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System of homogeneous equations satisfied by these specified vectors
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Given two vectors $x = beginpmatrixt\1endpmatrix$ and $y = beginpmatrixt^2\2tendpmatrix$ for $t > 0$. Find a system of homogeneous equations satisfied by these two vector functions.
What does this mean?
The first step of the problem was to check for linear dependence.
I took the determinant so that $2t^2 - t^2 = t^2$. Since $t^2$ cannot equal zero given that $t > 0$, then the set of vectors are linearly independent.
What does it mean after to find a system of homogenous equations that satisfies these two vector functions?
Should I be finding eigenvalues and eigenvectors in terms of $t$ and $r$?
Thank you for the guidance.
linear-algebra differential-equations homogeneous-equation
edited Jul 22 at 16:12
mrtaurho
740219
asked Jul 22 at 16:06
PERTURBATIONFLOW
396
 |Â
show 1 more comment
up vote
0
down vote
favorite
Given two vectors $x = beginpmatrixt\1endpmatrix$ and $y = beginpmatrixt^2\2tendpmatrix$ for $t > 0$. Find a system of homogeneous equations satisfied by these two vector functions.
What does this mean?
The first step of the problem was to check for linear dependence.
I took the determinant so that $2t^2 - t^2 = t^2$. Since $t^2$ cannot equal zero given that $t > 0$, then the set of vectors are linearly independent.
What does it mean after to find a system of homogenous equations that satisfies these two vector functions?
Should I be finding eigenvalues and eigenvectors in terms of $t$ and $r$?
Thank you for the guidance.
linear-algebra differential-equations homogeneous-equation
edited Jul 22 at 16:12
mrtaurho
740219
asked Jul 22 at 16:06
PERTURBATIONFLOW
396
Maybe as simple as I guessed first : $beginmatrixtcdot c_1 &+& t^2cdot c_2 & = & 0\ c_1 &+&2cdot tcdot c_2 & = & 0 endmatrix$?
– mrtaurho
Jul 22 at 16:15
Is it really that simple? hmmm
– PERTURBATIONFLOW
Jul 22 at 16:16
Yes, but it's better to use variables other than $x,y$.
– poyea
Jul 22 at 16:18
@poyea why do we need to choose variables other than x and y?
– PERTURBATIONFLOW
Jul 22 at 16:19
1
@PERTURBATIONFLOW Because your given vectors are already denoted as $x$ and $y$ and so it can be seen as as system were you multiply every single number/variable by a vector, which is not what we meant to do.
– mrtaurho
Jul 22 at 16:21
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given two vectors $x = beginpmatrixt\1endpmatrix$ and $y = beginpmatrixt^2\2tendpmatrix$ for $t > 0$. Find a system of homogeneous equations satisfied by these two vector functions.
What does this mean?
The first step of the problem was to check for linear dependence.
I took the determinant so that $2t^2 - t^2 = t^2$. Since $t^2$ cannot equal zero given that $t > 0$, then the set of vectors are linearly independent.
What does it mean after to find a system of homogenous equations that satisfies these two vector functions?
Should I be finding eigenvalues and eigenvectors in terms of $t$ and $r$?
Thank you for the guidance.
linear-algebra differential-equations homogeneous-equation
edited Jul 22 at 16:12
mrtaurho
740219
asked Jul 22 at 16:06
PERTURBATIONFLOW
396
Given two vectors $x = beginpmatrixt\1endpmatrix$ and $y = beginpmatrixt^2\2tendpmatrix$ for $t > 0$. Find a system of homogeneous equations satisfied by these two vector functions.
What does this mean?
The first step of the problem was to check for linear dependence.
I took the determinant so that $2t^2 - t^2 = t^2$. Since $t^2$ cannot equal zero given that $t > 0$, then the set of vectors are linearly independent.
What does it mean after to find a system of homogenous equations that satisfies these two vector functions?
Should I be finding eigenvalues and eigenvectors in terms of $t$ and $r$?
