Why is this method for solving linear equations systems using determinants works?Creating and solving large systems of equationsSolving non-linear (convex) systems of equationsTechniques for solving coupled differential equationsSolving systems of equations using matrices by row reductionNumber of equations required for eliminationName and explanation of a Numerical Analysis method for solving systems of non-linear equationsElementary Substitution in Solving Equations - Why it workssystems of equations with 3 variables using substitution methodsystem of equations using Gauss Jordan methodHahn-Banach From Systems of Linear Equations
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Why is this method for solving linear equations systems using determinants works?
Creating and solving large systems of equationsSolving non-linear (convex) systems of equationsTechniques for solving coupled differential equationsSolving systems of equations using matrices by row reductionNumber of equations required for eliminationName and explanation of a Numerical Analysis method for solving systems of non-linear equationsElementary Substitution in Solving Equations - Why it workssystems of equations with 3 variables using substitution methodsystem of equations using Gauss Jordan methodHahn-Banach From Systems of Linear Equations
$begingroup$
I'm new here and english is not my native language but I'll try to explain my question the best I can.
While studying methods for solving systems of linear equations I came across a method so called "Gauss algorithm" for which I doubt if it is it's real name or not, but I can't understand WHY it works.
I would really appreciate a proper explanation or maybe a hint for what is happening inside the method process that accounts for the logic behind it.
The process in question is the following and involves determinants:
Suppose we have the following system of linear equations, for instance
begincases
x+2y-z=-5 \2x-1y+2z=8 \3x+3y+4z=5
endcases
Then the result of appliying the method continues as follow
beginarrayccc
x & y & z & i.t. \
hline
1 & 2 & -1 & -5 \
2 & -1 & 2 & 8 \
3 & 3 & 4 & 5 \
hline
& -5 & 4 & 18 \
& -3 & 7 & 20 \
hline
& & -23 & -46 \
endarray
Where for example the coefficients $-5$ and $4$, and the independent term $18$, of the first row of the reduced system has been computed with the follow determinants:
beginequation
beginvmatrix
1 & 2 \
2 & -1 \
endvmatrix=-5
hspace1cm
beginvmatrix
1 & -1 \
2 & 2 \
endvmatrix=4
hspace1cm
beginvmatrix
1 & -5 \
2 & 8 \
endvmatrix=18
endequation
Where those determinants were computed using the first and the second row of the system of equations matrix. Then the other coefficients -3 and 7 and the independant term 20 were found by computing the respective determinants as above but using the first and third row of the system of equations matrix.
Appliying this method and if the system is compatible, then as above you can see that beginequation -23z = -46\
z=2endequation
and then by substitution in the subsequent former equations the other variables can be found.
Well that's the method I was studying and will appreciate to know more about it, it's a pity I don't know the name of it but maybe you can help.
Thanks in advance!
systems-of-equations
edited Apr 24 at 18:00
Yanko
8,7272830
asked Apr 24 at 17:12
MateoBMateoB
285
New contributor
MateoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm new here and english is not my native language but I'll try to explain my question the best I can.
While studying methods for solving systems of linear equations I came across a method so called "Gauss algorithm" for which I doubt if it is it's real name or not, but I can't understand WHY it works.
I would really appreciate a proper explanation or maybe a hint for what is happening inside the method process that accounts for the logic behind it.
The process in question is the following and involves determinants:
Suppose we have the following system of linear equations, for instance
begincases
x+2y-z=-5 \2x-1y+2z=8 \3x+3y+4z=5
endcases
Then the result of appliying the method continues as follow
beginarrayccc
x & y & z & i.t. \
hline
1 & 2 & -1 & -5 \
2 & -1 & 2 & 8 \
3 & 3 & 4 & 5 \
hline
& -5 & 4 & 18 \
& -3 & 7 & 20 \
hline
& & -23 & -46 \
endarray
Where for example the coefficients $-5$ and $4$, and the independent term $18$, of the first row of the reduced system has been computed with the follow determinants:
beginequation
beginvmatrix
1 & 2 \
2 & -1 \
endvmatrix=-5
hspace1cm
beginvmatrix
1 & -1 \
2 & 2 \
endvmatrix=4
hspace1cm
beginvmatrix
1 & -5 \
2 & 8 \
endvmatrix=18
endequation
Where those determinants were computed using the first and the second row of the system of equations matrix. Then the other coefficients -3 and 7 and the independant term 20 were found by computing the respective determinants as above but using the first and third row of the system of equations matrix.
