Gauss elimination to solve a system of linear equations.

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Question 1

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.

a)

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:

Gap 0

Why not.
I don't know.
Because you always do this.
To make sure that there is a 1 in the first row, first column position.

Expected answer:

Why not.
I don't know.
Because you always do this.
To make sure that there is a 1 in the first row, first column position.

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

Expected answer:

Score: 0/3.4

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b)

Now introduce zeros in the first column below the first entry by adding:
Expected answer: times the first row to the second row and
Expected answer: times the first row to the third row to get the matrix:

$\left( \begin{matrix} \phantom{.}\\ \phantom{.}\\ \phantom{.}\\ \phantom{.}\\ \end{matrix} \right.$	$\var{1}$	$\var{b}$	$\var{b*a-b}$	$\var{c3}$	$\left) \begin{matrix} \phantom{.}\\ \phantom{.}\\ \phantom{.}\\ \phantom{.}\\ \end{matrix} \right.$
	$\var{0}$	Expected answer:	Expected answer:	Expected answer:
	$\var{0}$	Expected answer:	Expected answer:	Expected answer:

Next multiply the second row by Expected answer: to get a 1 in the second entry in the second row.

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c)

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:
Expected answer: times the second row to the third row to get the matrix:

$\left( \begin{matrix} \phantom{.}\\ \phantom{.}\\ \phantom{.}\\ \phantom{.}\\ \end{matrix} \right.$	$\var{1}$	$\var{b}$	$\var{b*a-b}$	$\var{c3}$	$\left) \begin{matrix} \phantom{.}\\ \phantom{.}\\ \phantom{.}\\ \phantom{.}\\ \end{matrix} \right.$
	$\var{0}$	$\var{1}$	Expected answer:	Expected answer:
	$\var{0}$	$\var{0}$	Expected answer:	Expected answer:

From this you should find:

$z=\;\;$ Expected answer:

Score: 0/6

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d)

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:

$y=\;\;$ Expected answer:

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

Using this you can now find $x$ :

$x=\;\;$ Expected answer:

Score: 0/5.6

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Advice

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

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Numbas has failed

Gauss elimination to solve a system of linear equations.

a)

Gap 0

b)

c)

d)

Advice