Thank you for the guidance.
linear-algebra differential-equations homogeneous-equation
edited Jul 22 at 16:12
mrtaurho
740219
asked Jul 22 at 16:06
PERTURBATIONFLOW
396
edited Jul 22 at 16:12
mrtaurho
740219
edited Jul 22 at 16:12
mrtaurho
740219
edited Jul 22 at 16:12
mrtaurho
740219
740219
asked Jul 22 at 16:06
PERTURBATIONFLOW
396
asked Jul 22 at 16:06
PERTURBATIONFLOW
396
asked Jul 22 at 16:06
PERTURBATIONFLOW
396
396
Maybe as simple as I guessed first : $beginmatrixtcdot c_1 &+& t^2cdot c_2 & = & 0\ c_1 &+&2cdot tcdot c_2 & = & 0 endmatrix$?
– mrtaurho
Jul 22 at 16:15
Is it really that simple? hmmm
– PERTURBATIONFLOW
Jul 22 at 16:16
Yes, but it's better to use variables other than $x,y$.
– poyea
Jul 22 at 16:18
@poyea why do we need to choose variables other than x and y?
– PERTURBATIONFLOW
Jul 22 at 16:19
1
@PERTURBATIONFLOW Because your given vectors are already denoted as $x$ and $y$ and so it can be seen as as system were you multiply every single number/variable by a vector, which is not what we meant to do.
– mrtaurho
Jul 22 at 16:21
 |Â
show 1 more comment
Maybe as simple as I guessed first : $beginmatrixtcdot c_1 &+& t^2cdot c_2 & = & 0\ c_1 &+&2cdot tcdot c_2 & = & 0 endmatrix$?
– mrtaurho
Jul 22 at 16:15
Is it really that simple? hmmm
– PERTURBATIONFLOW
Jul 22 at 16:16
Yes, but it's better to use variables other than $x,y$.
– poyea
Jul 22 at 16:18
@poyea why do we need to choose variables other than x and y?
– PERTURBATIONFLOW
Jul 22 at 16:19
1
@PERTURBATIONFLOW Because your given vectors are already denoted as $x$ and $y$ and so it can be seen as as system were you multiply every single number/variable by a vector, which is not what we meant to do.
– mrtaurho
Jul 22 at 16:21
Maybe as simple as I guessed first : $beginmatrixtcdot c_1 &+& t^2cdot c_2 & = & 0\ c_1 &+&2cdot tcdot c_2 & = & 0 endmatrix$?
– mrtaurho
Jul 22 at 16:15
Maybe as simple as I guessed first : $beginmatrixtcdot c_1 &+& t^2cdot c_2 & = & 0\ c_1 &+&2cdot tcdot c_2 & = & 0 endmatrix$?
– mrtaurho
Jul 22 at 16:15
Is it really that simple? hmmm
– PERTURBATIONFLOW
Jul 22 at 16:16
Is it really that simple? hmmm
– PERTURBATIONFLOW
Jul 22 at 16:16
Yes, but it's better to use variables other than $x,y$.
– poyea
Jul 22 at 16:18
Yes, but it's better to use variables other than $x,y$.
– poyea
Jul 22 at 16:18
@poyea why do we need to choose variables other than x and y?
– PERTURBATIONFLOW
Jul 22 at 16:19
@poyea why do we need to choose variables other than x and y?
– PERTURBATIONFLOW
Jul 22 at 16:19
1
1
@PERTURBATIONFLOW Because your given vectors are already denoted as $x$ and $y$ and so it can be seen as as system were you multiply every single number/variable by a vector, which is not what we meant to do.
– mrtaurho
Jul 22 at 16:21
@PERTURBATIONFLOW Because your given vectors are already denoted as $x$ and $y$ and so it can be seen as as system were you multiply every single number/variable by a vector, which is not what we meant to do.
