Solving linear equations system using inverse matrix and finding back this matrix by using linear equations
$begingroup$
I stumbled into this question in the course of some experiment:
I had this system of linear equations:
m = 1a+2b+3c+4d
n = 2a+3b+4c+1d
o = 3a+4b+1c+2d
p = 4a+1b+2c+3d
I have no fixed values for m, n, o, p as my a, b, c, d are vectors, so m, n, o and p are output vectors.
So I calculated the following general case solution:
a2 = (-211m+152n-3o-22p)/-76
b2 = (10o+116m+p-76n)/18
c2 = (-17o+20n+3m+2p)/76
d2 = (7m-7n+o+p)/38
I wanted to check my results by comparing the vectors a and a2, b and b2 and so on, but I can't find the good values.
So I used an inverse matrix solver to solve the system, which gave me
a4=(-0,225*m)+(0,025*n)+(0,025*o)+(0,275*p)
b4=(0,025*m)+(0,025*n)+(0,275*o)+(-0,225*p)
c4=(0,025*m)+(0,275*n)+(-0,225*o)+(0,025*p)
d4=(0,275*m)+(-0,225*n)+(0,025*o)+(0,025*p)
These vectors, besides floating point mistakes, are found to be equal to originals.
My questions is:
How could I have reached this inverse matrix by using linear equations?
I am puzzled because I did this for a 3*3 system and it worked, as demonstrated below, still for a 4*4 system, it is a mess and it doesn't work, for some reason.
Can someone explain this to me?
Thanks
m = (1*a) + (2*b) + (3*c)
n = (2*a) + (3*b) + (1*c)
o = (3*a) + (1*b) + (2*c)
a4=((-5*m)+(1*n)+(7*o))/18
b4=((1*m)+(7*n)+(-5*o))/18
c4=((-7*m)+(5*n)+(-1*o))/-18
These coefficients match the inverse matrix.
-0.277777778 0.055555556 0.388888889
0.055555556 0.388888889 -0.277777778
-0.388888889 0.277777778 -0.055555556
linear-algebra matrix-equations numerical-linear-algebra
asked Dec 10 '18 at 12:37
Ando JuraiAndo Jurai
1063
$endgroup$
add a comment |
$begingroup$
I stumbled into this question in the course of some experiment:
I had this system of linear equations:
m = 1a+2b+3c+4d
n = 2a+3b+4c+1d
o = 3a+4b+1c+2d
p = 4a+1b+2c+3d
I have no fixed values for m, n, o, p as my a, b, c, d are vectors, so m, n, o and p are output vectors.
So I calculated the following general case solution:
a2 = (-211m+152n-3o-22p)/-76
b2 = (10o+116m+p-76n)/18
c2 = (-17o+20n+3m+2p)/76
d2 = (7m-7n+o+p)/38
I wanted to check my results by comparing the vectors a and a2, b and b2 and so on, but I can't find the good values.
So I used an inverse matrix solver to solve the system, which gave me
a4=(-0,225*m)+(0,025*n)+(0,025*o)+(0,275*p)
b4=(0,025*m)+(0,025*n)+(0,275*o)+(-0,225*p)
c4=(0,025*m)+(0,275*n)+(-0,225*o)+(0,025*p)
d4=(0,275*m)+(-0,225*n)+(0,025*o)+(0,025*p)
These vectors, besides floating point mistakes, are found to be equal to originals.
My questions is:
How could I have reached this inverse matrix by using linear equations?
I am puzzled because I did this for a 3*3 system and it worked, as demonstrated below, still for a 4*4 system, it is a mess and it doesn't work, for some reason.
Can someone explain this to me?
Thanks
m = (1*a) + (2*b) + (3*c)
n = (2*a) + (3*b) + (1*c)
o = (3*a) + (1*b) + (2*c)
a4=((-5*m)+(1*n)+(7*o))/18
b4=((1*m)+(7*n)+(-5*o))/18
c4=((-7*m)+(5*n)+(-1*o))/-18
These coefficients match the inverse matrix.
-0.277777778 0.055555556 0.388888889
0.055555556 0.388888889 -0.277777778
-0.388888889 0.277777778 -0.055555556
linear-algebra matrix-equations numerical-linear-algebra
asked Dec 10 '18 at 12:37
Ando JuraiAndo Jurai
1063
$endgroup$
$begingroup$
how did you compute your general case solution?
$endgroup$
– Exodd
Dec 10 '18 at 12:41
$begingroup$
I used Excel MINVERSE function. But for the 3*3 system, I actually did it first by linear equations (using substitution method). Problem being I fail miserably to do the same with the 4*4 system, and I can't understand why (which is even more miserable).
