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Check if vectors are the the fundamental set of solutions
Is thi set of vectors, $(2, 1), (3, 2), (1, 2)$, is linearly dependent or independent?How to solve the six elements equations below?I wonder whether the system of equations and inequations below have a solution.How to solve this system without WolframAlphaWriting the span of two vectors in constraint formBasis and dimension of a subspace of $mathcalM_2times 2(mathbbR)$Determining linear independence of a set of functionsTesting for linear independence (Vectors -> SOE)Are these vectors linearly independent? 3 vectors.Determine whether the sets spans in $R^2$
$begingroup$
$$begincases x_1+2x_2+x_3-x_4+x_5=0 \ 2x_1+x_2-x_3+2x_4-x_5=0 \ x_1+5x_2+4x_3-5x_4+4x_5=0 \ 4x_1+5x_2+x_3+x_5=0 endcases$$
$$begincases veca_1=(-5,4,3,3,-3) \ veca_2=(-7,5,-3,6,6) \ veca_3=(5,-4,6,-3,-6) endcases$$
Solution:
beginbmatrix
1&2&1&-1&1 \
2&1&-1&2&-1 \
1&5&4&-5&4 \
4&5&1&0&1
endbmatrix
Reducing to the row echelon form gives us
beginbmatrix
1&2&1&-1&1 \
0&3&-3&-4&-3
endbmatrix
Let $x_3 = c_1$, $x_4=c_2$, $x_5=c_3$.
$$begincases x_1+2x_2+c_1-c_2+c_3=0 \ 0x_1+3x_2-3c_1-4c_2-3c_3=0endcases$$
$$x_2=c_1+frac43c_2+c_3$$
$$x_1=-2c_1-frac83c_2-2c_3-2c_1+c_2-c_3=-4c_1-frac53c_2-3c_2$$
However, plugging the corresponding values of each vector into the resulting system does not work here. What did I miss?
linear-algebra systems-of-equations
asked Mar 21 at 23:19
Ben TerryBen Terry
1
$endgroup$
add a comment |
$begingroup$
$$begincases x_1+2x_2+x_3-x_4+x_5=0 \ 2x_1+x_2-x_3+2x_4-x_5=0 \ x_1+5x_2+4x_3-5x_4+4x_5=0 \ 4x_1+5x_2+x_3+x_5=0 endcases$$
$$begincases veca_1=(-5,4,3,3,-3) \ veca_2=(-7,5,-3,6,6) \ veca_3=(5,-4,6,-3,-6) endcases$$
Solution:
beginbmatrix
1&2&1&-1&1 \
2&1&-1&2&-1 \
1&5&4&-5&4 \
4&5&1&0&1
endbmatrix
Reducing to the row echelon form gives us
beginbmatrix
1&2&1&-1&1 \
0&3&-3&-4&-3
endbmatrix
Let $x_3 = c_1$, $x_4=c_2$, $x_5=c_3$.
$$begincases x_1+2x_2+c_1-c_2+c_3=0 \ 0x_1+3x_2-3c_1-4c_2-3c_3=0endcases$$
$$x_2=c_1+frac43c_2+c_3$$
$$x_1=-2c_1-frac83c_2-2c_3-2c_1+c_2-c_3=-4c_1-frac53c_2-3c_2$$
However, plugging the corresponding values of each vector into the resulting system does not work here. What did I miss?
linear-algebra systems-of-equations
asked Mar 21 at 23:19
Ben TerryBen Terry
1
$endgroup$
$begingroup$
Looks to me like you’ve made a sign error in your row-reduced matrix. That aside, you could instead check that the three vectors satisfy the original system and that they’re linearly independent. From the row-reduction, you know that the nullity is 3, so you’re done.
