// Numbas version: finer_feedback_settings {"name": "Determinant of a 3x3 matrix", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "variables": {"c": {"name": "c", "templateType": "randrange", "description": "", "group": "Ungrouped variables", "definition": "random(1..20#1)"}, "b": {"name": "b", "templateType": "randrange", "description": "", "group": "Ungrouped variables", "definition": "random(6..9#1)"}, "g": {"name": "g", "templateType": "anything", "description": "", "group": "Ungrouped variables", "definition": "1"}, "a": {"name": "a", "templateType": "randrange", "description": "", "group": "Ungrouped variables", "definition": "random(1..20#1)"}, "d": {"name": "d", "templateType": "randrange", "description": "", "group": "Ungrouped variables", "definition": "random(1..6#1)"}, "f": {"name": "f", "templateType": "randrange", "description": "", "group": "Ungrouped variables", "definition": "random(1..5#1)"}, "k": {"name": "k", "templateType": "anything", "description": "", "group": "Ungrouped variables", "definition": "{a}*{e1}*{d}+{b}*{c}*{f}-{a}*{c}*{g}-{b}*{e1}*{f}*{d}"}, "e1": {"name": "e1", "templateType": "randrange", "description": "", "group": "Ungrouped variables", "definition": "random(6..9#1)"}}, "metadata": {"description": "

This question tests learner's knowledge of the inverse matrix method for a 3x3 matrix.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "name": "Determinant of a 3x3 matrix", "parts": [{"variableReplacements": [], "showCorrectAnswer": true, "type": "gapfill", "gaps": [{"variableReplacements": [], "minValue": "{f}*{d}", "showCorrectAnswer": true, "type": "numberentry", "precisionType": "dp", "notationStyles": ["plain", "en", "si-en"], "precisionMessage": "You have not given your answer to the correct precision.", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "precision": 0, "correctAnswerFraction": false, "strictPrecision": false, "mustBeReducedPC": 0, "mustBeReduced": false, "maxValue": "{f}*{d}", "showPrecisionHint": false, "allowFractions": false, "scripts": {}, "marks": 1, "correctAnswerStyle": "plain"}, {"variableReplacements": [], "minValue": "{b}*{e1}", "showCorrectAnswer": true, "type": "numberentry", "notationStyles": ["plain", "en", "si-en"], "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "mustBeReducedPC": 0, "mustBeReduced": false, "maxValue": "{b}*{e1}", "allowFractions": false, "scripts": {}, "marks": 1, "correctAnswerStyle": "plain"}], "prompt": "

Enter the smaller of the two values

\\(x=\\) [[0]]

Enter the larger of the two values

\\(x=\\) [[1]]

", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "scripts": {}, "marks": 0}], "rulesets": {}, "variable_groups": [], "variablesTest": {"condition": "", "maxRuns": 100}, "tags": [], "extensions": [], "statement": "

Calculate the two values for \\(x\\) that satisfy the equation \\(|A|=\\var{k}\\)

where the matrix \\(A\\) is given by:

\\(\\mathbf{A}=\\begin{pmatrix} x&\\var{a}&\\var{b}\\\\ \\var{c}&x&\\var{d}\\\\\\var{e1}&\\var{f}&\\var{g} \\end{pmatrix}\\)

", "preamble": {"css": "", "js": ""}, "ungrouped_variables": ["a", "b", "c", "d", "e1", "f", "g", "k"], "advice": "

Given the matrix:

\\(\\mathbf{A}=\\begin{pmatrix} {a11}&{a12}&{a13}\\\\ {a21}&{a22}&{a23}\\\\{a31}&{a32}&{a33} \\end{pmatrix}\\)

The determinant of a 3x3 matrix is determined by the formula:

\\(|\\mathbf{A}|={a11}*\\begin{vmatrix}{a22}&{a23}\\\\{a32}&{a33}\\end{vmatrix}-{a12}*\\begin{vmatrix}{a21}&{a23}\\\\{a31}&{a33}\\end{vmatrix}+{a13}*\\begin{vmatrix}{a21}&{a22}\\\\{a31}&{a32}\\end{vmatrix}\\)

So in this example:

\\(\\mathbf{A}=\\begin{pmatrix}x&\\var{a}&\\var{b}\\\\ \\var{c}&x&\\var{d}\\\\\\var{e1}&\\var{f}&\\var{g} \\end{pmatrix}\\)

\\(|\\mathbf{A}|=x*\\begin{vmatrix}x&\\var{d}\\\\\\var{f}&\\var{g}\\end{vmatrix}-\\var{a}*\\begin{vmatrix}\\var{c}&\\var{d}\\\\\\var{e1}&\\var{g}\\end{vmatrix}+\\var{b}*\\begin{vmatrix}\\var{c}&x\\\\\\var{e1}&\\var{f}\\end{vmatrix}\\)

\\(|\\mathbf{A}|=x*(x*\\var{g}-\\var{d}*\\var{f})-\\var{a}*(\\var{c}*\\var{g}-\\var{e1}*\\var{d})+\\var{b}*(\\var{c}*\\var{f}-\\var{e1}x)\\)

\\(|\\mathbf{A}|=x(x-\\simplify{{d}{f}})-\\var{a}(\\simplify{{c}*{g}}-\\simplify{{e1}*{d}})+\\var{b}(\\simplify{{c}*{f}}-\\var{e1}x)\\)

\\(|\\mathbf{A}|=x^2-\\simplify{{d}*{f}}x-\\simplify{{a}*{c}*{g}}+\\simplify{{a}*{e1}*{d}}+\\simplify{{b}*{c}*{f}}-\\simplify{{b}*{e1}}x\\)

\\(|\\mathbf{A}|=x^2-\\simplify{{d}*{f}+{b}*{e1}}x+\\simplify{{a}*{e1}*{d}+{b}*{c}*{f}-{a}*{c}*{g}}\\)

If \\(|\\mathbf{A}|=\\var{k}\\)

\\(x^2-\\simplify{{d}*{f}+{b}*{e1}}x+\\simplify{{a}*{e1}*{d}+{b}*{c}*{f}-{a}*{c}*{g}}=\\var{k}\\)

\\(x^2-\\simplify{{d}*{f}+{b}*{e1}}x+\\simplify{{b}*{e1}*{f}*{d}}=0\\)

This can be solved by formula or by finding factors

\\((x-\\simplify{{f}*{d}})(x-\\simplify{{b}*{e1}})=0\\)

\\(x-\\simplify{{f}*{d}}=0\\,\\,\\,\\,\\,or\\,\\,\\,\\,\\,x-\\simplify{{b}*{e1}}=0\\)

\\(x=\\simplify{{f}*{d}}\\,\\,\\,\\,\\,or\\,\\,\\,\\,\\,x=\\simplify{{b}*{e1}}\\)

", "type": "question", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}