// Numbas version: exam_results_page_options {"name": "E1 Q1: Gaussian Elimination", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "E1 Q1: Gaussian Elimination", "tags": [], "metadata": {"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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Solve the following system of equations using Gaussian Elimination (reducing the matrix to REF).
\\[\\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} \\]

\n

Part a) Introduce zeros in the first column using the first row.

\n

Part 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.

\n

Part c) Solve for $y$ and $x$ using the second and first rows of the reduced matrix. 

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

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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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