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Is $S= (3,1,4),(3,4,1),(4,3,1),(3,3,1) $ spanning $mathbbR^3$?
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Is $S= (3,1,4),(3,4,1),(4,3,1),(3,3,1) $ spanning $mathbbR^3$ ?
My attempt:
If $S$ spanning $mathbbR^3$ then
$(x,y,z)=alpha_1 (3,1,4)+ alpha_2(3,4,1)+ alpha_3(4,3,1)+ alpha_4(3,3,1) $
$A=left[beginarray@cccc 3 & 3 & 4 & 3 & x\ 1 & 4 & 3 & 3 & y \ 4 & 1 & 1 & 1 & z endarrayright]$
when I tried to reduce $A$ to The Echelon form, I got that:
$A=left[beginarray@cccc 1 & 0 & 0 & 1over 24 & f_1(x,y,z)\ 0 & 1 & 0 & 33over 72 & f_2(x,y,z) \ 0 & 0 & 1 & 3over 8 & f_3(x,y,z) endarrayright]$
Now, Is this system has a infinite solutions ? Or no solution ?
“When The non-homogeneous system has a unique solution, infinite solutions, no solution ?â€Â
linear-algebra
edited Jul 25 at 7:54
Babelfish
408112
asked Jul 25 at 6:45
Dima
438214
add a comment |Â
up vote
0
down vote
favorite
Is $S= (3,1,4),(3,4,1),(4,3,1),(3,3,1) $ spanning $mathbbR^3$ ?
My attempt:
If $S$ spanning $mathbbR^3$ then
$(x,y,z)=alpha_1 (3,1,4)+ alpha_2(3,4,1)+ alpha_3(4,3,1)+ alpha_4(3,3,1) $
$A=left[beginarray@cccc 3 & 3 & 4 & 3 & x\ 1 & 4 & 3 & 3 & y \ 4 & 1 & 1 & 1 & z endarrayright]$
when I tried to reduce $A$ to The Echelon form, I got that:
$A=left[beginarray@cccc 1 & 0 & 0 & 1over 24 & f_1(x,y,z)\ 0 & 1 & 0 & 33over 72 & f_2(x,y,z) \ 0 & 0 & 1 & 3over 8 & f_3(x,y,z) endarrayright]$
Now, Is this system has a infinite solutions ? Or no solution ?
“When The non-homogeneous system has a unique solution, infinite solutions, no solution ?â€Â
linear-algebra
edited Jul 25 at 7:54
Babelfish
408112
asked Jul 25 at 6:45
Dima
438214
2
Your matrix has rank $3$....
– Lord Shark the Unknown
Jul 25 at 6:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is $S= (3,1,4),(3,4,1),(4,3,1),(3,3,1) $ spanning $mathbbR^3$ ?
My attempt:
If $S$ spanning $mathbbR^3$ then
$(x,y,z)=alpha_1 (3,1,4)+ alpha_2(3,4,1)+ alpha_3(4,3,1)+ alpha_4(3,3,1) $
$A=left[beginarray@cccc 3 & 3 & 4 & 3 & x\ 1 & 4 & 3 & 3 & y \ 4 & 1 & 1 & 1 & z endarrayright]$
when I tried to reduce $A$ to The Echelon form, I got that:
$A=left[beginarray@cccc 1 & 0 & 0 & 1over 24 & f_1(x,y,z)\ 0 & 1 & 0 & 33over 72 & f_2(x,y,z) \ 0 & 0 & 1 & 3over 8 & f_3(x,y,z) endarrayright]$
Now, Is this system has a infinite solutions ? Or no solution ?
“When The non-homogeneous system has a unique solution, infinite solutions, no solution ?â€Â
linear-algebra
edited Jul 25 at 7:54
Babelfish
408112
asked Jul 25 at 6:45
Dima
438214
Is $S= (3,1,4),(3,4,1),(4,3,1),(3,3,1) $ spanning $mathbbR^3$ ?
My attempt:
If $S$ spanning $mathbbR^3$ then
$(x,y,z)=alpha_1 (3,1,4)+ alpha_2(3,4,1)+ alpha_3(4,3,1)+ alpha_4(3,3,1) $
$A=left[beginarray@cccc 3 & 3 & 4 & 3 & x\ 1 & 4 & 3 & 3 & y \ 4 & 1 & 1 & 1 & z endarrayright]$
when I tried to reduce $A$ to The Echelon form, I got that:
$A=left[beginarray@cccc 1 & 0 & 0 & 1over 24 & f_1(x,y,z)\ 0 & 1 & 0 & 33over 72 & f_2(x,y,z) \ 0 & 0 & 1 & 3over 8 & f_3(x,y,z) endarrayright]$
Now, Is this system has a infinite solutions ? Or no solution ?
