// Numbas version: finer_feedback_settings {"name": "Hollie's copy of Invert a 3x3 matrix using row operations", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "a11", "a12", "a13", "a21", "a22", "a23", "a31", "a32", "a33", "b", "b24", "b25", "c", "c1", "c2", "c3", "f1", "f2", "f3", "g1", "g2", "g3", "s"], "name": "Hollie's copy of Invert a 3x3 matrix using row operations", "tags": [], "advice": "
All of the working is now shown
", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\nIntroduce zeros in the first column below the first entry by adding suitable multiples of the first row to rows 2 and 3.
\nInput all numbers as fractions or integers and not as decimals.
\n\n
\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n$\\var{a11}$ | \n$\\var{a12}$ | \n$\\var{a13}$ | \n$1$ | \n$0$ | \n$0$ | \n\\[\\left) \\begin{matrix} \\phantom{.} \\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n
$0$ | \n[[0]] | \n[[1]] | \n[[2]] | \n$1$ | \n$0$ | \n||
$0$ | \n[[3]] | \n[[4]] | \n[[5]] | \n$0$ | \n$1$ | \n
Now, if necessary, multiply the second row by a suitable number so that the second entry in the second row is 1.
\n ", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"allowFractions": false, "variableReplacements": [], "maxValue": "{-f2*g2}", "minValue": "{-f2*g2}", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": "0.75", "type": "numberentry"}, {"allowFractions": false, "variableReplacements": [], "maxValue": "{-a*f2*g3}", "minValue": "{-a*f2*g3}", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": "0.75", "type": "numberentry"}, {"allowFractions": false, "variableReplacements": [], "maxValue": "{ -f2*f1*a}", "minValue": "{ -f2*f1*a}", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": "0.75", "type": "numberentry"}, {"allowFractions": false, "variableReplacements": [], "maxValue": "{f3*g2*c*b*(1-a)}", "minValue": "{f3*g2*c*b*(1-a)}", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": "0.75", "type": "numberentry"}, {"allowFractions": false, "variableReplacements": [], "maxValue": "{f3*g3*(1-a^2*b*c+a*b*c)}", "minValue": "{f3*g3*(1-a^2*b*c+a*b*c)}", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": "0.75", "type": "numberentry"}, {"allowFractions": false, "variableReplacements": [], "maxValue": "{-f3*f1*a*c}", "minValue": "{-f3*f1*a*c}", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": "0.75", "type": "numberentry"}], "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "gapfill"}, {"prompt": "\nNow using this matrix, introduce a zero in the second column below the second entry of the second column by:
\nAdding [[0]] times the second row to the third row to get the matrix:
\n\n
\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n$\\var{a11}$ | \n$\\var{a12}$ | \n$\\var{a13}$ | \n$1$ | \n$0$ | \n$0$ | \n\\[\\left) \\begin{matrix} \\phantom{.} \\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n
$0$ | \n$1$ | \n[[1]] | \n[[2]] | \n[[3]] | \n$0$ | \n||
$0$ | \n$0$ | \n[[4]] | \n[[5]] | \n[[6]] | \n$1$ | \n
Now, if necessary, multiply the third row by a suitable constant so that the third entry in the third column is 1.
\nWith this matrix, use the third row to introduce zeros into the second and first entries in the third column by adding suitable multiples of the third row to the second and first rows.
\nMultiply third row by [[0]]and add to the second row.
\nMultiply third row by [[1]]and add to the first row.
\nUsing this new matrix there is one more operation needed.
\nMultiplying the second row by $\\var{-a12}$ and adding to the first row to obtain the inverse matrix appearing on the right hand side.
\n\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n$1$ | \n$0$ | \n$0$ | \n[[2]] | \n[[3]] | \n[[4]] | \n\\[\\left) \\begin{matrix} \\phantom{.} \\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n
$0$ | \n$1$ | \n$0$ | \n[[5]] | \n[[6]] | \n[[7]] | \n||
$0$ | \n$0$ | \n$1$ | \n[[8]] | \n[[9]] | \n[[10]] | \n
Find the inverse of the following matrix:
\\[A = \\left(\\begin{array}{rrr} \\var{a11} & \\var{a12} & \\var{a13}\\\\ \\var{a21} & \\var{a22} & \\var{a23}\\\\ \\var{a31} & \\var{a32} & \\var{a33}\\\\ \\end{array}\\right)\\]
Form the $3 \\times 6$ augmented matrix $B$ by placing $I_3$ to the right of $A$ as below:
\\[B = \\left(\\begin{array}{rrr|ccc} \\var{a11} & \\var{a12} & \\var{a13} &\\var{1}&\\var{0}&\\var{0}\\\\ \\var{a21} & \\var{a22} & \\var{a23}&\\var{0}&\\var{1}&\\var{0}\\\\ \\var{a31} & \\var{a32} & \\var{a33}&\\var{0}&\\var{0}&\\var{1}\\\\ \\end{array}\\right)\\]
In subsequent parts work with this matrix using row operations and introduce the identity matrix on the left hand side with the inverse of A eventually appearing on the right hand side.
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", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "contributors": [{"name": "Hollie Tarr", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1176/"}]}]}], "contributors": [{"name": "Hollie Tarr", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1176/"}]}