// Numbas version: exam_results_page_options {"name": "Simon's copy of Solve simultaneous equations by finding inverse matrix,", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"extensions": [], "tags": [], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "
Putting a pair of linear equations into matrix notation and then solving by finding the inverse of the coefficient matrix.
"}, "ungrouped_variables": ["ma", "a00", "a01", "a10", "a11", "mb", "ma_inverse", "x", "y"], "statement": "Rewrite the following system of equations as a matrix equation
\n\\[ \\mathbf{Av} = \\mathbf{b} \\]
\nfor a matrix $\\mathbf{A}$ and column vectors $\\mathbf{v}$ and $\\mathbf{b}$.
\n\\begin{align}
\\simplify[std]{ {ma[0][0]}x + {ma[0][1]}y} &= \\var{mb[0][0]} \\\\
\\simplify[std]{ {ma[1][0]}x + {ma[1][1]}y} &= \\var{mb[1][0]}
\\end{align}
Input all numbers as fractions or integers and not as decimals.
", "variablesTest": {"condition": "", "maxRuns": 100}, "variable_groups": [], "parts": [{"showCorrectAnswer": true, "prompt": "$\\mathbf{A} = $ [[0]]
\n[[1]] | \n
[[2]] | \n
$\\mathbf{b} = $ [[3]]
Find the inverse of $\\mathbf{A}$.
\n$\\mathbf{A}^{-1} = $ [[0]]
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\n$\\mathbf{A}^{-1}\\mathbf{b} = $ [[0]]
", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "sortAnswers": false, "type": "gapfill", "scripts": {}, "marks": 0, "unitTests": [], "gaps": [{"showCorrectAnswer": true, "numRows": "2", "allowResize": false, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "type": "matrix", "tolerance": 0, "correctAnswer": "ma_inverse*mb", "scripts": {}, "marks": 1, "markPerCell": false, "unitTests": [], "correctAnswerFractions": true, "allowFractions": true, "numColumns": 1, "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "showFeedbackIcon": true}], "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "showFeedbackIcon": true}, {"showCorrectAnswer": true, "prompt": "Finally, solve the equations.
\n$x = $ [[0]]
\n$y = $ [[1]]
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\nNo entry is 0.
", "name": "ma"}, "y": {"definition": "(ma_inverse*mb)[1][0]", "group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "y"}, "a11": {"definition": "random(-9..9 except [0,a01,-a01])", "group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "a11"}}, "advice": "The equations can be written in the matrix form
\n\\[ \\var{ma}\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\var{mb} \\]
\n$\\mathrm{det}(\\mathbf{A}) = \\simplify[]{ {ma[0][0]}*{ma[1][1]} - {ma[0][1]}*{ma[1][0]}} = \\var{det(ma)} \\neq 0$, so $\\mathbf{A}$ is invertible.
\n\\[ \\mathbf{A}^{-1} = \\frac{1}{\\var{det(ma)}} \\begin{pmatrix} \\var{ma[1][1]}&\\var{-ma[0][1]} \\\\ \\var{-ma[1][0]}&\\var{ma[0][0]} \\end{pmatrix} = \\simplify[fractionnumbers]{{ma_inverse}} \\]
\nWe have
\n\\begin{align}
\\mathbf{A}^{-1}\\mathbf{b} &= \\simplify[fractionnumbers]{{ma_inverse}*{mb}} \\\\
&= \\simplify[fractionnumbers]{{ma_inverse*mb}}
\\end{align}
Given the equation $\\mathbf{Av}=\\mathbf{b}$ we can make $\\mathbf{v}$ the subject by pre-multiplying each side by $\\mathbf{A}^{-1}$:
\n\\begin{align}
\\mathbf{A}^{-1}\\mathbf{A}\\mathbf{v} &= \\mathbf{A}^{-1}\\mathbf{b} \\\\
\\mathbf{v} &= \\mathbf{A}^{-1}\\mathbf{b} \\\\ \\\\
\\end{align}
Hence,
\n\\[ \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\simplify[fractionnumbers]{{ma_inverse*mb}} \\]
\nThat is,
\n\\begin{align}
x &= \\simplify[fractionnumbers]{{x}}, \\\\ \\\\
y &= \\simplify[fractionnumbers]{{y}}
\\end{align}