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Solving a system of three linear equations in 3 unknowns using Gaussian Elimination (or Gauss-Jordan algorithm) in 5 stages. Solutions are all integers. Set up so that sometimes it has infinitely many solutions (one free variable), sometimes unique solution. Scaffolded so meant for formative. The variable d determines the cases (d=1: unique solution, d-0: infinitely many solutions). The other variables are set up so that no entries become zero for some randomisations but not others.
", "licence": "Creative Commons Attribution 4.0 International"}, "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&+\\simplify[!zeroFactor,unitFactor]{{d}*z}&=&\\var{c1}
\\end{eqnarray*} \\]
Part a) 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 column below the second entry in the second row using the second row.
Part d) Determine whether the system has no solutions, a unique solution or infinitely many solutions.
Part e) If the system has solutions, calculate the reduced row echelon form of the matrix.
Part f) Read off the general solution from the reduced row echelon form, introducing free variables if necessary.
For parts a) to c), look at the revealed answers for this question. All the information needed is there.
\nPart d) We have \\(3\\) leading \\(1\\)s in the matrix in part c), one in each row and each column. Therefore we have a unique solution. As the third row is completely zero, we have no leading \\(1\\) in the third column, and \\(z\\) can be any value, so we have a free variable. This means we have infinitely many solutions.
\nPart e) As the third row is completely zero, we cannot clear the top two entries in the third column. We can clear the two top entries in the third colum using the third row. Since we have a leading \\(1\\) in the second entry in the second row, we can clear the second entry in the first row. Since the second entry in the first row is \\(\\var{b}\\neq 0\\), the matrix is definitely not yet in reduced row echelon form.
For the rest of this part, look at the revealed answers.
Part f) As the third row is a zero row, we get \\(z=t\\) as a free variable: \\(z\\) is not restricted at all. This then also appears in the general solutions for \\(x\\) and \\(y\\): \\(x-\\var{b}z=\\var{x}\\), so \\(x=\\var{x}+\\var{b}t\\), and \\(y+\\var{a}z=\\var{a*c3-c2}\\), so \\(y=\\var{a*c3-c2}-\\var{a}t\\).
You can read off the unique solution from the final matrix.
particular solution of x if d=0, or the unique solution of x if d=1.
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\n\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{,}\\end{matrix} \\right.\\] | \n[[0]] | \n[[1]] | \n[[2]] | \n\\[\\left. \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\middle| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n[[3]] | \n\\[\\left) \\begin{matrix} \\phantom{.} \\\\ \\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{.}\\\\\\phantom{.} \\end{matrix} \\right.\\] | \n$\\var{1}$ | \n$\\var{b}$ | \n$\\var{b*a-b}$ | \n\\[\\left. \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\middle| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n$\\var{c3}$ | \n\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\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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In this part we introduce a $0$ in the second column below the second entry in the second column (which is the leading \\(1\\) of the second row) by adding:
[[0]] times the second row to the third row to get the matrix:
\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\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{.}\\\\ \\phantom{.}\\end{matrix} \\middle| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n$\\var{c3}$ | \n\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\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
You seem to have got the sign wrong.
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[[0]]
Continue with this part if the system has a unique solution or infinitely many solutions.
\nTo read off the solutions to \\(x\\), \\(y\\) and \\(z\\) most easily, we now want to get the matrix into reduced row echelon form.
\nWhich steps do you need to perform to get the matrix into reduced row echelon form?
[[0]]
Write out the reduced row echelon form:
\n\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n$\\var{1}$ | \n[[1]] | \n[[2]] | \n\\[\\left. \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\middle| \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n[[3]] | \n\\[\\left) \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\end{matrix} \\right.\\] | \n
$\\var{0}$ | \n$\\var{1}$ | \n[[6]] | \n[[4]] | \n|||
$\\var{0}$ | \n$\\var{0}$ | \n[[7]] | \n[[5]] | \n
Finally read off the general solution from the reduced row echelon form. If your solution has a free variable, then use \\(t\\) in the answer. If it has two free variables, use \\(t\\) for \\(z\\) and \\(s\\) for the other free variable.
\n\\(x=\\ \\) [[2]]
\n\\(y=\\ \\) [[1]]
\n\\(z=\\ \\)[[0]]
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