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3x33 matrix with determinant = a11

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

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

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

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"

First set up the augmented matrix:

\n

\\(\\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:

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
{a11}123
4567
890
\n

\n

\\(\\left(\\begin{array}{rrr|c}\\var{a11}&\\var{a12}&\\var{a13}&\\var{r}\\\\0&1&[[1]]&[[2]]\\\\0&[[3]]&[[4]]&[[5]]\\\\\\end{array}\\right)\\) 

\n

Use appropriate row operations to get zeroes above and below the diagonal in the second column:

\n

\\(\\left(\\begin{array}{rrr|c} \\var{a11}&0&[[6]]&[[7]]\\\\0&1&[[8]]&[[9]]\\\\0&0&[[10]]&[[11]]\\\\\\end{array}\\right)\\) 

\n

Use appropriate row operations to get zeroes above and below the diagonal in the third column:

\n

\\(\\left(\\begin{array}{rrr|c} \\var{a11}&0&0&[[12]]\\\\0&1&0&[[13]]\\\\0&0&\\var{c34}&[[14]]\\\\\\end{array}\\right)\\)

\n

Hence

\n

\\(x=\\) [[15]]

\n

\\(y=\\) [[16]]

\n

\\(z=\\) [[0]]

\n

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The correct row operations in the first iteration are

\n

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

\n

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

\n

\n

The correct row operations in the second iteration are

\n

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

\n

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

\n

\n

The correct row operations in the third iteration are

\n

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

\n

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

\n

\n

", "type": "question", "contributors": [{"name": "Jim Kelly", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2145/"}]}]}], "contributors": [{"name": "Jim Kelly", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2145/"}]}