– mrtaurho
Jul 22 at 16:21
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
You aren't asked to find a system of equations satisfying these two vector functions--what would it mean to satisfy a function? Rather, you're asked to find a system of equations satisfied by these two functions, meaning that these functions are solutions to said system.
For the sake of simplicity, let's start by looking for a system of first-order linear homogeneous equations to do the job. That is, we're looking for a system of the form: $$begincasesv_1'+a_1,1(t)v_1+a_1,2(t)v_2=0\v_2'+a_2,1(t)v_1+a_2,2(t)v_2=0endcasestag$star$$$
What we need, then, is for $(star)$ to hold when $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t).$
Subbing $(v_1,v_2)=(t,1)$ into $(star)$ gives $$begincases1+a_1,1(t)t+a_1,2(t)=0\a_2,1(t)t+a_2,2(t)=0,endcases$$ and subbing in $(t^2,2t)$ gives $$begincases2t+a_1,1(t)t^2+2a_1,2(t)t=0\2+a_2,1(t)t^2+2a_2,2(t)t=0.endcases$$ This gives us a system of $4$ equations to solve, which is good, because we have $4$ unknowns $a_i,j:=a_i,j(t)$ to figure out. Our system is: $$begincases1+a_1,1t+a_1,2=0\a_2,1t+a_2,2=0\2t+a_1,1t^2+2a_1,2t=0\2+a_2,1t^2+2a_2,2t=0endcases$$
By the second of these equations, we can substitute $$a_2,2=-a_2,1ttag1$$ into the rest of our system, which then yields: $$begincases1+a_1,1t+a_1,2=0\2t+a_1,1t^2+2a_1,2t=0\2-a_2,1t^2=0endcases$$
By the last of these equations together with $(1),$ we require $$a_2,1(t):=frac2t^2,a_2,2(t):=frac-2ttag2$$
On the other hand, we clearly require $$a_1,2=-1-a_1,1ttag3$$ from the first equation of our starting system. Substituting into the third equation then yields $$-a_1,1t^2=0,$$ so by this equation and $(3)$ we must have $$a_1,1(t):=0,a_1,2(t)=-1.tag4$$
At this point, we have that $(star)$ holds $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t),$ as desired.
There's another way we could go about this, though. Observe that your two vectors are both of the form $bigl(f(t),f'(t)bigr).$ So, one obvious homogeneous equation that they both must satisfy is $v_1'-v_2=0,$ which is exactly $v_1'+a_1,1v_1+a_1,2v_2=0,$ though with less work necessary.
At that point, we just need an equation of the form $$v_2'+c_1(t)v_1+c_2(t)v_2=0.$$ Subbing our desired solution vectors in gives us the system $$begincases0+c_1t+2c_2t=0\2+c_1t^2+2c_2t=0,endcases$$ which is readily solved to get $c_1=a_2,1$ and $c_2=a_2,2$ as above.
answered Jul 29 at 15:56
Cameron Buie
83.5k771153
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You aren't asked to find a system of equations satisfying these two vector functions--what would it mean to satisfy a function? Rather, you're asked to find a system of equations satisfied by these two functions, meaning that these functions are solutions to said system.
