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

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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]]

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Now write down the entries of the matrix you will use for Gaussian Elimination, remember to include the constants as the last column.

\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.\$ [[1]] [[2]] [[3]] [[4]] \$\\left) \\begin{matrix} \\phantom{.} \\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\$ [[5]] [[6]] [[7]] [[8]] [[9]] [[10]] [[11]] [[12]]
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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.

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 \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.\$ $\\var{1}$ $\\var{b}$ $\\var{b*a-b}$ $\\var{c3}$ \$\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\$ $\\var{0}$ [[2]] [[3]] [[4]] $\\var{0}$ [[5]] [[6]] [[7]]
\n \n \n \n

Next multiply the second row by [[8]] to get a 1 in the second entry in the second row.

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

From this you should find:

\n \n \n \n

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

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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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 coumn below the second entry in the second row using the second row.
Also need to solve for $z$ using the last row of the reduced matrix.
Part d) Solve for $y$ and $x$ using the second and first rows of the reduced matrix.

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15/7/2015:

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", "description": "

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

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