“When The non-homogeneous system has a unique solution, infinite solutions, no solution ?â€Â
linear-algebra
edited Jul 25 at 7:54
Babelfish
408112
asked Jul 25 at 6:45
Dima
438214
edited Jul 25 at 7:54
Babelfish
408112
edited Jul 25 at 7:54
Babelfish
408112
edited Jul 25 at 7:54
Babelfish
408112
408112
asked Jul 25 at 6:45
Dima
438214
asked Jul 25 at 6:45
Dima
438214
asked Jul 25 at 6:45
Dima
438214
438214
2
Your matrix has rank $3$....
– Lord Shark the Unknown
Jul 25 at 6:47
add a comment |Â
2
Your matrix has rank $3$....
– Lord Shark the Unknown
Jul 25 at 6:47
2
2
Your matrix has rank $3$....
– Lord Shark the Unknown
Jul 25 at 6:47
Your matrix has rank $3$....
– Lord Shark the Unknown
Jul 25 at 6:47
add a comment |Â
1 Answer
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1
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accepted
Your system has infinite solutions. You can for example choose the parameter $alpha_4$ arbitrarily in $mathbbR$. Then you get unique values for $alpha_1=f_1(x,y,z)-fracalpha_424$, $alpha_2=f_2(x,y,z)-frac33alpha_472$ etc.
So, for each vector $(x,y,z)$, you obtain infinitely many combinations in $S$.
This is the case since the first three vectors $b_1,b_2,b_3$ of $S=,b_1,b_2,b_3,b_4,$ already span $mathbbR^3$. If you ignore the 4th vector, you obtain the echelon form
$$tildeA = left[beginarray@ccc1&0&0&tildef_1(x,y,z)\0&1&0&tildef_2(x,y,z)\0&0&1&tildef_3(x,y,z)endarrayright]$$
giving you unique values for $alpha_1,alpha_2, alpha_3$ without any choices.
In fact, those three vectors are a basis for $mathbbR^3$, since they span $mathbbR^3$ and their cardinality is $3$, which is the dimension of $mathbbR^3$.
Alternatively, you may check that the first three vectors are linearly independent, meaning that if $alpha_1 b_1 + alpha_2 b_2 + alpha_3 b_3=0$, you can prove that $alpha_1=alpha_2=alpha_3=0$. Then you obtain that $b_1,b_2,b_3$ is a basis of $mathbbR^3$, so they have to span $mathbbR^3$, too.
Your third option is to show, that you can generate the three standard basis vectors $e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)$ with $S$. You probably know that $e_1,e_2,e_3$ span $mathbbR^3$, so $S$ spanns $mathbbR^3$ aswell, if you can generate the standard basis.
answered Jul 25 at 7:23
Babelfish
408112
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your system has infinite solutions. You can for example choose the parameter $alpha_4$ arbitrarily in $mathbbR$. Then you get unique values for $alpha_1=f_1(x,y,z)-fracalpha_424$, $alpha_2=f_2(x,y,z)-frac33alpha_472$ etc.
So, for each vector $(x,y,z)$, you obtain infinitely many combinations in $S$.
This is the case since the first three vectors $b_1,b_2,b_3$ of $S=,b_1,b_2,b_3,b_4,$ already span $mathbbR^3$. If you ignore the 4th vector, you obtain the echelon form
$$tildeA = left[beginarray@ccc1&0&0&tildef_1(x,y,z)\0&1&0&tildef_2(x,y,z)\0&0&1&tildef_3(x,y,z)endarrayright]$$
giving you unique values for $alpha_1,alpha_2, alpha_3$ without any choices.
In fact, those three vectors are a basis for $mathbbR^3$, since they span $mathbbR^3$ and their cardinality is $3$, which is the dimension of $mathbbR^3$.
Alternatively, you may check that the first three vectors are linearly independent, meaning that if $alpha_1 b_1 + alpha_2 b_2 + alpha_3 b_3=0$, you can prove that $alpha_1=alpha_2=alpha_3=0$. Then you obtain that $b_1,b_2,b_3$ is a basis of $mathbbR^3$, so they have to span $mathbbR^3$, too.
Your third option is to show, that you can generate the three standard basis vectors $e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)$ with $S$. You probably know that $e_1,e_2,e_3$ span $mathbbR^3$, so $S$ spanns $mathbbR^3$ aswell, if you can generate the standard basis.
answered Jul 25 at 7:23
Babelfish
408112
add a comment |Â
up vote
1
down vote
accepted
Your system has infinite solutions. You can for example choose the parameter $alpha_4$ arbitrarily in $mathbbR$. Then you get unique values for $alpha_1=f_1(x,y,z)-fracalpha_424$, $alpha_2=f_2(x,y,z)-frac33alpha_472$ etc.
So, for each vector $(x,y,z)$, you obtain infinitely many combinations in $S$.