For the sake of simplicity, let's start by looking for a system of first-order linear homogeneous equations to do the job. That is, we're looking for a system of the form: $$begincasesv_1'+a_1,1(t)v_1+a_1,2(t)v_2=0\v_2'+a_2,1(t)v_1+a_2,2(t)v_2=0endcasestag$star$$$
What we need, then, is for $(star)$ to hold when $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t).$
Subbing $(v_1,v_2)=(t,1)$ into $(star)$ gives $$begincases1+a_1,1(t)t+a_1,2(t)=0\a_2,1(t)t+a_2,2(t)=0,endcases$$ and subbing in $(t^2,2t)$ gives $$begincases2t+a_1,1(t)t^2+2a_1,2(t)t=0\2+a_2,1(t)t^2+2a_2,2(t)t=0.endcases$$ This gives us a system of $4$ equations to solve, which is good, because we have $4$ unknowns $a_i,j:=a_i,j(t)$ to figure out. Our system is: $$begincases1+a_1,1t+a_1,2=0\a_2,1t+a_2,2=0\2t+a_1,1t^2+2a_1,2t=0\2+a_2,1t^2+2a_2,2t=0endcases$$
By the second of these equations, we can substitute $$a_2,2=-a_2,1ttag1$$ into the rest of our system, which then yields: $$begincases1+a_1,1t+a_1,2=0\2t+a_1,1t^2+2a_1,2t=0\2-a_2,1t^2=0endcases$$
By the last of these equations together with $(1),$ we require $$a_2,1(t):=frac2t^2,a_2,2(t):=frac-2ttag2$$
On the other hand, we clearly require $$a_1,2=-1-a_1,1ttag3$$ from the first equation of our starting system. Substituting into the third equation then yields $$-a_1,1t^2=0,$$ so by this equation and $(3)$ we must have $$a_1,1(t):=0,a_1,2(t)=-1.tag4$$
At this point, we have that $(star)$ holds $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t),$ as desired.
There's another way we could go about this, though. Observe that your two vectors are both of the form $bigl(f(t),f'(t)bigr).$ So, one obvious homogeneous equation that they both must satisfy is $v_1'-v_2=0,$ which is exactly $v_1'+a_1,1v_1+a_1,2v_2=0,$ though with less work necessary.
At that point, we just need an equation of the form $$v_2'+c_1(t)v_1+c_2(t)v_2=0.$$ Subbing our desired solution vectors in gives us the system $$begincases0+c_1t+2c_2t=0\2+c_1t^2+2c_2t=0,endcases$$ which is readily solved to get $c_1=a_2,1$ and $c_2=a_2,2$ as above.
answered Jul 29 at 15:56
Cameron Buie
83.5k771153
add a comment |Â
up vote
0
down vote
accepted
You aren't asked to find a system of equations satisfying these two vector functions--what would it mean to satisfy a function? Rather, you're asked to find a system of equations satisfied by these two functions, meaning that these functions are solutions to said system.
For the sake of simplicity, let's start by looking for a system of first-order linear homogeneous equations to do the job. That is, we're looking for a system of the form: $$begincasesv_1'+a_1,1(t)v_1+a_1,2(t)v_2=0\v_2'+a_2,1(t)v_1+a_2,2(t)v_2=0endcasestag$star$$$
What we need, then, is for $(star)$ to hold when $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t).$
Subbing $(v_1,v_2)=(t,1)$ into $(star)$ gives $$begincases1+a_1,1(t)t+a_1,2(t)=0\a_2,1(t)t+a_2,2(t)=0,endcases$$ and subbing in $(t^2,2t)$ gives $$begincases2t+a_1,1(t)t^2+2a_1,2(t)t=0\2+a_2,1(t)t^2+2a_2,2(t)t=0.endcases$$ This gives us a system of $4$ equations to solve, which is good, because we have $4$ unknowns $a_i,j:=a_i,j(t)$ to figure out. Our system is: $$begincases1+a_1,1t+a_1,2=0\a_2,1t+a_2,2=0\2t+a_1,1t^2+2a_1,2t=0\2+a_2,1t^2+2a_2,2t=0endcases$$
By the second of these equations, we can substitute $$a_2,2=-a_2,1ttag1$$ into the rest of our system, which then yields: $$begincases1+a_1,1t+a_1,2=0\2t+a_1,1t^2+2a_1,2t=0\2-a_2,1t^2=0endcases$$
By the last of these equations together with $(1),$ we require $$a_2,1(t):=frac2t^2,a_2,2(t):=frac-2ttag2$$
On the other hand, we clearly require $$a_1,2=-1-a_1,1ttag3$$ from the first equation of our starting system. Substituting into the third equation then yields $$-a_1,1t^2=0,$$ so by this equation and $(3)$ we must have $$a_1,1(t):=0,a_1,2(t)=-1.tag4$$
At this point, we have that $(star)$ holds $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t),$ as desired.