$endgroup$
– amd
Mar 21 at 23:42
add a comment |
$begingroup$
$$begincases x_1+2x_2+x_3-x_4+x_5=0 \ 2x_1+x_2-x_3+2x_4-x_5=0 \ x_1+5x_2+4x_3-5x_4+4x_5=0 \ 4x_1+5x_2+x_3+x_5=0 endcases$$
$$begincases veca_1=(-5,4,3,3,-3) \ veca_2=(-7,5,-3,6,6) \ veca_3=(5,-4,6,-3,-6) endcases$$
Solution:
beginbmatrix
1&2&1&-1&1 \
2&1&-1&2&-1 \
1&5&4&-5&4 \
4&5&1&0&1
endbmatrix
Reducing to the row echelon form gives us
beginbmatrix
1&2&1&-1&1 \
0&3&-3&-4&-3
endbmatrix
Let $x_3 = c_1$, $x_4=c_2$, $x_5=c_3$.
$$begincases x_1+2x_2+c_1-c_2+c_3=0 \ 0x_1+3x_2-3c_1-4c_2-3c_3=0endcases$$
$$x_2=c_1+frac43c_2+c_3$$
$$x_1=-2c_1-frac83c_2-2c_3-2c_1+c_2-c_3=-4c_1-frac53c_2-3c_2$$
However, plugging the corresponding values of each vector into the resulting system does not work here. What did I miss?
linear-algebra systems-of-equations
asked Mar 21 at 23:19
Ben TerryBen Terry
1
$endgroup$
$$begincases x_1+2x_2+x_3-x_4+x_5=0 \ 2x_1+x_2-x_3+2x_4-x_5=0 \ x_1+5x_2+4x_3-5x_4+4x_5=0 \ 4x_1+5x_2+x_3+x_5=0 endcases$$
$$begincases veca_1=(-5,4,3,3,-3) \ veca_2=(-7,5,-3,6,6) \ veca_3=(5,-4,6,-3,-6) endcases$$
Solution:
beginbmatrix
1&2&1&-1&1 \
2&1&-1&2&-1 \
1&5&4&-5&4 \
4&5&1&0&1
endbmatrix
Reducing to the row echelon form gives us
beginbmatrix
1&2&1&-1&1 \
0&3&-3&-4&-3
endbmatrix
Let $x_3 = c_1$, $x_4=c_2$, $x_5=c_3$.
$$begincases x_1+2x_2+c_1-c_2+c_3=0 \ 0x_1+3x_2-3c_1-4c_2-3c_3=0endcases$$
$$x_2=c_1+frac43c_2+c_3$$
$$x_1=-2c_1-frac83c_2-2c_3-2c_1+c_2-c_3=-4c_1-frac53c_2-3c_2$$
However, plugging the corresponding values of each vector into the resulting system does not work here. What did I miss?
linear-algebra systems-of-equations
linear-algebra systems-of-equations
asked Mar 21 at 23:19
Ben TerryBen Terry
1
asked Mar 21 at 23:19
Ben TerryBen Terry
1
asked Mar 21 at 23:19
Ben TerryBen Terry
1
asked Mar 21 at 23:19
Ben TerryBen Terry
1
asked Mar 21 at 23:19
Ben TerryBen Terry
1
1
$begingroup$
Looks to me like you’ve made a sign error in your row-reduced matrix. That aside, you could instead check that the three vectors satisfy the original system and that they’re linearly independent. From the row-reduction, you know that the nullity is 3, so you’re done.
$endgroup$
– amd
Mar 21 at 23:42
add a comment |
$begingroup$
Looks to me like you’ve made a sign error in your row-reduced matrix. That aside, you could instead check that the three vectors satisfy the original system and that they’re linearly independent. From the row-reduction, you know that the nullity is 3, so you’re done.
$endgroup$
– amd
Mar 21 at 23:42
$begingroup$
Looks to me like you’ve made a sign error in your row-reduced matrix. That aside, you could instead check that the three vectors satisfy the original system and that they’re linearly independent. From the row-reduction, you know that the nullity is 3, so you’re done.
$endgroup$
– amd
Mar 21 at 23:42
$begingroup$
Looks to me like you’ve made a sign error in your row-reduced matrix. That aside, you could instead check that the three vectors satisfy the original system and that they’re linearly independent. From the row-reduction, you know that the nullity is 3, so you’re done.