Appliying this method and if the system is compatible, then as above you can see that beginequation -23z = -46\
z=2endequation
and then by substitution in the subsequent former equations the other variables can be found.
Well that's the method I was studying and will appreciate to know more about it, it's a pity I don't know the name of it but maybe you can help.
Thanks in advance!
systems-of-equations
edited Apr 24 at 18:00
Yanko
8,7272830
asked Apr 24 at 17:12
MateoBMateoB
285
New contributor
MateoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Gauss is a real name of a mathamtician who at some point developed the tools for solving linear equations. I'm not sure if this algorithm is his, but as the answer shows it relies on his ideas.
$endgroup$
– Yanko
Apr 24 at 17:56
1
$begingroup$
There's another algorithm, called "Gauss elimination" (en.wikipedia.org/wiki/Gaussian_elimination) which doesn't use determinants. It is much simpler to understand.
$endgroup$
– Yanko
Apr 24 at 17:59
1
$begingroup$
The use of 2x2 determinants here is pointless IMO. All it does is make a very simple algorithm more confusing.
$endgroup$
– alephzero
Apr 24 at 22:43
add a comment |
$begingroup$
I'm new here and english is not my native language but I'll try to explain my question the best I can.
While studying methods for solving systems of linear equations I came across a method so called "Gauss algorithm" for which I doubt if it is it's real name or not, but I can't understand WHY it works.
I would really appreciate a proper explanation or maybe a hint for what is happening inside the method process that accounts for the logic behind it.
The process in question is the following and involves determinants:
Suppose we have the following system of linear equations, for instance
begincases
x+2y-z=-5 \2x-1y+2z=8 \3x+3y+4z=5
endcases
Then the result of appliying the method continues as follow
beginarrayccc
x & y & z & i.t. \
hline
1 & 2 & -1 & -5 \
2 & -1 & 2 & 8 \
3 & 3 & 4 & 5 \
hline
& -5 & 4 & 18 \
& -3 & 7 & 20 \
hline
& & -23 & -46 \
endarray
Where for example the coefficients $-5$ and $4$, and the independent term $18$, of the first row of the reduced system has been computed with the follow determinants:
beginequation
beginvmatrix
1 & 2 \
2 & -1 \
endvmatrix=-5
hspace1cm
beginvmatrix
1 & -1 \
2 & 2 \
endvmatrix=4
hspace1cm
beginvmatrix
1 & -5 \
2 & 8 \
endvmatrix=18
endequation
Where those determinants were computed using the first and the second row of the system of equations matrix. Then the other coefficients -3 and 7 and the independant term 20 were found by computing the respective determinants as above but using the first and third row of the system of equations matrix.
Appliying this method and if the system is compatible, then as above you can see that beginequation -23z = -46\
z=2endequation
and then by substitution in the subsequent former equations the other variables can be found.
Well that's the method I was studying and will appreciate to know more about it, it's a pity I don't know the name of it but maybe you can help.
Thanks in advance!
systems-of-equations
edited Apr 24 at 18:00
Yanko
8,7272830
asked Apr 24 at 17:12
MateoBMateoB
285
New contributor
MateoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm new here and english is not my native language but I'll try to explain my question the best I can.
While studying methods for solving systems of linear equations I came across a method so called "Gauss algorithm" for which I doubt if it is it's real name or not, but I can't understand WHY it works.
I would really appreciate a proper explanation or maybe a hint for what is happening inside the method process that accounts for the logic behind it.
