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Write the augmented matrix corresponding to this system of equations

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\][[0]][[1]][[2]]\\[\\left| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\][[3]]\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]
[[4]][[5]][[6]][[7]]
[[8]][[9]][[10]][[11]]
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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:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]$\\var{1}$$\\var{b}$$\\var{b*a-b}$\\[\\left| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]$\\var{c3}$\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]
$\\var{0}$[[2]][[3]][[4]]
$\\var{0}$[[5]][[6]][[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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(Note that from the last part, you may have multiplied the second row by a suitable number to get a $1$ in the second entry in the second row.)

\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:

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]$\\var{1}$$\\var{b}$$\\var{b*a-b}$\\[\\left| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]$\\var{c3}$\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]
$\\var{0}$$\\var{1}$[[1]][[2]]
$\\var{0}$$\\var{0}$[[3]][[4]]
\n

From this you should find:

\n

$z=\\;\\;$[[5]]

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From the second row of the reduced matrix you find an equation involving only $y$ and $z$ and using your value for $z$ we find:

\n \n \n \n

$y=\\;\\;$[[0]]

\n \n \n \n

Then using the first row we have the equation :
\\[\\simplify[all]{x+ {b}y+{b*a-b}z={c3}}\\]

\n \n \n \n

Using this you can now find $x$:

\n \n \n \n

$x=\\;\\;$[[1]]

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Look at the revealed answers for this question. All the information needed is there.

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Solving a system of three linear equations in 3 unknowns using Gauss Elimination in 4 stages. Solutions are all integral.

"}, "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "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} \\]

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