$endgroup$
– Ando Jurai
Dec 10 '18 at 12:43
$begingroup$
Yes it does. Actually I also came up with the same solution after giving it a new go. I can't explain what didn't work before, because I found other solutions with no mistakes in coefficient calculations. I think I found (randomly) the good combination which was "homogenous" here, with the same constant to divide all the matrix, while it was pretty messy before. Did you found it by just solving the system by elimination?
$endgroup$
– Ando Jurai
Dec 10 '18 at 17:20
add a comment |
$begingroup$
I stumbled into this question in the course of some experiment:
I had this system of linear equations:
m = 1a+2b+3c+4d
n = 2a+3b+4c+1d
o = 3a+4b+1c+2d
p = 4a+1b+2c+3d
I have no fixed values for m, n, o, p as my a, b, c, d are vectors, so m, n, o and p are output vectors.
So I calculated the following general case solution:
a2 = (-211m+152n-3o-22p)/-76
b2 = (10o+116m+p-76n)/18
c2 = (-17o+20n+3m+2p)/76
d2 = (7m-7n+o+p)/38
I wanted to check my results by comparing the vectors a and a2, b and b2 and so on, but I can't find the good values.
So I used an inverse matrix solver to solve the system, which gave me
a4=(-0,225*m)+(0,025*n)+(0,025*o)+(0,275*p)
b4=(0,025*m)+(0,025*n)+(0,275*o)+(-0,225*p)
c4=(0,025*m)+(0,275*n)+(-0,225*o)+(0,025*p)
d4=(0,275*m)+(-0,225*n)+(0,025*o)+(0,025*p)
These vectors, besides floating point mistakes, are found to be equal to originals.
My questions is:
How could I have reached this inverse matrix by using linear equations?
I am puzzled because I did this for a 3*3 system and it worked, as demonstrated below, still for a 4*4 system, it is a mess and it doesn't work, for some reason.
Can someone explain this to me?
Thanks
m = (1*a) + (2*b) + (3*c)
n = (2*a) + (3*b) + (1*c)
o = (3*a) + (1*b) + (2*c)
a4=((-5*m)+(1*n)+(7*o))/18
b4=((1*m)+(7*n)+(-5*o))/18
c4=((-7*m)+(5*n)+(-1*o))/-18
These coefficients match the inverse matrix.
-0.277777778 0.055555556 0.388888889
0.055555556 0.388888889 -0.277777778
-0.388888889 0.277777778 -0.055555556
linear-algebra matrix-equations numerical-linear-algebra
asked Dec 10 '18 at 12:37
Ando JuraiAndo Jurai
1063
$endgroup$
I stumbled into this question in the course of some experiment:
I had this system of linear equations:
m = 1a+2b+3c+4d
n = 2a+3b+4c+1d
o = 3a+4b+1c+2d
p = 4a+1b+2c+3d
I have no fixed values for m, n, o, p as my a, b, c, d are vectors, so m, n, o and p are output vectors.
So I calculated the following general case solution:
a2 = (-211m+152n-3o-22p)/-76
b2 = (10o+116m+p-76n)/18
c2 = (-17o+20n+3m+2p)/76
d2 = (7m-7n+o+p)/38
I wanted to check my results by comparing the vectors a and a2, b and b2 and so on, but I can't find the good values.
So I used an inverse matrix solver to solve the system, which gave me
a4=(-0,225*m)+(0,025*n)+(0,025*o)+(0,275*p)
b4=(0,025*m)+(0,025*n)+(0,275*o)+(-0,225*p)
c4=(0,025*m)+(0,275*n)+(-0,225*o)+(0,025*p)
d4=(0,275*m)+(-0,225*n)+(0,025*o)+(0,025*p)
These vectors, besides floating point mistakes, are found to be equal to originals.
My questions is:
How could I have reached this inverse matrix by using linear equations?
I am puzzled because I did this for a 3*3 system and it worked, as demonstrated below, still for a 4*4 system, it is a mess and it doesn't work, for some reason.
Can someone explain this to me?
Thanks
m = (1*a) + (2*b) + (3*c)
n = (2*a) + (3*b) + (1*c)
o = (3*a) + (1*b) + (2*c)
a4=((-5*m)+(1*n)+(7*o))/18
b4=((1*m)+(7*n)+(-5*o))/18
c4=((-7*m)+(5*n)+(-1*o))/-18
These coefficients match the inverse matrix.