The process in question is the following and involves determinants:
Suppose we have the following system of linear equations, for instance
begincases
x+2y-z=-5 \2x-1y+2z=8 \3x+3y+4z=5
endcases
Then the result of appliying the method continues as follow
beginarrayccc
x & y & z & i.t. \
hline
1 & 2 & -1 & -5 \
2 & -1 & 2 & 8 \
3 & 3 & 4 & 5 \
hline
& -5 & 4 & 18 \
& -3 & 7 & 20 \
hline
& & -23 & -46 \
endarray
Where for example the coefficients $-5$ and $4$, and the independent term $18$, of the first row of the reduced system has been computed with the follow determinants:
beginequation
beginvmatrix
1 & 2 \
2 & -1 \
endvmatrix=-5
hspace1cm
beginvmatrix
1 & -1 \
2 & 2 \
endvmatrix=4
hspace1cm
beginvmatrix
1 & -5 \
2 & 8 \
endvmatrix=18
endequation
Where those determinants were computed using the first and the second row of the system of equations matrix. Then the other coefficients -3 and 7 and the independant term 20 were found by computing the respective determinants as above but using the first and third row of the system of equations matrix.
Appliying this method and if the system is compatible, then as above you can see that beginequation -23z = -46\
z=2endequation
and then by substitution in the subsequent former equations the other variables can be found.
Well that's the method I was studying and will appreciate to know more about it, it's a pity I don't know the name of it but maybe you can help.
Thanks in advance!
systems-of-equations
systems-of-equations
edited Apr 24 at 18:00
Yanko
8,7272830
asked Apr 24 at 17:12
MateoBMateoB
285
New contributor
MateoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 24 at 18:00
Yanko
8,7272830
asked Apr 24 at 17:12
MateoBMateoB
285
New contributor
MateoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 24 at 18:00
Yanko
8,7272830
edited Apr 24 at 18:00
Yanko
8,7272830
edited Apr 24 at 18:00
Yanko
8,7272830
8,7272830
asked Apr 24 at 17:12
MateoBMateoB
285
New contributor
MateoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 24 at 17:12
MateoBMateoB
285
asked Apr 24 at 17:12
MateoBMateoB
285
285
New contributor
MateoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
MateoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
MateoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Gauss is a real name of a mathamtician who at some point developed the tools for solving linear equations. I'm not sure if this algorithm is his, but as the answer shows it relies on his ideas.
$endgroup$
– Yanko
Apr 24 at 17:56
1
$begingroup$
There's another algorithm, called "Gauss elimination" (en.wikipedia.org/wiki/Gaussian_elimination) which doesn't use determinants. It is much simpler to understand.
$endgroup$
– Yanko
Apr 24 at 17:59
1
$begingroup$
The use of 2x2 determinants here is pointless IMO. All it does is make a very simple algorithm more confusing.
$endgroup$
– alephzero
Apr 24 at 22:43
add a comment |
1
$begingroup$
Gauss is a real name of a mathamtician who at some point developed the tools for solving linear equations. I'm not sure if this algorithm is his, but as the answer shows it relies on his ideas.
$endgroup$
– Yanko
Apr 24 at 17:56
1
$begingroup$
There's another algorithm, called "Gauss elimination" (en.wikipedia.org/wiki/Gaussian_elimination) which doesn't use determinants. It is much simpler to understand.
$endgroup$
– Yanko
Apr 24 at 17:59
1
$begingroup$
The use of 2x2 determinants here is pointless IMO. All it does is make a very simple algorithm more confusing.
$endgroup$
– alephzero
Apr 24 at 22:43
1
1
$begingroup$
Gauss is a real name of a mathamtician who at some point developed the tools for solving linear equations. I'm not sure if this algorithm is his, but as the answer shows it relies on his ideas.
$endgroup$
– Yanko
Apr 24 at 17:56
$begingroup$
Gauss is a real name of a mathamtician who at some point developed the tools for solving linear equations. I'm not sure if this algorithm is his, but as the answer shows it relies on his ideas.
$endgroup$
– Yanko
Apr 24 at 17:56
1
1
$begingroup$
There's another algorithm, called "Gauss elimination" (en.wikipedia.org/wiki/Gaussian_elimination) which doesn't use determinants. It is much simpler to understand.
