Solve A linear equation system b=0 Rstudio
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i would solve a linear equation system like this:
x_1*3+x_2*4+x_3*5+x_4*6+x_6*2=0
x_1*21+x_2*23+x_3*45+x_4*37*+x_6*0=0
x_1*340+x_2*24+x_3*25+x_4*31+x_6*0=0
x_1*32+x_2*45+x_3*5+x_4*6+x_7*2=0
x_1*9+x_2*11+x_3*13+x_4*49+x_7*0=0
x_1*5+x_2*88+x_3*100+x_4*102+X_7*2=0
[x_1][x_2][x_3] [x_4] [,5]
[1,] 3 4 5 6 2
[2,] 21 23 45 37 0
[3,] 340 24 25 31 0
[4,] 32 45 5 6 2
[5,] 9 11 13 49 0
[6,] 5 88 100 102 2
i use solve this linear homogeneous equation system with MASS::null(t(M),
but the problem is that find x_1....x_4, but x_5 find only one solution but i need different three value that is x_5,1,x_5,2 and x_5,3.
value of matrix are random, and they can be changed
r equation equation-solving linear-equation
asked Nov 16 '18 at 12:23
antonio nuzzoantonio nuzzo
177
|
show 1 more comment
i would solve a linear equation system like this:
x_1*3+x_2*4+x_3*5+x_4*6+x_6*2=0
x_1*21+x_2*23+x_3*45+x_4*37*+x_6*0=0
x_1*340+x_2*24+x_3*25+x_4*31+x_6*0=0
x_1*32+x_2*45+x_3*5+x_4*6+x_7*2=0
x_1*9+x_2*11+x_3*13+x_4*49+x_7*0=0
x_1*5+x_2*88+x_3*100+x_4*102+X_7*2=0
[x_1][x_2][x_3] [x_4] [,5]
[1,] 3 4 5 6 2
[2,] 21 23 45 37 0
[3,] 340 24 25 31 0
[4,] 32 45 5 6 2
[5,] 9 11 13 49 0
[6,] 5 88 100 102 2
i use solve this linear homogeneous equation system with MASS::null(t(M),
but the problem is that find x_1....x_4, but x_5 find only one solution but i need different three value that is x_5,1,x_5,2 and x_5,3.
value of matrix are random, and they can be changed
r equation equation-solving linear-equation
asked Nov 16 '18 at 12:23
antonio nuzzoantonio nuzzo
177
Look at thesolvefunction.
– snaut
Nov 16 '18 at 12:42
@snaut, is not good, because for a Ax=0 return trivial solution
– antonio nuzzo
Nov 16 '18 at 12:49
Ok, sorry I missunderstood what you need. So you want the whole subspace of solutions?
– snaut
Nov 16 '18 at 12:52
yes, but the problem of my system is that there are x_1....x_4 unknows and x_5,1 x_5,2 x_5,3,if the last unknows(x_5) was only one i could have solve the system, but with different x_5 unknows i don't know how solve it.
– antonio nuzzo
Nov 16 '18 at 12:59
Can you explain your notation? I didn't thinkx_5,1was a legal variable name.
– Carl Witthoft
Nov 16 '18 at 14:19
|
show 1 more comment
i would solve a linear equation system like this:
x_1*3+x_2*4+x_3*5+x_4*6+x_6*2=0
x_1*21+x_2*23+x_3*45+x_4*37*+x_6*0=0
x_1*340+x_2*24+x_3*25+x_4*31+x_6*0=0
x_1*32+x_2*45+x_3*5+x_4*6+x_7*2=0
x_1*9+x_2*11+x_3*13+x_4*49+x_7*0=0
x_1*5+x_2*88+x_3*100+x_4*102+X_7*2=0
[x_1][x_2][x_3] [x_4] [,5]
[1,] 3 4 5 6 2
[2,] 21 23 45 37 0
[3,] 340 24 25 31 0
[4,] 32 45 5 6 2
[5,] 9 11 13 49 0
[6,] 5 88 100 102 2
i use solve this linear homogeneous equation system with MASS::null(t(M),
but the problem is that find x_1....x_4, but x_5 find only one solution but i need different three value that is x_5,1,x_5,2 and x_5,3.
