// Numbas version: exam_results_page_options {"name": "Inverse of 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": [{"name": "Inverse of a 3x3 matrix using row operations", "statement": "

Find the inverse of the matrix A by applying row operations:

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

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This question asks learners to use row operations to find the inverse of a 3x3 matrix.

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The correct row operations in the first iteration are

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new row2 = old row2- \\(\\var{k}*\\)row1

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new row3 = old row3- \\(\\var{k1}*\\)row1

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The correct row operations in the second iteration are

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new row1 = old row1 - \\(\\var{a12}*\\)row2

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new row3 = old row3 - \\(\\simplify{{a32}-{k1}*{a12}}*\\)row2

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The correct row operations in the third iteration are

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new row1 = old row1 - \\(\\simplify{{a13}-{a12}*({a23}-{k}*{a13})}*\\)row3

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new row2 = old row2 \\(-(\\simplify{{a23}-{k}*{a13}})*\\)row3

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3x33 matrix with determinant = a11

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First set up the augmented matrix:

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\\(\\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)\\)

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Use appropriate row operations to get zeroes below the diagonal in the first column:

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\\(\\left(\\begin{array}{rrr|ccc} \\var{a11}&\\var{a12}&\\var{a13}&\\var{1}&\\var{0}&\\var{0}\\\\0&1&?&?&?&?\\\\0&?&?&?&?&?\\\\\\end{array}\\right)\\) =  [[0]]

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Taking this matrix apply the appropriate row operations to get zeroes above and below the diagonal in column 2 as shown below:

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\\(\\left(\\begin{array}{rrr|ccc} \\var{a11}&0&?&\\var{1}&?&?\\\\0&1&?&?&?&?\\\\0&0&?&?&?&?\\\\\\end{array}\\right)\\) = [[0]]

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Next take this matrix and apply the appropriate row operations to get zeroes above the diagonal in column 3 as shown below:

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\\(\\left(\\begin{array}{rrr|ccc} \\var{a11}&0&0&?&?&?\\\\0&1&0&?&?&?\\\\0&0&1&?&?&?\\\\\\end{array}\\right)\\) = [[0]]

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By carrying out one final row operation input all the entries in \\(A^{-1}\\), correct to two decimal places.

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\\(A^{-1}\\) = [[0]]

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