This is the case since the first three vectors $b_1,b_2,b_3$ of $S=,b_1,b_2,b_3,b_4,$ already span $mathbbR^3$. If you ignore the 4th vector, you obtain the echelon form
$$tildeA = left[beginarray@ccc1&0&0&tildef_1(x,y,z)\0&1&0&tildef_2(x,y,z)\0&0&1&tildef_3(x,y,z)endarrayright]$$
giving you unique values for $alpha_1,alpha_2, alpha_3$ without any choices.
In fact, those three vectors are a basis for $mathbbR^3$, since they span $mathbbR^3$ and their cardinality is $3$, which is the dimension of $mathbbR^3$.
Alternatively, you may check that the first three vectors are linearly independent, meaning that if $alpha_1 b_1 + alpha_2 b_2 + alpha_3 b_3=0$, you can prove that $alpha_1=alpha_2=alpha_3=0$. Then you obtain that $b_1,b_2,b_3$ is a basis of $mathbbR^3$, so they have to span $mathbbR^3$, too.
Your third option is to show, that you can generate the three standard basis vectors $e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)$ with $S$. You probably know that $e_1,e_2,e_3$ span $mathbbR^3$, so $S$ spanns $mathbbR^3$ aswell, if you can generate the standard basis.
answered Jul 25 at 7:23
Babelfish
408112
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your system has infinite solutions. You can for example choose the parameter $alpha_4$ arbitrarily in $mathbbR$. Then you get unique values for $alpha_1=f_1(x,y,z)-fracalpha_424$, $alpha_2=f_2(x,y,z)-frac33alpha_472$ etc.
So, for each vector $(x,y,z)$, you obtain infinitely many combinations in $S$.
This is the case since the first three vectors $b_1,b_2,b_3$ of $S=,b_1,b_2,b_3,b_4,$ already span $mathbbR^3$. If you ignore the 4th vector, you obtain the echelon form
$$tildeA = left[beginarray@ccc1&0&0&tildef_1(x,y,z)\0&1&0&tildef_2(x,y,z)\0&0&1&tildef_3(x,y,z)endarrayright]$$
giving you unique values for $alpha_1,alpha_2, alpha_3$ without any choices.
In fact, those three vectors are a basis for $mathbbR^3$, since they span $mathbbR^3$ and their cardinality is $3$, which is the dimension of $mathbbR^3$.
Alternatively, you may check that the first three vectors are linearly independent, meaning that if $alpha_1 b_1 + alpha_2 b_2 + alpha_3 b_3=0$, you can prove that $alpha_1=alpha_2=alpha_3=0$. Then you obtain that $b_1,b_2,b_3$ is a basis of $mathbbR^3$, so they have to span $mathbbR^3$, too.
Your third option is to show, that you can generate the three standard basis vectors $e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)$ with $S$. You probably know that $e_1,e_2,e_3$ span $mathbbR^3$, so $S$ spanns $mathbbR^3$ aswell, if you can generate the standard basis.
answered Jul 25 at 7:23
Babelfish
408112
Your system has infinite solutions. You can for example choose the parameter $alpha_4$ arbitrarily in $mathbbR$. Then you get unique values for $alpha_1=f_1(x,y,z)-fracalpha_424$, $alpha_2=f_2(x,y,z)-frac33alpha_472$ etc.
So, for each vector $(x,y,z)$, you obtain infinitely many combinations in $S$.
This is the case since the first three vectors $b_1,b_2,b_3$ of $S=,b_1,b_2,b_3,b_4,$ already span $mathbbR^3$. If you ignore the 4th vector, you obtain the echelon form
$$tildeA = left[beginarray@ccc1&0&0&tildef_1(x,y,z)\0&1&0&tildef_2(x,y,z)\0&0&1&tildef_3(x,y,z)endarrayright]$$
giving you unique values for $alpha_1,alpha_2, alpha_3$ without any choices.
In fact, those three vectors are a basis for $mathbbR^3$, since they span $mathbbR^3$ and their cardinality is $3$, which is the dimension of $mathbbR^3$.
Alternatively, you may check that the first three vectors are linearly independent, meaning that if $alpha_1 b_1 + alpha_2 b_2 + alpha_3 b_3=0$, you can prove that $alpha_1=alpha_2=alpha_3=0$. Then you obtain that $b_1,b_2,b_3$ is a basis of $mathbbR^3$, so they have to span $mathbbR^3$, too.
Your third option is to show, that you can generate the three standard basis vectors $e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)$ with $S$. You probably know that $e_1,e_2,e_3$ span $mathbbR^3$, so $S$ spanns $mathbbR^3$ aswell, if you can generate the standard basis.
answered Jul 25 at 7:23
Babelfish
408112
answered Jul 25 at 7:23
Babelfish
408112
answered Jul 25 at 7:23
Babelfish
408112
answered Jul 25 at 7:23
Babelfish
408112
408112
add a comment |Â
add a comment |Â
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2
Your matrix has rank $3$....
– Lord Shark the Unknown
Jul 25 at 6:47