$endgroup$
– Yanko
Apr 24 at 17:59
$begingroup$
There's another algorithm, called "Gauss elimination" (en.wikipedia.org/wiki/Gaussian_elimination) which doesn't use determinants. It is much simpler to understand.
$endgroup$
– Yanko
Apr 24 at 17:59
1
1
$begingroup$
The use of 2x2 determinants here is pointless IMO. All it does is make a very simple algorithm more confusing.
$endgroup$
– alephzero
Apr 24 at 22:43
$begingroup$
The use of 2x2 determinants here is pointless IMO. All it does is make a very simple algorithm more confusing.
$endgroup$
– alephzero
Apr 24 at 22:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The computation of the determinants is equivalent to solving step by step the system of equations in the following way:
- Multiply the first equation by $2$, which is the coefficient of $x$ in the second equation;
- Multiply the second one by $1$, which is the coefficient of $x$ in the first equation; and then
- Subtract the first from the second equation.
This will result in the fourth equation.
Afterwards we can do the same for the first and third equations to get the fifth equation:
- Multiply the first equation by $3$, which is the coefficient of $x$ in the third equation;
- Multiply the third one by $1$, which is the coefficient of $x$ in the first equation; and then
- Subtract the first from the third equation.
Now we can do a similar approach to the fourth and fifth equation to get the sixth equation:
- Multiply the fourth equation by $-3$, which is the coefficient of $y$ in the fifth equation;
- Multiply the fifth one by $-5$, which is the coefficient of $y$ in the fourth equation; and then
- Subtract the fourth from the fifth equation.
And voilà, we have the sixth equation.
answered Apr 24 at 17:43
ErtxiemErtxiem
1,122312
$endgroup$
add a comment |
$begingroup$
The method is called Gauss elimination.
It works like this.
Consider some equations, and we only record some of the coefficients of them.
... [A] ... B ... | E
... C ... D ... | F
There are further lines not shown, and the data above sorresponds to something like
$$
beginaligned
dots +boxedax+dots+bz &=e ,\
dots +cx+dots+dz &=f ,
endaligned
$$
Now we declare a value $ane 0$ under the coefficients to be the "pivot", and put it in a box.
The pivot line is divided imaginary by the pivot $a$, thus getting $dots +x+dots+(b/a)z =(e/a)$, and then used to eliminate all coefficients below (and in a total elimination also above) the pivot. So multiply the above equation by $-c$, and add this to the one involving $cx$ to eliminate $c$. One obtains, in the formal description, in the "next elimination block":
... [A] ... B ... | E
... 0 ... D - BC/A ... | F - EC/A
and note that $D-BC/A$ is the determinant $AD-BC$ divided by the pivot $A$. This is the native Gauss elimination. Numerically,
people tend to chose $A$ with maximal absolute value.
The above scheme uses a variation, all new rows are multiplied by the pivot.
So:
... [A] ... B ... | E
... 0 ... AD - BC ... | AF - EC
This is it!
In our case, also always copying the pivot line:
$$
beginarrayccc
x & y & z & i.t. \
hline
boxed1 & 2 & -1 & -5 \
2 & -1 & 2 & 8 \
3 & 3 & 4 & 5 \
hline
1 & 2 & -1 & -5 \
& boxed-5 & 4 & 18 \
& -3 & 7 & 20 \
hline
1 & 2 & -1 & -5 \
& -5 & 4 & 18 \
& & boxed-23 & -46 \
endarray
$$
answered Apr 24 at 18:19
dan_fuleadan_fulea
7,3231513
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The computation of the determinants is equivalent to solving step by step the system of equations in the following way:
- Multiply the first equation by $2$, which is the coefficient of $x$ in the second equation;
- Multiply the second one by $1$, which is the coefficient of $x$ in the first equation; and then
- Subtract the first from the second equation.
This will result in the fourth equation.
Afterwards we can do the same for the first and third equations to get the fifth equation:
- Multiply the first equation by $3$, which is the coefficient of $x$ in the third equation;
- Multiply the third one by $1$, which is the coefficient of $x$ in the first equation; and then
- Subtract the first from the third equation.