-0.277777778 0.055555556 0.388888889
0.055555556 0.388888889 -0.277777778
-0.388888889 0.277777778 -0.055555556
linear-algebra matrix-equations numerical-linear-algebra
linear-algebra matrix-equations numerical-linear-algebra
asked Dec 10 '18 at 12:37
Ando JuraiAndo Jurai
1063
asked Dec 10 '18 at 12:37
Ando JuraiAndo Jurai
1063
asked Dec 10 '18 at 12:37
Ando JuraiAndo Jurai
1063
asked Dec 10 '18 at 12:37
Ando JuraiAndo Jurai
1063
asked Dec 10 '18 at 12:37
Ando JuraiAndo Jurai
1063
1063
$begingroup$
how did you compute your general case solution?
$endgroup$
– Exodd
Dec 10 '18 at 12:41
$begingroup$
I used Excel MINVERSE function. But for the 3*3 system, I actually did it first by linear equations (using substitution method). Problem being I fail miserably to do the same with the 4*4 system, and I can't understand why (which is even more miserable).
$endgroup$
– Ando Jurai
Dec 10 '18 at 12:43
$begingroup$
Yes it does. Actually I also came up with the same solution after giving it a new go. I can't explain what didn't work before, because I found other solutions with no mistakes in coefficient calculations. I think I found (randomly) the good combination which was "homogenous" here, with the same constant to divide all the matrix, while it was pretty messy before. Did you found it by just solving the system by elimination?
$endgroup$
– Ando Jurai
Dec 10 '18 at 17:20
add a comment |
$begingroup$
how did you compute your general case solution?
$endgroup$
– Exodd
Dec 10 '18 at 12:41
$begingroup$
I used Excel MINVERSE function. But for the 3*3 system, I actually did it first by linear equations (using substitution method). Problem being I fail miserably to do the same with the 4*4 system, and I can't understand why (which is even more miserable).
$endgroup$
– Ando Jurai
Dec 10 '18 at 12:43
$begingroup$
Yes it does. Actually I also came up with the same solution after giving it a new go. I can't explain what didn't work before, because I found other solutions with no mistakes in coefficient calculations. I think I found (randomly) the good combination which was "homogenous" here, with the same constant to divide all the matrix, while it was pretty messy before. Did you found it by just solving the system by elimination?
$endgroup$
– Ando Jurai
Dec 10 '18 at 17:20
$begingroup$
how did you compute your general case solution?
$endgroup$
– Exodd
Dec 10 '18 at 12:41
$begingroup$
how did you compute your general case solution?
$endgroup$
– Exodd
Dec 10 '18 at 12:41
$begingroup$
I used Excel MINVERSE function. But for the 3*3 system, I actually did it first by linear equations (using substitution method). Problem being I fail miserably to do the same with the 4*4 system, and I can't understand why (which is even more miserable).
$endgroup$
– Ando Jurai
Dec 10 '18 at 12:43
$begingroup$
I used Excel MINVERSE function. But for the 3*3 system, I actually did it first by linear equations (using substitution method). Problem being I fail miserably to do the same with the 4*4 system, and I can't understand why (which is even more miserable).
$endgroup$
– Ando Jurai
Dec 10 '18 at 12:43
$begingroup$
Yes it does. Actually I also came up with the same solution after giving it a new go. I can't explain what didn't work before, because I found other solutions with no mistakes in coefficient calculations. I think I found (randomly) the good combination which was "homogenous" here, with the same constant to divide all the matrix, while it was pretty messy before. Did you found it by just solving the system by elimination?
$endgroup$
– Ando Jurai
Dec 10 '18 at 17:20
$begingroup$
Yes it does. Actually I also came up with the same solution after giving it a new go. I can't explain what didn't work before, because I found other solutions with no mistakes in coefficient calculations. I think I found (randomly) the good combination which was "homogenous" here, with the same constant to divide all the matrix, while it was pretty messy before. Did you found it by just solving the system by elimination?
$endgroup$
– Ando Jurai
Dec 10 '18 at 17:20
add a comment |
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$begingroup$
how did you compute your general case solution?
$endgroup$
– Exodd
Dec 10 '18 at 12:41
$begingroup$
I used Excel MINVERSE function. But for the 3*3 system, I actually did it first by linear equations (using substitution method). Problem being I fail miserably to do the same with the 4*4 system, and I can't understand why (which is even more miserable).
$endgroup$
– Ando Jurai
Dec 10 '18 at 12:43
$begingroup$
Yes it does. Actually I also came up with the same solution after giving it a new go. I can't explain what didn't work before, because I found other solutions with no mistakes in coefficient calculations. I think I found (randomly) the good combination which was "homogenous" here, with the same constant to divide all the matrix, while it was pretty messy before. Did you found it by just solving the system by elimination?
$endgroup$
– Ando Jurai
Dec 10 '18 at 17:20