \\var{a[0][2]} &\\var{1}&\\var{0}&\\var{0}\\\\ \\var{a[1][0]} & \\var{a[1][1]} & \\var{a[1][2]}&\\var{0}&\\var{1}&\\var{0}\\\\ \\var{a[2][0]} & \\var{a[2][1]} & \\var{a[2][2]}&\\var{0}&\\var{0}&\\var{1}\\\\ \\end{array}\\right)\\]

In subsequent parts work with this matrix using row operations to introduce the identity matrix on the left hand side, with the inverse of $\\mathrm{A}$ eventually appearing on the right hand side.

Input all numbers as fractions or integers and not as decimals.

", "variable_groups": [{"variables": ["a"], "name": "matrix A"}, {"variables": ["step0", "zero_2_1", "zero_3_1", "one_2_2", "step1", "zero_3_2", "one_3_3", "step2", "zero_1_3", "zero_2_3", "step3", "zero_1_2", "step4", "inverse", "one_3_3_message"], "name": "Steps"}], "progress": "in-progress", "variablesTest": {"maxRuns": 100, "condition": "f[1]<>g[1]"}, "type": "question", "variables": {"a": {"definition": "matrix([\n [f[0]*g[0], f[0]*rb*g[1], f[0]*(rb*ra-rb)*g[2]],\n [f[1]*ra*g[0], f[1]*(ra*rb-1)*g[1], f[1]*(ra^2*rb-ra-ra*rb)*g[2]],\n [f[2]*ra*rc*g[0],f[2]*rc*rb*g[1], f[2]*g[2]]\n])", "templateType": "anything", "group": "matrix A", "name": "a", "description": "

This matrix is designed so its determinant is $-1$, meaning its inverse has integer components.

Additionally, when you introduce zeros in the second and third rows, you should end up with a $1$ in each of the leading columns, so you don't have to do a division.

"}, "zero_3_1": {"definition": "-(step0[2][0]/step0[0][0])", "templateType": "anything", "group": "Steps", "name": "zero_3_1", "description": ""}, "inverse": {"definition": "matrix([\n list(step4[0])[3..6],\n list(step4[1])[3..6],\n list(step4[2])[3..6]\n])", "templateType": "anything", "group": "Steps", "name": "inverse", "description": "

The inverse of $\\mathrm{A}$

"}, "zero_3_2": {"definition": "-step1[2][1]/step1[1][1]", "templateType": "anything", "group": "Steps", "name": "zero_3_2", "description": ""}, "g": {"definition": "[f[0]]+repeat(random(-1,1),2)", "templateType": "anything", "group": "Ungrouped variables", "name": "g", "description": ""}, "f": {"definition": "repeat(random(-1,1),3)", "templateType": "anything", "group": "Ungrouped variables", "name": "f", "description": ""}, "zero_2_1": {"definition": "-(step0[1][0]/step0[0][0])", "templateType": "anything", "group": "Steps", "name": "zero_2_1", "description": ""}, "step0": {"definition": "matrix([\n list(a[0])+[1,0,0],\n list(a[1])+[0,1,0],\n list(a[2])+[0,0,1]\n])", "templateType": "anything", "group": "Steps", "name": "step0", "description": ""}, "rb": {"definition": "//5\nrandom(2..6)", "templateType": "anything", "group": "Ungrouped variables", "name": "rb", "description": "

A random number

"}, "step4": {"definition": "matrix([\n step3[0]+zero_1_2*step3[1],\n step3[1],\n step3[2]\n])", "templateType": "anything", "group": "Steps", "name": "step4", "description": "

Make sure row 1 has a zero in the second column.

"}, "one_3_3_message": {"definition": "\"Note that the third column in the third row is $\\\\var{step1[2][2]+zero_3_2*step1[1][2]}$ and not $1$, so divide the third row by $\\\\var{one_3_3}$.\"", "templateType": "anything", "group": "Steps", "name": "one_3_3_message", "description": "

To be displayed if the third row needs to be scaled in step 2.

"}, "step3": {"definition": "matrix([\n step2[0]+zero_1_3*step2[2],\n step2[1]+zero_2_3*step2[2],\n step2[2]\n])", "templateType": "anything", "group": "Steps", "name": "step3", "description": "

Make sure rows 1 and 2 have a zero in the third column

"}, "step2": {"definition": "matrix([\n step1[0],\n step1[1],\n (step1[2]+zero_3_2*step1[1])*one_3_3\n])", "templateType": "anything", "group": "Steps", "name": "step2", "description": "

Make sure row 3 has a zero in the second column, and row 3 has a 1 in the third column.

"}, "step1": {"definition": "matrix([\n step0[0],\n (step0[1]+zero_2_1*step0[0])*one_2_2,\n step0[2]+zero_3_1*step0[0]\n])", "templateType": "anything", "group": "Steps", "name": "step1", "description": "

Make sure rows 2 and 3 have a 0 in the first column, and row 2 has a 1 in the second column.

"}, "rc": {"definition": "//2\nrandom(1,2,3)", "templateType": "anything", "group": "Ungrouped variables", "name": "rc", "description": "

A random number

"}, "one_2_2": {"definition": "1/(step0[1][1]+step0[0][1]*zero_2_1)", "templateType": "anything", "group": "Steps", "name": "one_2_2", "description": ""}, "zero_1_2": {"definition": "-(step3[0][1]/step3[1][1])", "templateType": "anything", "group": "Steps", "name": "zero_1_2", "description": ""}, "one_3_3": {"definition": "(1/(step1[2][2]+step1[1][2]*zero_3_2))", "templateType": "anything", "group": "Steps", "name": "one_3_3", "description": ""}, "zero_1_3": {"definition": "-(step2[0][2]/step2[2][2])", "templateType": "anything", "group": "Steps", "name": "zero_1_3", "description": ""}, "zero_2_3": {"definition": "-(step2[1][2]/step2[2][2])", "templateType": "anything", "group": "Steps", "name": "zero_2_3", "description": ""}, "ra": {"definition": "//4\nrandom(2..6)", "templateType": "anything", "group": "Ungrouped variables", "name": "ra", "description": "

A random number

"}}, "metadata": {"notes": "

January 2015

Rewritten by CP to use the matrix entry part type.

Why do we zero the second column in two different stages? The algorithm is clearer to me if it goes:

zero rows 2 and 3 column 1 using row 1
make sure there's a 1 in the second column of row 2
zero rows 1 and 3 column 2 using row 2
make sure there's a 1 in the third column of row 3
zero rows 1 and 2 column 3 using row 3

5/07/2012:

Added tags.

Changed grammar in the question.

Question appears to be working correctly.

14/07/2012:

Need to align columns where input takes place through the stages.

", "description": "

$A$ a $3 \\times 3$ matrix. Using row operations on the augmented matrix $\\left(A | I_3\\right)$ reduce to $\\left(I_3 | A^{-1}\\right)$.

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}]}], "contributors": [{"name": "Gareth Woods", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/978/"}], "extensions": [], "custom_part_types": [], "resources": []}