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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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\\[\\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} \\]
Part a) Introduce zeros in the first column using the first row.
\nPart b) Introduce zeros in the second coumn below the second entry in the second row using the second row, then solve for $z$ using the last row of the reduced matrix.
\nPart c) Solve for $y$ and $x$ using the second and first rows of the reduced matrix.
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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]]
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