value of matrix are random, and they can be changed
r equation equation-solving linear-equation
asked Nov 16 '18 at 12:23
antonio nuzzoantonio nuzzo
177
i would solve a linear equation system like this:
x_1*3+x_2*4+x_3*5+x_4*6+x_6*2=0
x_1*21+x_2*23+x_3*45+x_4*37*+x_6*0=0
x_1*340+x_2*24+x_3*25+x_4*31+x_6*0=0
x_1*32+x_2*45+x_3*5+x_4*6+x_7*2=0
x_1*9+x_2*11+x_3*13+x_4*49+x_7*0=0
x_1*5+x_2*88+x_3*100+x_4*102+X_7*2=0
[x_1][x_2][x_3] [x_4] [,5]
[1,] 3 4 5 6 2
[2,] 21 23 45 37 0
[3,] 340 24 25 31 0
[4,] 32 45 5 6 2
[5,] 9 11 13 49 0
[6,] 5 88 100 102 2
i use solve this linear homogeneous equation system with MASS::null(t(M),
but the problem is that find x_1....x_4, but x_5 find only one solution but i need different three value that is x_5,1,x_5,2 and x_5,3.
value of matrix are random, and they can be changed
r equation equation-solving linear-equation
r equation equation-solving linear-equation
asked Nov 16 '18 at 12:23
antonio nuzzoantonio nuzzo
177
asked Nov 16 '18 at 12:23
antonio nuzzoantonio nuzzo
177
edited Nov 16 '18 at 14:29
antonio nuzzo
asked Nov 16 '18 at 12:23
antonio nuzzoantonio nuzzo
177
asked Nov 16 '18 at 12:23
antonio nuzzoantonio nuzzo
177
asked Nov 16 '18 at 12:23
antonio nuzzoantonio nuzzo
177
177
Look at thesolvefunction.
– snaut
Nov 16 '18 at 12:42
@snaut, is not good, because for a Ax=0 return trivial solution
– antonio nuzzo
Nov 16 '18 at 12:49
Ok, sorry I missunderstood what you need. So you want the whole subspace of solutions?
– snaut
Nov 16 '18 at 12:52
yes, but the problem of my system is that there are x_1....x_4 unknows and x_5,1 x_5,2 x_5,3,if the last unknows(x_5) was only one i could have solve the system, but with different x_5 unknows i don't know how solve it.
– antonio nuzzo
Nov 16 '18 at 12:59
Can you explain your notation? I didn't thinkx_5,1was a legal variable name.
– Carl Witthoft
Nov 16 '18 at 14:19
|
show 1 more comment
Look at thesolvefunction.
– snaut
Nov 16 '18 at 12:42
@snaut, is not good, because for a Ax=0 return trivial solution
– antonio nuzzo
Nov 16 '18 at 12:49
Ok, sorry I missunderstood what you need. So you want the whole subspace of solutions?
– snaut
Nov 16 '18 at 12:52
yes, but the problem of my system is that there are x_1....x_4 unknows and x_5,1 x_5,2 x_5,3,if the last unknows(x_5) was only one i could have solve the system, but with different x_5 unknows i don't know how solve it.
– antonio nuzzo
Nov 16 '18 at 12:59
Can you explain your notation? I didn't thinkx_5,1was a legal variable name.
– Carl Witthoft
Nov 16 '18 at 14:19
Look at the
solve function.– snaut
Nov 16 '18 at 12:42
Look at the
solve function.– snaut
Nov 16 '18 at 12:42
@snaut, is not good, because for a Ax=0 return trivial solution
– antonio nuzzo
Nov 16 '18 at 12:49
@snaut, is not good, because for a Ax=0 return trivial solution
– antonio nuzzo
Nov 16 '18 at 12:49
Ok, sorry I missunderstood what you need. So you want the whole subspace of solutions?
– snaut
Nov 16 '18 at 12:52
Ok, sorry I missunderstood what you need. So you want the whole subspace of solutions?
– snaut
Nov 16 '18 at 12:52
yes, but the problem of my system is that there are x_1....x_4 unknows and x_5,1 x_5,2 x_5,3,if the last unknows(x_5) was only one i could have solve the system, but with different x_5 unknows i don't know how solve it.
– antonio nuzzo
Nov 16 '18 at 12:59
yes, but the problem of my system is that there are x_1....x_4 unknows and x_5,1 x_5,2 x_5,3,if the last unknows(x_5) was only one i could have solve the system, but with different x_5 unknows i don't know how solve it.