$endgroup$
– amd
Mar 21 at 23:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$
left(
beginarrayrrrr
1 & 0 & 0 &0 \
-1 & 0 & 1 &0 \
3 &-1 &-1 &0 \
-5 & 0 & 1& 1 \
endarray
right)
left(
beginarrayrrrrr
1& 2 &1& -1 & 1\
2& 1 &-1 & 2 &-1\
1 &5 & 4 &-5 & 4\
4 &5& 1 & 0& 1\
endarray
right) =
left(
beginarrayrrrrr
1& 2 &1& -1 & 1\
0& 3 &3 & -4 &3\
0 &0 & 0 &0 & 0\
0 &0& 0 & 0& 0\
endarray
right)
$$
answered Mar 22 at 0:28
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$
left(
beginarrayrrrr
1 & 0 & 0 &0 \
-1 & 0 & 1 &0 \
3 &-1 &-1 &0 \
-5 & 0 & 1& 1 \
endarray
right)
left(
beginarrayrrrrr
1& 2 &1& -1 & 1\
2& 1 &-1 & 2 &-1\
1 &5 & 4 &-5 & 4\
4 &5& 1 & 0& 1\
endarray
right) =
left(
beginarrayrrrrr
1& 2 &1& -1 & 1\
0& 3 &3 & -4 &3\
0 &0 & 0 &0 & 0\
0 &0& 0 & 0& 0\
endarray
right)
$$
answered Mar 22 at 0:28
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
$$
left(
beginarrayrrrr
1 & 0 & 0 &0 \
-1 & 0 & 1 &0 \
3 &-1 &-1 &0 \
-5 & 0 & 1& 1 \
endarray
right)
left(
beginarrayrrrrr
1& 2 &1& -1 & 1\
2& 1 &-1 & 2 &-1\
1 &5 & 4 &-5 & 4\
4 &5& 1 & 0& 1\
endarray
right) =
left(
beginarrayrrrrr
1& 2 &1& -1 & 1\
0& 3 &3 & -4 &3\
0 &0 & 0 &0 & 0\
0 &0& 0 & 0& 0\
endarray
right)
$$
answered Mar 22 at 0:28
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
$$
left(
beginarrayrrrr
1 & 0 & 0 &0 \
-1 & 0 & 1 &0 \
3 &-1 &-1 &0 \
-5 & 0 & 1& 1 \
endarray
right)
left(
beginarrayrrrrr
1& 2 &1& -1 & 1\
2& 1 &-1 & 2 &-1\
1 &5 & 4 &-5 & 4\
4 &5& 1 & 0& 1\
endarray
right) =
left(
beginarrayrrrrr
1& 2 &1& -1 & 1\
0& 3 &3 & -4 &3\
0 &0 & 0 &0 & 0\
0 &0& 0 & 0& 0\
endarray
right)
$$
answered Mar 22 at 0:28
Will JagyWill Jagy
104k5102201
$endgroup$
$$
left(
beginarrayrrrr
1 & 0 & 0 &0 \
-1 & 0 & 1 &0 \
3 &-1 &-1 &0 \
-5 & 0 & 1& 1 \
endarray
right)
left(
beginarrayrrrrr
1& 2 &1& -1 & 1\
2& 1 &-1 & 2 &-1\
1 &5 & 4 &-5 & 4\
4 &5& 1 & 0& 1\
endarray
right) =
left(
beginarrayrrrrr
1& 2 &1& -1 & 1\
0& 3 &3 & -4 &3\
0 &0 & 0 &0 & 0\
0 &0& 0 & 0& 0\
endarray
right)
$$
answered Mar 22 at 0:28
Will JagyWill Jagy
104k5102201
answered Mar 22 at 0:28
Will JagyWill Jagy
104k5102201
answered Mar 22 at 0:28
Will JagyWill Jagy
104k5102201
answered Mar 22 at 0:28
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
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$begingroup$
Looks to me like you’ve made a sign error in your row-reduced matrix. That aside, you could instead check that the three vectors satisfy the original system and that they’re linearly independent. From the row-reduction, you know that the nullity is 3, so you’re done.
$endgroup$
– amd
Mar 21 at 23:42