Determine which of the following vectors p(t) is a linear combination
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Determine which of the following vectors p(t) is a linear combination of:$$p_1=t^2-t$$ $$p_2=t^{2}-2t+1$$ $$p_3=-t^{2}+1$$
a) $$p(t)=3t^{2}-3+1$$
b) $$p(t)=t^{2}-t+1$$
c) $$p(t)=t+1$$
d) $$p(t)=2t^{2}-t-1$$
Ok. This is why I'm doing... For example to check the first one, the letter a, i'm using the scalars $$x,y,z$$.
$$p(t)=xp_1+yp_2+zp_3$$
$$3t^2-3t+1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$
$$3t^2-3t+1=xt^2-xt+yt^2-2ty+y-t^2z+z$$
$$3t^2-3t+1=xt^2+yt^2-t^2z-xt-2ty+y+z$$
$$3t^2-3t+1=(x+y-z)t^2-(x+2y)t+(y+z)$$
Then I get and this equation system:
$$x+y-z=3$$
$$x+2y=3$$
$$y+z=1$$
which doesn't have a solution and then I conclude that $$p(t)$$ is not a linear combination of the vectors $$p_1,p_2,p_3$$.
It is my analysis right? I did the same with the rest and when having the equation system non of them have solution so I conclude that non of them are linear combination. But I want to be sure that what I'm doing is right.
I'm teaching myself linear algebra and any help from the math community will be appreciated.
linear-algebra vector-spaces
asked Nov 22 at 19:43
gi2302
103
add a comment |
up vote
1
down vote
favorite
Determine which of the following vectors p(t) is a linear combination of:$$p_1=t^2-t$$ $$p_2=t^{2}-2t+1$$ $$p_3=-t^{2}+1$$
a) $$p(t)=3t^{2}-3+1$$
b) $$p(t)=t^{2}-t+1$$
c) $$p(t)=t+1$$
d) $$p(t)=2t^{2}-t-1$$
Ok. This is why I'm doing... For example to check the first one, the letter a, i'm using the scalars $$x,y,z$$.
$$p(t)=xp_1+yp_2+zp_3$$
$$3t^2-3t+1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$
$$3t^2-3t+1=xt^2-xt+yt^2-2ty+y-t^2z+z$$
$$3t^2-3t+1=xt^2+yt^2-t^2z-xt-2ty+y+z$$
$$3t^2-3t+1=(x+y-z)t^2-(x+2y)t+(y+z)$$
Then I get and this equation system:
$$x+y-z=3$$
$$x+2y=3$$
$$y+z=1$$
which doesn't have a solution and then I conclude that $$p(t)$$ is not a linear combination of the vectors $$p_1,p_2,p_3$$.
It is my analysis right? I did the same with the rest and when having the equation system non of them have solution so I conclude that non of them are linear combination. But I want to be sure that what I'm doing is right.
I'm teaching myself linear algebra and any help from the math community will be appreciated.
linear-algebra vector-spaces
asked Nov 22 at 19:43
gi2302
103
You approach is correct, but there is a quicker way of eliminating (a), (b) and (c) (see the answer from hamam_Abdallah) and you must have made a slip in your calculations in (d): $2t^2 - t - 1 = (t^2 - t) - (-t^2 + 1) = p_1 - p_3$
– Rob Arthan
Nov 22 at 19:58
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Determine which of the following vectors p(t) is a linear combination of:$$p_1=t^2-t$$ $$p_2=t^{2}-2t+1$$ $$p_3=-t^{2}+1$$
a) $$p(t)=3t^{2}-3+1$$
b) $$p(t)=t^{2}-t+1$$
c) $$p(t)=t+1$$
d) $$p(t)=2t^{2}-t-1$$
Ok. This is why I'm doing... For example to check the first one, the letter a, i'm using the scalars $$x,y,z$$.
$$p(t)=xp_1+yp_2+zp_3$$
$$3t^2-3t+1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$
$$3t^2-3t+1=xt^2-xt+yt^2-2ty+y-t^2z+z$$
$$3t^2-3t+1=xt^2+yt^2-t^2z-xt-2ty+y+z$$
$$3t^2-3t+1=(x+y-z)t^2-(x+2y)t+(y+z)$$
Then I get and this equation system:
$$x+y-z=3$$
$$x+2y=3$$
$$y+z=1$$
which doesn't have a solution and then I conclude that $$p(t)$$ is not a linear combination of the vectors $$p_1,p_2,p_3$$.