– antonio nuzzo
Nov 16 '18 at 12:59
Can you explain your notation? I didn't think
x_5,1 was a legal variable name.– Carl Witthoft
Nov 16 '18 at 14:19
Can you explain your notation? I didn't think
x_5,1 was a legal variable name.– Carl Witthoft
Nov 16 '18 at 14:19
|
show 1 more comment
2 Answers
2
active
oldest
votes
Ok, had to reactivate my rusted linear algebra knowledge, you can do this by using the Singular Value Decomposition, if all elements of the diagonal part of the SVD are non zero, only the trivial solutions exists:
solution_space <- function(A){
my_svd <- svd(A)
if(all(my_svd$d != 0)){
return(rep(0, ncol(A)))
} else {
return(my_svd$u[,my_svd$d == 0, drop=F])
}
}
A %*% solution_space(A)
You can try the code with these matrices:
A <- matrix(c(1,1,0,1,1,0,0,0,1), 3, 3)
A <- matrix(c(1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1), 4, 4)
A <- matrix(c(1,1,0,1,1,0,0,0,0), 3, 3)
answered Nov 16 '18 at 13:50
snautsnaut
1,0761020
add a comment |
With the update which shows you've got 5 equations with 7 unknowns, it's obvious that there is a multidimensional surface of solutions.
I fear I don't have code to calculate that surface, but let me toot my own horn and offer the ktsolve package. For any given set of inputs from your { $x_1, x_2, ...x_7$ } [ah rats no latex markdown] , enter a collection of known values and ktsolve will run a back-solver (usually BB ) to find the unknowns.
So if you can feed your problem a selected set of any two of {X_5, X_6, X_7}, you can find all five of the other values.
answered Nov 16 '18 at 14:45
Carl WitthoftCarl Witthoft
16.1k63458
add a comment |
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2 Answers
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2 Answers
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active
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Ok, had to reactivate my rusted linear algebra knowledge, you can do this by using the Singular Value Decomposition, if all elements of the diagonal part of the SVD are non zero, only the trivial solutions exists:
solution_space <- function(A){
my_svd <- svd(A)
if(all(my_svd$d != 0)){
return(rep(0, ncol(A)))
} else {
return(my_svd$u[,my_svd$d == 0, drop=F])
}
}
A %*% solution_space(A)
You can try the code with these matrices:
A <- matrix(c(1,1,0,1,1,0,0,0,1), 3, 3)
A <- matrix(c(1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1), 4, 4)
A <- matrix(c(1,1,0,1,1,0,0,0,0), 3, 3)
answered Nov 16 '18 at 13:50
snautsnaut
1,0761020
add a comment |
Ok, had to reactivate my rusted linear algebra knowledge, you can do this by using the Singular Value Decomposition, if all elements of the diagonal part of the SVD are non zero, only the trivial solutions exists:
solution_space <- function(A){
my_svd <- svd(A)
if(all(my_svd$d != 0)){
return(rep(0, ncol(A)))
} else {
return(my_svd$u[,my_svd$d == 0, drop=F])
}
}
A %*% solution_space(A)
You can try the code with these matrices:
A <- matrix(c(1,1,0,1,1,0,0,0,1), 3, 3)
A <- matrix(c(1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1), 4, 4)
A <- matrix(c(1,1,0,1,1,0,0,0,0), 3, 3)
answered Nov 16 '18 at 13:50
snautsnaut
1,0761020
add a comment |
Ok, had to reactivate my rusted linear algebra knowledge, you can do this by using the Singular Value Decomposition, if all elements of the diagonal part of the SVD are non zero, only the trivial solutions exists:
solution_space <- function(A){
my_svd <- svd(A)
if(all(my_svd$d != 0)){
return(rep(0, ncol(A)))
} else {
return(my_svd$u[,my_svd$d == 0, drop=F])
}
}
A %*% solution_space(A)
You can try the code with these matrices:
A <- matrix(c(1,1,0,1,1,0,0,0,1), 3, 3)
A <- matrix(c(1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1), 4, 4)
A <- matrix(c(1,1,0,1,1,0,0,0,0), 3, 3)
answered Nov 16 '18 at 13:50
snautsnaut
1,0761020
Ok, had to reactivate my rusted linear algebra knowledge, you can do this by using the Singular Value Decomposition, if all elements of the diagonal part of the SVD are non zero, only the trivial solutions exists:
solution_space <- function(A){
my_svd <- svd(A)
if(all(my_svd$d != 0)){
return(rep(0, ncol(A)))
} else {
return(my_svd$u[,my_svd$d == 0, drop=F])
}
}
A %*% solution_space(A)
You can try the code with these matrices:
A <- matrix(c(1,1,0,1,1,0,0,0,1), 3, 3)
A <- matrix(c(1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1), 4, 4)
A <- matrix(c(1,1,0,1,1,0,0,0,0), 3, 3)
answered Nov 16 '18 at 13:50
snautsnaut
1,0761020
answered Nov 16 '18 at 13:50
snautsnaut
1,0761020
answered Nov 16 '18 at 13:50
snautsnaut
1,0761020
answered Nov 16 '18 at 13:50
snautsnaut
1,0761020
1,0761020
add a comment |
add a comment |
With the update which shows you've got 5 equations with 7 unknowns, it's obvious that there is a multidimensional surface of solutions.
