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

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First set up the augmented matrix:

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\\(\\left(\\begin{array}{rrr|c} \\var{a11}&\\var{a12}&\\var{a13}&\\var{r}\\\\\\var{a21}&\\var{a22}&\\var{a23}&\\var{s}\\\\\\var{a31}&\\var{a32}&\\var{a33}&\\var{t}\\\\\\end{array}\\right)\\)

\n

Use appropriate row operations to get zeroes below the diagonal in the first column as shown in the matrix below on the left and hence enter the correct values into the matrix on the right.

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\\(\\left(\\begin{array}{rrr|c} \\var{a11}&\\var{a12}&\\var{a13}&\\var{r}\\\\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 in the matrix below on the left and hence enter the correct values into the matrix on the right.

\n

\\(\\left(\\begin{array}{rrr|c} \\var{a11}&0&?&?\\\\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 in the matrix below on the left and hence enter the correct values into the matrix on the right.

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

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Perform one final row operation and enter your solutions:

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\\(x=\\) [[0]]

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\\(y=\\) [[1]]

\n

\\(z=\\) [[2]]

\n

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Solve the following syastem of equations using row operations.

\n

\\(\\var{a11}x+\\var{a12}y+\\var{a13}z=\\var{r}\\)

\n

\\(\\var{a21}x+\\var{a22}y+\\var{a23}z=\\var{s}\\)

\n

\\(\\var{a31}x+\\var{a32}y+\\var{a33}z=\\var{t}\\)

\n

\n

\n

", "preamble": {"css": "", "js": ""}, "ungrouped_variables": ["a11", "a12", "a13", "k", "a21", "a22", "a23", "k1", "a31", "a33", "a32", "k2", "k3", "x1", "y1", "b24", "z1", "r", "s", "t", "b23", "b32", "b33", "b34", "c13", "c14", "c33", "c34", "d14", "d24", "d34"], "advice": "

The correct row operations in the first iteration are

\n

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

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

\n

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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

\n

\n

The correct row operations in the third iteration are

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new row1 = old row1 - \\(\\simplify{{a13}-{a12}*({a23}-{k}*{a13})}*\\)row3\\(/\\var{c33}\\)

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new row2 = old row2 \\(-(\\simplify{{a23}-{k}*{a13}})*\\)row3\\(/\\var{c33}\\)

\n

\n

", "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/"}]}