It is my analysis right? I did the same with the rest and when having the equation system non of them have solution so I conclude that non of them are linear combination. But I want to be sure that what I'm doing is right.
I'm teaching myself linear algebra and any help from the math community will be appreciated.
linear-algebra vector-spaces
asked Nov 22 at 19:43
gi2302
103
Determine which of the following vectors p(t) is a linear combination of:$$p_1=t^2-t$$ $$p_2=t^{2}-2t+1$$ $$p_3=-t^{2}+1$$
a) $$p(t)=3t^{2}-3+1$$
b) $$p(t)=t^{2}-t+1$$
c) $$p(t)=t+1$$
d) $$p(t)=2t^{2}-t-1$$
Ok. This is why I'm doing... For example to check the first one, the letter a, i'm using the scalars $$x,y,z$$.
$$p(t)=xp_1+yp_2+zp_3$$
$$3t^2-3t+1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$
$$3t^2-3t+1=xt^2-xt+yt^2-2ty+y-t^2z+z$$
$$3t^2-3t+1=xt^2+yt^2-t^2z-xt-2ty+y+z$$
$$3t^2-3t+1=(x+y-z)t^2-(x+2y)t+(y+z)$$
Then I get and this equation system:
$$x+y-z=3$$
$$x+2y=3$$
$$y+z=1$$
which doesn't have a solution and then I conclude that $$p(t)$$ is not a linear combination of the vectors $$p_1,p_2,p_3$$.
It is my analysis right? I did the same with the rest and when having the equation system non of them have solution so I conclude that non of them are linear combination. But I want to be sure that what I'm doing is right.
I'm teaching myself linear algebra and any help from the math community will be appreciated.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Nov 22 at 19:43
gi2302
103
asked Nov 22 at 19:43
gi2302
103
asked Nov 22 at 19:43
gi2302
103
asked Nov 22 at 19:43
gi2302
103
asked Nov 22 at 19:43
gi2302
103
103
You approach is correct, but there is a quicker way of eliminating (a), (b) and (c) (see the answer from hamam_Abdallah) and you must have made a slip in your calculations in (d): $2t^2 - t - 1 = (t^2 - t) - (-t^2 + 1) = p_1 - p_3$
– Rob Arthan
Nov 22 at 19:58
add a comment |
You approach is correct, but there is a quicker way of eliminating (a), (b) and (c) (see the answer from hamam_Abdallah) and you must have made a slip in your calculations in (d): $2t^2 - t - 1 = (t^2 - t) - (-t^2 + 1) = p_1 - p_3$
– Rob Arthan
Nov 22 at 19:58
You approach is correct, but there is a quicker way of eliminating (a), (b) and (c) (see the answer from hamam_Abdallah) and you must have made a slip in your calculations in (d): $2t^2 - t - 1 = (t^2 - t) - (-t^2 + 1) = p_1 - p_3$
– Rob Arthan
Nov 22 at 19:58
You approach is correct, but there is a quicker way of eliminating (a), (b) and (c) (see the answer from hamam_Abdallah) and you must have made a slip in your calculations in (d): $2t^2 - t - 1 = (t^2 - t) - (-t^2 + 1) = p_1 - p_3$
– Rob Arthan
Nov 22 at 19:58
add a comment |
1 Answer
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We have
$$p_1(1)=p_2(1)=p_3(1)=0$$
thus you will have
$$p(1)=alpha p_1(1)+beta p_2(1)+gamma p_3(1)=0$$
the only one satisfying this condition is the last :(d).
$$p(t)=2t^2-t-1=p_1(t)-p_3(t)$$
answered Nov 22 at 19:50
hamam_Abdallah
37.7k21634
I'm sorry could you elaborate. Why $$p(1)=0$$?
