Look at the revealed answers for this question. All the information needed is there.
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\n\n| \\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n[[0]] | \n[[1]] | \n[[2]] | \n\\[\\left| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n[[3]] | \n\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n
| [[4]] | \n[[5]] | \n[[6]] | \n[[7]] | \n|||
| [[8]] | \n[[9]] | \n[[10]] | \n[[11]] | \n
Now introduce zeros in the first column below the first entry by adding:
[[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$\\var{1}$ | \n$\\var{b}$ | \n$\\var{b*a-b}$ | \n\\[\\left| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n$\\var{c3}$ | \n\\[\\left) \\begin{matrix} \\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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\n
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$\\var{1}$ | \n$\\var{b}$ | \n$\\var{b*a-b}$ | \n\\[\\left| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n$\\var{c3}$ | \n\\[\\left) \\begin{matrix} \\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
From this you should find:
\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]]
\n \n \n \n \n "}], "metadata": {"description": "Solving a system of three linear equations in 3 unknowns using Gauss Elimination in 4 stages. Solutions are all integral.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "tags": [], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "name": "Praneetha's copy of Gaussian Elimination", "extensions": [], "ungrouped_variables": ["a", "b", "c", "c1", "c2", "c3", "y", "x", "z"], "preamble": {"css": "", "js": ""}, "statement": "Solve the system of equations using Gaussian Elimination
\\[\\begin{eqnarray} &x&+\\;&\\var{b}y&+\\;\\var{b*a-b}z&=&\\var{c3}\\\\ &\\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} \\end{eqnarray} \\]