Now we can do a similar approach to the fourth and fifth equation to get the sixth equation:
- Multiply the fourth equation by $-3$, which is the coefficient of $y$ in the fifth equation;
- Multiply the fifth one by $-5$, which is the coefficient of $y$ in the fourth equation; and then
- Subtract the fourth from the fifth equation.
And voilà, we have the sixth equation.
answered Apr 24 at 17:43
ErtxiemErtxiem
1,122312
$endgroup$
add a comment |
$begingroup$
The computation of the determinants is equivalent to solving step by step the system of equations in the following way:
- Multiply the first equation by $2$, which is the coefficient of $x$ in the second equation;
- Multiply the second one by $1$, which is the coefficient of $x$ in the first equation; and then
- Subtract the first from the second equation.
This will result in the fourth equation.
Afterwards we can do the same for the first and third equations to get the fifth equation:
- Multiply the first equation by $3$, which is the coefficient of $x$ in the third equation;
- Multiply the third one by $1$, which is the coefficient of $x$ in the first equation; and then
- Subtract the first from the third equation.
Now we can do a similar approach to the fourth and fifth equation to get the sixth equation:
- Multiply the fourth equation by $-3$, which is the coefficient of $y$ in the fifth equation;
- Multiply the fifth one by $-5$, which is the coefficient of $y$ in the fourth equation; and then
- Subtract the fourth from the fifth equation.
And voilà, we have the sixth equation.
answered Apr 24 at 17:43
ErtxiemErtxiem
1,122312
$endgroup$
add a comment |
$begingroup$
The computation of the determinants is equivalent to solving step by step the system of equations in the following way:
- Multiply the first equation by $2$, which is the coefficient of $x$ in the second equation;
- Multiply the second one by $1$, which is the coefficient of $x$ in the first equation; and then
- Subtract the first from the second equation.
This will result in the fourth equation.
Afterwards we can do the same for the first and third equations to get the fifth equation:
- Multiply the first equation by $3$, which is the coefficient of $x$ in the third equation;
- Multiply the third one by $1$, which is the coefficient of $x$ in the first equation; and then
- Subtract the first from the third equation.
Now we can do a similar approach to the fourth and fifth equation to get the sixth equation:
- Multiply the fourth equation by $-3$, which is the coefficient of $y$ in the fifth equation;
- Multiply the fifth one by $-5$, which is the coefficient of $y$ in the fourth equation; and then
- Subtract the fourth from the fifth equation.
And voilà, we have the sixth equation.
answered Apr 24 at 17:43
ErtxiemErtxiem
1,122312
$endgroup$
The computation of the determinants is equivalent to solving step by step the system of equations in the following way:
- Multiply the first equation by $2$, which is the coefficient of $x$ in the second equation;
- Multiply the second one by $1$, which is the coefficient of $x$ in the first equation; and then
- Subtract the first from the second equation.
This will result in the fourth equation.
Afterwards we can do the same for the first and third equations to get the fifth equation:
- Multiply the first equation by $3$, which is the coefficient of $x$ in the third equation;
- Multiply the third one by $1$, which is the coefficient of $x$ in the first equation; and then
- Subtract the first from the third equation.
Now we can do a similar approach to the fourth and fifth equation to get the sixth equation:
- Multiply the fourth equation by $-3$, which is the coefficient of $y$ in the fifth equation;
- Multiply the fifth one by $-5$, which is the coefficient of $y$ in the fourth equation; and then
- Subtract the fourth from the fifth equation.
And voilà, we have the sixth equation.
answered Apr 24 at 17:43
ErtxiemErtxiem
1,122312
answered Apr 24 at 17:43
ErtxiemErtxiem
1,122312
answered Apr 24 at 17:43
ErtxiemErtxiem
1,122312
answered Apr 24 at 17:43
ErtxiemErtxiem
1,122312
1,122312
add a comment |
add a comment |
$begingroup$
The method is called Gauss elimination.
It works like this.
Consider some equations, and we only record some of the coefficients of them.