There's another way we could go about this, though. Observe that your two vectors are both of the form $bigl(f(t),f'(t)bigr).$ So, one obvious homogeneous equation that they both must satisfy is $v_1'-v_2=0,$ which is exactly $v_1'+a_1,1v_1+a_1,2v_2=0,$ though with less work necessary.
At that point, we just need an equation of the form $$v_2'+c_1(t)v_1+c_2(t)v_2=0.$$ Subbing our desired solution vectors in gives us the system $$begincases0+c_1t+2c_2t=0\2+c_1t^2+2c_2t=0,endcases$$ which is readily solved to get $c_1=a_2,1$ and $c_2=a_2,2$ as above.
answered Jul 29 at 15:56
Cameron Buie
83.5k771153
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You aren't asked to find a system of equations satisfying these two vector functions--what would it mean to satisfy a function? Rather, you're asked to find a system of equations satisfied by these two functions, meaning that these functions are solutions to said system.
For the sake of simplicity, let's start by looking for a system of first-order linear homogeneous equations to do the job. That is, we're looking for a system of the form: $$begincasesv_1'+a_1,1(t)v_1+a_1,2(t)v_2=0\v_2'+a_2,1(t)v_1+a_2,2(t)v_2=0endcasestag$star$$$
What we need, then, is for $(star)$ to hold when $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t).$
Subbing $(v_1,v_2)=(t,1)$ into $(star)$ gives $$begincases1+a_1,1(t)t+a_1,2(t)=0\a_2,1(t)t+a_2,2(t)=0,endcases$$ and subbing in $(t^2,2t)$ gives $$begincases2t+a_1,1(t)t^2+2a_1,2(t)t=0\2+a_2,1(t)t^2+2a_2,2(t)t=0.endcases$$ This gives us a system of $4$ equations to solve, which is good, because we have $4$ unknowns $a_i,j:=a_i,j(t)$ to figure out. Our system is: $$begincases1+a_1,1t+a_1,2=0\a_2,1t+a_2,2=0\2t+a_1,1t^2+2a_1,2t=0\2+a_2,1t^2+2a_2,2t=0endcases$$
By the second of these equations, we can substitute $$a_2,2=-a_2,1ttag1$$ into the rest of our system, which then yields: $$begincases1+a_1,1t+a_1,2=0\2t+a_1,1t^2+2a_1,2t=0\2-a_2,1t^2=0endcases$$
By the last of these equations together with $(1),$ we require $$a_2,1(t):=frac2t^2,a_2,2(t):=frac-2ttag2$$
On the other hand, we clearly require $$a_1,2=-1-a_1,1ttag3$$ from the first equation of our starting system. Substituting into the third equation then yields $$-a_1,1t^2=0,$$ so by this equation and $(3)$ we must have $$a_1,1(t):=0,a_1,2(t)=-1.tag4$$
At this point, we have that $(star)$ holds $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t),$ as desired.
There's another way we could go about this, though. Observe that your two vectors are both of the form $bigl(f(t),f'(t)bigr).$ So, one obvious homogeneous equation that they both must satisfy is $v_1'-v_2=0,$ which is exactly $v_1'+a_1,1v_1+a_1,2v_2=0,$ though with less work necessary.
At that point, we just need an equation of the form $$v_2'+c_1(t)v_1+c_2(t)v_2=0.$$ Subbing our desired solution vectors in gives us the system $$begincases0+c_1t+2c_2t=0\2+c_1t^2+2c_2t=0,endcases$$ which is readily solved to get $c_1=a_2,1$ and $c_2=a_2,2$ as above.
answered Jul 29 at 15:56
Cameron Buie
83.5k771153
You aren't asked to find a system of equations satisfying these two vector functions--what would it mean to satisfy a function? Rather, you're asked to find a system of equations satisfied by these two functions, meaning that these functions are solutions to said system.