– gi2302
Nov 22 at 19:54
@gi2302 because $p=ap_1+bp_2+cp_3$.
– hamam_Abdallah
Nov 22 at 19:55
This is what I have on letter d). What I'm missing.. $$p(t)=xp_1+yp_2+zp_3$$ $$2t^2-t-1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$ $$2t^2-t-1=xt^2-xt+yt^2-2ty+y-t^2z+z$$ $$2t^2-t-1=xt^2+yt^2-t^2z-xt-2ty+y+z$$ $$2t^2-t-1=(x+y-z)t^2-(x+2y)t-(-y-z)$$ Then I get and this equation system: $$x+y-z=2$$ $$x+2y=1$$ $$-y-z=1$$ I don't get a solution for this system.. :-(
– gi2302
Nov 22 at 20:10
@gi2302 $(1,0,-1)$ is a solution.
– hamam_Abdallah
Nov 22 at 20:29
you're right... I probably missed something while solving the system. Thank you very much!
– gi2302
Nov 22 at 20:32
add a comment |
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1 Answer
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1 Answer
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We have
$$p_1(1)=p_2(1)=p_3(1)=0$$
thus you will have
$$p(1)=alpha p_1(1)+beta p_2(1)+gamma p_3(1)=0$$
the only one satisfying this condition is the last :(d).
$$p(t)=2t^2-t-1=p_1(t)-p_3(t)$$
answered Nov 22 at 19:50
hamam_Abdallah
37.7k21634
I'm sorry could you elaborate. Why $$p(1)=0$$?
– gi2302
Nov 22 at 19:54
@gi2302 because $p=ap_1+bp_2+cp_3$.
– hamam_Abdallah
Nov 22 at 19:55
This is what I have on letter d). What I'm missing.. $$p(t)=xp_1+yp_2+zp_3$$ $$2t^2-t-1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$ $$2t^2-t-1=xt^2-xt+yt^2-2ty+y-t^2z+z$$ $$2t^2-t-1=xt^2+yt^2-t^2z-xt-2ty+y+z$$ $$2t^2-t-1=(x+y-z)t^2-(x+2y)t-(-y-z)$$ Then I get and this equation system: $$x+y-z=2$$ $$x+2y=1$$ $$-y-z=1$$ I don't get a solution for this system.. :-(
– gi2302
Nov 22 at 20:10
@gi2302 $(1,0,-1)$ is a solution.
– hamam_Abdallah
Nov 22 at 20:29
you're right... I probably missed something while solving the system. Thank you very much!
– gi2302
Nov 22 at 20:32
add a comment |
up vote
0
down vote
We have
$$p_1(1)=p_2(1)=p_3(1)=0$$
thus you will have
$$p(1)=alpha p_1(1)+beta p_2(1)+gamma p_3(1)=0$$
the only one satisfying this condition is the last :(d).
$$p(t)=2t^2-t-1=p_1(t)-p_3(t)$$
answered Nov 22 at 19:50
hamam_Abdallah
37.7k21634
I'm sorry could you elaborate. Why $$p(1)=0$$?
– gi2302
Nov 22 at 19:54
@gi2302 because $p=ap_1+bp_2+cp_3$.
– hamam_Abdallah
Nov 22 at 19:55
This is what I have on letter d). What I'm missing.. $$p(t)=xp_1+yp_2+zp_3$$ $$2t^2-t-1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$ $$2t^2-t-1=xt^2-xt+yt^2-2ty+y-t^2z+z$$ $$2t^2-t-1=xt^2+yt^2-t^2z-xt-2ty+y+z$$ $$2t^2-t-1=(x+y-z)t^2-(x+2y)t-(-y-z)$$ Then I get and this equation system: $$x+y-z=2$$ $$x+2y=1$$ $$-y-z=1$$ I don't get a solution for this system.. :-(
– gi2302
Nov 22 at 20:10
@gi2302 $(1,0,-1)$ is a solution.
