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Solving a system of three linear equations in 3 unknowns using Gaussian Elimination (or Gauss-Jordan algorithm) in 5 stages. Solutions are all integers. Introductory question where the numbers come out quite nice with not much dividing. Set-up is meant for formative assessment. Adapated from a question copied from Newcastle.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Solve the system of equations using Gaussian Elimination:
\\[\\begin{eqnarray*} &x^{\\var{a}}y^{\\var{a*b-1}}z^{\\var{a^2*b-a-a*b}}&=&e^{\\var{c2}}\\\\ &x^{\\var{a*c}}y^{\\var{c*b}}z&=&e^{\\var{c1}}\\\\ &xy^{\\var{b}}z^{\\var{b*a-b}}&=&e^{\\var{c3}} \\end{eqnarray*} \\]
Part a) Recast the system as a system of linear equations using logarithms
Part b) Rearrange the order of the equations and represent this as a system of equations using a matrix.
Part c) Introduce zeros in the first column using the first row.
Part d) Introduce zeros in the second column below the second entry in the second row using the second row.
Part e) Introduce zeros in the third column above the third entry in the third row using the third row.
Part f) Introduce zeros in the second column above the second entry in the second row using the second row.
Read off the solution to the linear system, and determine the final solution of the original system
Look at the revealed answers for this question. All the information needed is there.
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\nFor each non-linear equation in the system, write down the linear equation you determine using logarithms
\n$x^{\\var{a}}y^{\\var{a*b-1}}z^{\\var{a^2*b-a-a*b}}=e^{\\var{c2}}$ | \n$\\longrightarrow$ | \n[[0]] | \n
$x^{\\var{a*c}}y^{\\var{c*b}}z=e^{\\var{c1}}$ | \n$\\longrightarrow$ | \n[[1]] | \n
$xy^{\\var{b}}z^{\\var{b*a-b}}=e^{\\var{c3}}$ | \n$\\longrightarrow$ | \n[[2]] | \n
Re-arrange the rows so that the third row becomes the first row, the first the second and the second the third.
WHY? Choose one of the following:
[[0]]
Now write down the entries of the matrix you will use for Gaussian Elimination, remember to include the constants as the last column.
\n\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{,}\\end{matrix} \\right.\\] | \n[[1]] | \n[[2]] | \n[[3]] | \n\\[\\left. \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\middle| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n[[4]] | \n\\[\\left) \\begin{matrix} \\phantom{.} \\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.} \\end{matrix} \\right.\\] | \n
[[5]] | \n[[6]] | \n[[7]] | \n[[8]] | \n|||
[[9]] | \n[[10]] | \n[[11]] | \n[[12]] | \n
To make sure that there is a 1 in the first row, first column position.
", "Because you always do this.
", "Why not.
", "I don't know.
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[[0]] times the first row to the second row and
[[1]] times the first row to the third row to get the matrix:
\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\\\phantom{.} \\end{matrix} \\right.\\] | \n$\\var{1}$ | \n$\\var{b}$ | \n$\\var{b*a-b}$ | \n\\[\\left. \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\middle| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n$\\var{c3}$ | \n\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\\\phantom{.} \\end{matrix} \\right.\\] | \n
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$\\var{0}$ | \n[[5]] | \n[[6]] | \n[[7]] | \n
Next multiply the second row by [[8]] to get a 1 in the second entry in the second row.
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In this part we introduce a $0$ in the second column below the second entry in the second column (which is the leading \\(1\\) of the second row) by adding:
[[0]] times the second row to the third row to get the matrix:
\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n$\\var{1}$ | \n$\\var{b}$ | \n$\\var{b*a-b}$ | \n\\[\\left. \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\middle| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n$\\var{c3}$ | \n\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n
$\\var{0}$ | \n$\\var{1}$ | \n[[1]] | \n[[2]] | \n|||
$\\var{0}$ | \n$\\var{0}$ | \n[[3]] | \n[[4]] | \n
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\nWe first introduce $0$s in the third column above the leading one in the third row by adding:
[[0]] times the third row to the second row and
[[1]] times the third row to the first row to get the matrix:
\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n$\\var{1}$ | \n$\\var{b}$ | \n$\\var{0}$ | \n\\[\\left. \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\middle| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n[[2]] | \n\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n
$\\var{0}$ | \n$\\var{1}$ | \n\\(\\var{0}\\) | \n[[3]] | \n|||
$\\var{0}$ | \n$\\var{0}$ | \n\\(\\var{1}\\) | \n[[4]] | \n
Finally, we introduce a $0$ in the second column above the leading \\(1\\) in the second row by adding:
[[5]] times the second row to the first row to get the matrix:
\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n$\\var{1}$ | \n\\(\\var{0}\\) | \n$\\var{0}$ | \n\\[\\left. \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\middle| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n[[6]] | \n\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n
$\\var{0}$ | \n$\\var{1}$ | \n\\(\\var{0}\\) | \n[[3]] | \n|||
$\\var{0}$ | \n$\\var{0}$ | \n\\(\\var{1}\\) | \n[[4]] | \n
From this you can read off the solution for $X$, $Y$ and $Z$ (the gaps will fill automatically from your answers in the augmented matrix):
\n\\(X=\\ \\) [[6]]
\n\\(Y=\\ \\) [[3]]
\n\\(Z=\\ \\) [[4]]
\nThe solution to the original system is then given by reversing the substitutions
\n[[6]]$=\\ln x\\longrightarrow x = \\exp(\\,$[[6]]$)$
\n[[3]]$=\\ln x\\longrightarrow x = \\exp(\\,$[[3]]$)$
\n[[4]]$=\\ln x\\longrightarrow x = \\exp(\\,$[[4]]$)$
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