... [A] ... B ... | E
... C ... D ... | F
There are further lines not shown, and the data above sorresponds to something like
$$
beginaligned
dots +boxedax+dots+bz &=e ,\
dots +cx+dots+dz &=f ,
endaligned
$$
Now we declare a value $ane 0$ under the coefficients to be the "pivot", and put it in a box.
The pivot line is divided imaginary by the pivot $a$, thus getting $dots +x+dots+(b/a)z =(e/a)$, and then used to eliminate all coefficients below (and in a total elimination also above) the pivot. So multiply the above equation by $-c$, and add this to the one involving $cx$ to eliminate $c$. One obtains, in the formal description, in the "next elimination block":
... [A] ... B ... | E
... 0 ... D - BC/A ... | F - EC/A
and note that $D-BC/A$ is the determinant $AD-BC$ divided by the pivot $A$. This is the native Gauss elimination. Numerically,
people tend to chose $A$ with maximal absolute value.
The above scheme uses a variation, all new rows are multiplied by the pivot.
So:
... [A] ... B ... | E
... 0 ... AD - BC ... | AF - EC
This is it!
In our case, also always copying the pivot line:
$$
beginarrayccc
x & y & z & i.t. \
hline
boxed1 & 2 & -1 & -5 \
2 & -1 & 2 & 8 \
3 & 3 & 4 & 5 \
hline
1 & 2 & -1 & -5 \
& boxed-5 & 4 & 18 \
& -3 & 7 & 20 \
hline
1 & 2 & -1 & -5 \
& -5 & 4 & 18 \
& & boxed-23 & -46 \
endarray
$$
answered Apr 24 at 18:19
dan_fuleadan_fulea
7,3231513
$endgroup$
add a comment |
$begingroup$
The method is called Gauss elimination.
It works like this.
Consider some equations, and we only record some of the coefficients of them.
... [A] ... B ... | E
... C ... D ... | F
There are further lines not shown, and the data above sorresponds to something like
$$
beginaligned
dots +boxedax+dots+bz &=e ,\
dots +cx+dots+dz &=f ,
endaligned
$$
Now we declare a value $ane 0$ under the coefficients to be the "pivot", and put it in a box.
The pivot line is divided imaginary by the pivot $a$, thus getting $dots +x+dots+(b/a)z =(e/a)$, and then used to eliminate all coefficients below (and in a total elimination also above) the pivot. So multiply the above equation by $-c$, and add this to the one involving $cx$ to eliminate $c$. One obtains, in the formal description, in the "next elimination block":
... [A] ... B ... | E
... 0 ... D - BC/A ... | F - EC/A
and note that $D-BC/A$ is the determinant $AD-BC$ divided by the pivot $A$. This is the native Gauss elimination. Numerically,
people tend to chose $A$ with maximal absolute value.
The above scheme uses a variation, all new rows are multiplied by the pivot.
So:
... [A] ... B ... | E
... 0 ... AD - BC ... | AF - EC
This is it!
In our case, also always copying the pivot line:
$$
beginarrayccc
x & y & z & i.t. \
hline
boxed1 & 2 & -1 & -5 \
2 & -1 & 2 & 8 \
3 & 3 & 4 & 5 \
hline
1 & 2 & -1 & -5 \
& boxed-5 & 4 & 18 \
& -3 & 7 & 20 \
hline
1 & 2 & -1 & -5 \
& -5 & 4 & 18 \
& & boxed-23 & -46 \
endarray
$$
answered Apr 24 at 18:19
dan_fuleadan_fulea
7,3231513
$endgroup$
add a comment |
$begingroup$
The method is called Gauss elimination.
It works like this.
Consider some equations, and we only record some of the coefficients of them.
... [A] ... B ... | E
... C ... D ... | F
There are further lines not shown, and the data above sorresponds to something like
$$
beginaligned
dots +boxedax+dots+bz &=e ,\
dots +cx+dots+dz &=f ,
endaligned
$$
Now we declare a value $ane 0$ under the coefficients to be the "pivot", and put it in a box.
