Mixing
How do you tell the amount of solution of this row reduced matrix?
$begingroup$
The first four correspond to the variables x, y, z and u and the fifth column corresponds to the right-sides.
The thing i'm asking is, how many solutions does this system of questions have and how do you find that out? Not sure how to think of it because there's no "Free variable"
matrices
$endgroup$
migrated from mathematica.stackexchange.com Jan 28 at 12:41
This question came from our site for users of Wolfram Mathematica.
add a comment |
$begingroup$
The first four correspond to the variables x, y, z and u and the fifth column corresponds to the right-sides.
The thing i'm asking is, how many solutions does this system of questions have and how do you find that out? Not sure how to think of it because there's no "Free variable"
matrices
$endgroup$
migrated from mathematica.stackexchange.com Jan 28 at 12:41
This question came from our site for users of Wolfram Mathematica.
3
$begingroup$
(1) This is not a Mathematica question. (2) Dimension of solution space = #vars - #pivots.
$endgroup$
– Daniel Lichtblau
Jan 25 at 17:44
add a comment |
$begingroup$
The first four correspond to the variables x, y, z and u and the fifth column corresponds to the right-sides.
The thing i'm asking is, how many solutions does this system of questions have and how do you find that out? Not sure how to think of it because there's no "Free variable"
matrices
$endgroup$
The first four correspond to the variables x, y, z and u and the fifth column corresponds to the right-sides.
The thing i'm asking is, how many solutions does this system of questions have and how do you find that out? Not sure how to think of it because there's no "Free variable"
matrices
matrices
asked Jan 25 at 16:03
Tom Sjöberg
migrated from mathematica.stackexchange.com Jan 28 at 12:41
This question came from our site for users of Wolfram Mathematica.
migrated from mathematica.stackexchange.com Jan 28 at 12:41
This question came from our site for users of Wolfram Mathematica.
3
$begingroup$
(1) This is not a Mathematica question. (2) Dimension of solution space = #vars - #pivots.
$endgroup$
– Daniel Lichtblau
Jan 25 at 17:44
add a comment |
3
$begingroup$
(1) This is not a Mathematica question. (2) Dimension of solution space = #vars - #pivots.
$endgroup$
– Daniel Lichtblau
Jan 25 at 17:44
3
3
$begingroup$
(1) This is not a Mathematica question. (2) Dimension of solution space = #vars - #pivots.
$endgroup$
– Daniel Lichtblau
Jan 25 at 17:44
$begingroup$
(1) This is not a Mathematica question. (2) Dimension of solution space = #vars - #pivots.
$endgroup$
– Daniel Lichtblau
Jan 25 at 17:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write down the equations and you get:
$$
x+frac{88}{87}u=frac5{29}\
y+frac{317}{87}u=frac{77}{29}\
z-frac{184}{87}u=-frac{21}{29}
$$
You see that you can easily get a solution depending on $u$ and $u$ is a free variable:
$$
x=frac5{29}-frac{88}{87}u\
y=frac{77}{29}-frac{317}{87}u\
z=-frac{21}{29}+frac{184}{87}u
$$
answered Jan 28 at 13:05
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write down the equations and you get:
$$
x+frac{88}{87}u=frac5{29}\
y+frac{317}{87}u=frac{77}{29}\
z-frac{184}{87}u=-frac{21}{29}
$$
You see that you can easily get a solution depending on $u$ and $u$ is a free variable:
$$
x=frac5{29}-frac{88}{87}u\
y=frac{77}{29}-frac{317}{87}u\
z=-frac{21}{29}+frac{184}{87}u
$$
answered Jan 28 at 13:05
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
add a comment |
$begingroup$
Write down the equations and you get:
$$
x+frac{88}{87}u=frac5{29}\
y+frac{317}{87}u=frac{77}{29}\
z-frac{184}{87}u=-frac{21}{29}
$$
You see that you can easily get a solution depending on $u$ and $u$ is a free variable:
$$
x=frac5{29}-frac{88}{87}u\
y=frac{77}{29}-frac{317}{87}u\
z=-frac{21}{29}+frac{184}{87}u
$$
answered Jan 28 at 13:05
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
add a comment |
$begingroup$
Write down the equations and you get:
$$
x+frac{88}{87}u=frac5{29}\
y+frac{317}{87}u=frac{77}{29}\
z-frac{184}{87}u=-frac{21}{29}
$$
You see that you can easily get a solution depending on $u$ and $u$ is a free variable:
$$
x=frac5{29}-frac{88}{87}u\
y=frac{77}{29}-frac{317}{87}u\
z=-frac{21}{29}+frac{184}{87}u
$$
answered Jan 28 at 13:05
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
Write down the equations and you get:
$$
x+frac{88}{87}u=frac5{29}\
y+frac{317}{87}u=frac{77}{29}\
z-frac{184}{87}u=-frac{21}{29}
$$
You see that you can easily get a solution depending on $u$ and $u$ is a free variable:
$$
x=frac5{29}-frac{88}{87}u\
y=frac{77}{29}-frac{317}{87}u\
z=-frac{21}{29}+frac{184}{87}u
$$
answered Jan 28 at 13:05
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 28 at 13:05
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 28 at 13:05
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 28 at 13:05
Mundron SchmidtMundron Schmidt
7,5042729
7,5042729
add a comment |
add a comment |
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3
$begingroup$
(1) This is not a Mathematica question. (2) Dimension of solution space = #vars - #pivots.
$endgroup$
– Daniel Lichtblau
Jan 25 at 17:44