I fear I don't have code to calculate that surface, but let me toot my own horn and offer the ktsolve package. For any given set of inputs from your { $x_1, x_2, ...x_7$ } [ah rats no latex markdown] , enter a collection of known values and ktsolve will run a back-solver (usually BB ) to find the unknowns.
So if you can feed your problem a selected set of any two of {X_5, X_6, X_7}, you can find all five of the other values.
answered Nov 16 '18 at 14:45
Carl WitthoftCarl Witthoft
16.1k63458
add a comment |
With the update which shows you've got 5 equations with 7 unknowns, it's obvious that there is a multidimensional surface of solutions.
I fear I don't have code to calculate that surface, but let me toot my own horn and offer the ktsolve package. For any given set of inputs from your { $x_1, x_2, ...x_7$ } [ah rats no latex markdown] , enter a collection of known values and ktsolve will run a back-solver (usually BB ) to find the unknowns.
So if you can feed your problem a selected set of any two of {X_5, X_6, X_7}, you can find all five of the other values.
answered Nov 16 '18 at 14:45
Carl WitthoftCarl Witthoft
16.1k63458
add a comment |
With the update which shows you've got 5 equations with 7 unknowns, it's obvious that there is a multidimensional surface of solutions.
I fear I don't have code to calculate that surface, but let me toot my own horn and offer the ktsolve package. For any given set of inputs from your { $x_1, x_2, ...x_7$ } [ah rats no latex markdown] , enter a collection of known values and ktsolve will run a back-solver (usually BB ) to find the unknowns.
So if you can feed your problem a selected set of any two of {X_5, X_6, X_7}, you can find all five of the other values.
answered Nov 16 '18 at 14:45
Carl WitthoftCarl Witthoft
16.1k63458
With the update which shows you've got 5 equations with 7 unknowns, it's obvious that there is a multidimensional surface of solutions.
I fear I don't have code to calculate that surface, but let me toot my own horn and offer the ktsolve package. For any given set of inputs from your { $x_1, x_2, ...x_7$ } [ah rats no latex markdown] , enter a collection of known values and ktsolve will run a back-solver (usually BB ) to find the unknowns.
So if you can feed your problem a selected set of any two of {X_5, X_6, X_7}, you can find all five of the other values.
answered Nov 16 '18 at 14:45
Carl WitthoftCarl Witthoft
16.1k63458
answered Nov 16 '18 at 14:45
Carl WitthoftCarl Witthoft
16.1k63458
answered Nov 16 '18 at 14:45
Carl WitthoftCarl Witthoft
16.1k63458
answered Nov 16 '18 at 14:45
Carl WitthoftCarl Witthoft
16.1k63458
16.1k63458
add a comment |
add a comment |
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Look at the
solvefunction.– snaut
Nov 16 '18 at 12:42
@snaut, is not good, because for a Ax=0 return trivial solution
– antonio nuzzo
Nov 16 '18 at 12:49
Ok, sorry I missunderstood what you need. So you want the whole subspace of solutions?
– snaut
Nov 16 '18 at 12:52
yes, but the problem of my system is that there are x_1....x_4 unknows and x_5,1 x_5,2 x_5,3,if the last unknows(x_5) was only one i could have solve the system, but with different x_5 unknows i don't know how solve it.
– antonio nuzzo
Nov 16 '18 at 12:59
Can you explain your notation? I didn't think
x_5,1was a legal variable name.– Carl Witthoft
Nov 16 '18 at 14:19