The pivot line is divided imaginary by the pivot $a$, thus getting $dots +x+dots+(b/a)z =(e/a)$, and then used to eliminate all coefficients below (and in a total elimination also above) the pivot. So multiply the above equation by $-c$, and add this to the one involving $cx$ to eliminate $c$. One obtains, in the formal description, in the "next elimination block":
... [A] ... B ... | E
... 0 ... D - BC/A ... | F - EC/A
and note that $D-BC/A$ is the determinant $AD-BC$ divided by the pivot $A$. This is the native Gauss elimination. Numerically,
people tend to chose $A$ with maximal absolute value.
The above scheme uses a variation, all new rows are multiplied by the pivot.
So:
... [A] ... B ... | E
... 0 ... AD - BC ... | AF - EC
This is it!
In our case, also always copying the pivot line:
$$
beginarrayccc
x & y & z & i.t. \
hline
boxed1 & 2 & -1 & -5 \
2 & -1 & 2 & 8 \
3 & 3 & 4 & 5 \
hline
1 & 2 & -1 & -5 \
& boxed-5 & 4 & 18 \
& -3 & 7 & 20 \
hline
1 & 2 & -1 & -5 \
& -5 & 4 & 18 \
& & boxed-23 & -46 \
endarray
$$
answered Apr 24 at 18:19
dan_fuleadan_fulea
7,3231513
$endgroup$
The method is called Gauss elimination.
It works like this.
Consider some equations, and we only record some of the coefficients of them.
... [A] ... B ... | E
... C ... D ... | F
There are further lines not shown, and the data above sorresponds to something like
$$
beginaligned
dots +boxedax+dots+bz &=e ,\
dots +cx+dots+dz &=f ,
endaligned
$$
Now we declare a value $ane 0$ under the coefficients to be the "pivot", and put it in a box.
The pivot line is divided imaginary by the pivot $a$, thus getting $dots +x+dots+(b/a)z =(e/a)$, and then used to eliminate all coefficients below (and in a total elimination also above) the pivot. So multiply the above equation by $-c$, and add this to the one involving $cx$ to eliminate $c$. One obtains, in the formal description, in the "next elimination block":
... [A] ... B ... | E
... 0 ... D - BC/A ... | F - EC/A
and note that $D-BC/A$ is the determinant $AD-BC$ divided by the pivot $A$. This is the native Gauss elimination. Numerically,
people tend to chose $A$ with maximal absolute value.
The above scheme uses a variation, all new rows are multiplied by the pivot.
So:
... [A] ... B ... | E
... 0 ... AD - BC ... | AF - EC
This is it!
In our case, also always copying the pivot line:
$$
beginarrayccc
x & y & z & i.t. \
hline
boxed1 & 2 & -1 & -5 \
2 & -1 & 2 & 8 \
3 & 3 & 4 & 5 \
hline
1 & 2 & -1 & -5 \
& boxed-5 & 4 & 18 \
& -3 & 7 & 20 \
hline
1 & 2 & -1 & -5 \
& -5 & 4 & 18 \
& & boxed-23 & -46 \
endarray
$$
answered Apr 24 at 18:19
dan_fuleadan_fulea
7,3231513
answered Apr 24 at 18:19
dan_fuleadan_fulea
7,3231513
answered Apr 24 at 18:19
dan_fuleadan_fulea
7,3231513
answered Apr 24 at 18:19
dan_fuleadan_fulea
7,3231513
7,3231513
add a comment |
add a comment |
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1
$begingroup$
Gauss is a real name of a mathamtician who at some point developed the tools for solving linear equations. I'm not sure if this algorithm is his, but as the answer shows it relies on his ideas.
$endgroup$
– Yanko
Apr 24 at 17:56
1
$begingroup$
There's another algorithm, called "Gauss elimination" (en.wikipedia.org/wiki/Gaussian_elimination) which doesn't use determinants. It is much simpler to understand.
$endgroup$
– Yanko
Apr 24 at 17:59
1
$begingroup$
The use of 2x2 determinants here is pointless IMO. All it does is make a very simple algorithm more confusing.
$endgroup$
– alephzero
Apr 24 at 22:43