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Solving a system of three linear equations in 3 unknowns using Gauss Elimination in 4 stages. Solutions are all integral.
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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{.}\\\\ \\end{matrix} \\right.\\] | \n[[1]] | \n[[2]] | \n[[3]] | \n[[4]] | \n\\[\\left) \\begin{matrix} \\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{.}\\\\ \\end{matrix} \\right.\\] | \n \n$\\var{1}$ | \n \n$\\var{b}$ | \n \n$\\var{b*a-b}$ | \n \n$\\var{c3}$ | \n \n\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n \n
$\\var{0}$ | \n \n[[2]] | \n \n[[3]] | \n \n[[4]] | \n \n||
$\\var{0}$ | \n \n[[5]] | \n \n[[6]] | \n \n[[7]] | \n \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 by adding:
[[0]] times the second row to the third row to get the matrix:
\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n \n$\\var{1}$ | \n \n$\\var{b}$ | \n \n$\\var{b*a-b}$ | \n \n$\\var{c3}$ | \n \n\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n \n
$\\var{0}$ | \n \n$\\var{1}$ | \n \n[[1]] | \n \n[[2]] | \n \n||
$\\var{0}$ | \n \n$\\var{0}$ | \n \n[[3]] | \n \n[[4]] | \n \n
From this you should find:
\n \n \n \n$z=\\;\\;$[[5]]
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\n \n \n \n$y=\\;\\;$[[0]]
\n \n \n \nThen using the first row we have the equation :
\\[\\simplify[all]{x+ {b}y+{b*a-b}z={c3}}\\]
Using this you can now find $x$:
\n \n \n \n$x=\\;\\;$[[1]]
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", "functions": {}, "tags": [], "variables": {"c3": {"name": "c3", "templateType": "anything", "definition": "random(1..3)", "group": "Ungrouped variables", "description": ""}, "y": {"name": "y", "templateType": "anything", "definition": "a*c3-c2-a*z", "group": "Ungrouped variables", "description": ""}, "c2": {"name": "c2", "templateType": "anything", "definition": "random(1..3)", "group": "Ungrouped variables", "description": ""}, "c1": {"name": "c1", "templateType": "anything", "definition": "random(1..5)", "group": "Ungrouped variables", "description": ""}, "a": {"name": "a", "templateType": "anything", "definition": "random(2..6)", "group": "Ungrouped variables", "description": ""}, "c": {"name": "c", "templateType": "anything", "definition": "random(1..3)", "group": "Ungrouped variables", "description": ""}, "b": {"name": "b", "templateType": "anything", "definition": "random(2..6)", "group": "Ungrouped variables", "description": ""}, "z": {"name": "z", "templateType": "anything", "definition": "c1+c2*c*(b-a*b)+c3*c*(a^2*b-a-a*b)", "group": "Ungrouped variables", "description": ""}, "x": {"name": "x", "templateType": "anything", "definition": "c3-(b*a-b)*z-b*y", "group": "Ungrouped variables", "description": ""}}, "statement": "Solve the system of equations using Gauss 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
Part d) Solve for $y$ and $x$ using the second and first rows of the reduced matrix.