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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*} &\\var{a}x&+\\;&\\var{a*b-1}y&+\\;\\var{a^2*b-a-a*b}z&=&\\var{c2}\\\\ &\\var{a*c}x&+\\;&\\var{c*b}y&+\\;z&=&\\var{c1}\\\\ &x&+\\;&\\var{b}y&+\\;\\var{b*a-b}z&=&\\var{c3} \\end{eqnarray*} \\]
Part a) Rearrange the order of the equations and represent this as a system of equations using a matrix.
Part b) Introduce zeros in the first column using the first row.
Part c) Introduce zeros in the second column below the second entry in the second row using the second row.
Part d) Introduce zeros in the third column above the third entry in the third row using the third row.
Part e) Introduce zeros in the second column above the second entry in the second row using the second row.
Also read off the solution to the system.
Look at the revealed answers for this question. All the information needed is there.
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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
$\\var{0}$ | \n[[2]] | \n[[3]] | \n[[4]] | \n|||
$\\var{0}$ | \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
You seem to have got the sign wrong.
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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 third row to the second 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 (the gaps will fill automatically from your answers in the augmented matrix):
\n\\(x=\\ \\) [[6]]
\n\\(y=\\ \\) [[3]]
\n\\(z=\\ \\) [[4]]
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