For the sake of simplicity, let's start by looking for a system of first-order linear homogeneous equations to do the job. That is, we're looking for a system of the form: $$begincasesv_1'+a_1,1(t)v_1+a_1,2(t)v_2=0\v_2'+a_2,1(t)v_1+a_2,2(t)v_2=0endcasestag$star$$$
What we need, then, is for $(star)$ to hold when $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t).$
Subbing $(v_1,v_2)=(t,1)$ into $(star)$ gives $$begincases1+a_1,1(t)t+a_1,2(t)=0\a_2,1(t)t+a_2,2(t)=0,endcases$$ and subbing in $(t^2,2t)$ gives $$begincases2t+a_1,1(t)t^2+2a_1,2(t)t=0\2+a_2,1(t)t^2+2a_2,2(t)t=0.endcases$$ This gives us a system of $4$ equations to solve, which is good, because we have $4$ unknowns $a_i,j:=a_i,j(t)$ to figure out. Our system is: $$begincases1+a_1,1t+a_1,2=0\a_2,1t+a_2,2=0\2t+a_1,1t^2+2a_1,2t=0\2+a_2,1t^2+2a_2,2t=0endcases$$
By the second of these equations, we can substitute $$a_2,2=-a_2,1ttag1$$ into the rest of our system, which then yields: $$begincases1+a_1,1t+a_1,2=0\2t+a_1,1t^2+2a_1,2t=0\2-a_2,1t^2=0endcases$$
By the last of these equations together with $(1),$ we require $$a_2,1(t):=frac2t^2,a_2,2(t):=frac-2ttag2$$
On the other hand, we clearly require $$a_1,2=-1-a_1,1ttag3$$ from the first equation of our starting system. Substituting into the third equation then yields $$-a_1,1t^2=0,$$ so by this equation and $(3)$ we must have $$a_1,1(t):=0,a_1,2(t)=-1.tag4$$
At this point, we have that $(star)$ holds $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t),$ as desired.
There's another way we could go about this, though. Observe that your two vectors are both of the form $bigl(f(t),f'(t)bigr).$ So, one obvious homogeneous equation that they both must satisfy is $v_1'-v_2=0,$ which is exactly $v_1'+a_1,1v_1+a_1,2v_2=0,$ though with less work necessary.
At that point, we just need an equation of the form $$v_2'+c_1(t)v_1+c_2(t)v_2=0.$$ Subbing our desired solution vectors in gives us the system $$begincases0+c_1t+2c_2t=0\2+c_1t^2+2c_2t=0,endcases$$ which is readily solved to get $c_1=a_2,1$ and $c_2=a_2,2$ as above.
answered Jul 29 at 15:56
Cameron Buie
83.5k771153
edited Aug 3 at 23:43
answered Jul 29 at 15:56
Cameron Buie
83.5k771153
answered Jul 29 at 15:56
Cameron Buie
83.5k771153
answered Jul 29 at 15:56
Cameron Buie
83.5k771153
83.5k771153
add a comment |Â
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Maybe as simple as I guessed first : $beginmatrixtcdot c_1 &+& t^2cdot c_2 & = & 0\ c_1 &+&2cdot tcdot c_2 & = & 0 endmatrix$?
– mrtaurho
Jul 22 at 16:15
Is it really that simple? hmmm
– PERTURBATIONFLOW
Jul 22 at 16:16
Yes, but it's better to use variables other than $x,y$.
– poyea
Jul 22 at 16:18
@poyea why do we need to choose variables other than x and y?
– PERTURBATIONFLOW
Jul 22 at 16:19
1
@PERTURBATIONFLOW Because your given vectors are already denoted as $x$ and $y$ and so it can be seen as as system were you multiply every single number/variable by a vector, which is not what we meant to do.
– mrtaurho
Jul 22 at 16:21