– hamam_Abdallah
Nov 22 at 20:29
you're right... I probably missed something while solving the system. Thank you very much!
– gi2302
Nov 22 at 20:32
add a comment |
up vote
0
down vote
up vote
0
down vote
We have
$$p_1(1)=p_2(1)=p_3(1)=0$$
thus you will have
$$p(1)=alpha p_1(1)+beta p_2(1)+gamma p_3(1)=0$$
the only one satisfying this condition is the last :(d).
$$p(t)=2t^2-t-1=p_1(t)-p_3(t)$$
answered Nov 22 at 19:50
hamam_Abdallah
37.7k21634
We have
$$p_1(1)=p_2(1)=p_3(1)=0$$
thus you will have
$$p(1)=alpha p_1(1)+beta p_2(1)+gamma p_3(1)=0$$
the only one satisfying this condition is the last :(d).
$$p(t)=2t^2-t-1=p_1(t)-p_3(t)$$
answered Nov 22 at 19:50
hamam_Abdallah
37.7k21634
edited Nov 22 at 20:26
answered Nov 22 at 19:50
hamam_Abdallah
37.7k21634
answered Nov 22 at 19:50
hamam_Abdallah
37.7k21634
answered Nov 22 at 19:50
hamam_Abdallah
37.7k21634
37.7k21634
I'm sorry could you elaborate. Why $$p(1)=0$$?
– gi2302
Nov 22 at 19:54
@gi2302 because $p=ap_1+bp_2+cp_3$.
– hamam_Abdallah
Nov 22 at 19:55
This is what I have on letter d). What I'm missing.. $$p(t)=xp_1+yp_2+zp_3$$ $$2t^2-t-1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$ $$2t^2-t-1=xt^2-xt+yt^2-2ty+y-t^2z+z$$ $$2t^2-t-1=xt^2+yt^2-t^2z-xt-2ty+y+z$$ $$2t^2-t-1=(x+y-z)t^2-(x+2y)t-(-y-z)$$ Then I get and this equation system: $$x+y-z=2$$ $$x+2y=1$$ $$-y-z=1$$ I don't get a solution for this system.. :-(
– gi2302
Nov 22 at 20:10
@gi2302 $(1,0,-1)$ is a solution.
– hamam_Abdallah
Nov 22 at 20:29
you're right... I probably missed something while solving the system. Thank you very much!
– gi2302
Nov 22 at 20:32
add a comment |
I'm sorry could you elaborate. Why $$p(1)=0$$?
– gi2302
Nov 22 at 19:54
@gi2302 because $p=ap_1+bp_2+cp_3$.
– hamam_Abdallah
Nov 22 at 19:55
This is what I have on letter d). What I'm missing.. $$p(t)=xp_1+yp_2+zp_3$$ $$2t^2-t-1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$ $$2t^2-t-1=xt^2-xt+yt^2-2ty+y-t^2z+z$$ $$2t^2-t-1=xt^2+yt^2-t^2z-xt-2ty+y+z$$ $$2t^2-t-1=(x+y-z)t^2-(x+2y)t-(-y-z)$$ Then I get and this equation system: $$x+y-z=2$$ $$x+2y=1$$ $$-y-z=1$$ I don't get a solution for this system.. :-(
– gi2302
Nov 22 at 20:10
@gi2302 $(1,0,-1)$ is a solution.
– hamam_Abdallah
Nov 22 at 20:29
you're right... I probably missed something while solving the system. Thank you very much!
– gi2302
Nov 22 at 20:32
I'm sorry could you elaborate. Why $$p(1)=0$$?
– gi2302
Nov 22 at 19:54
I'm sorry could you elaborate. Why $$p(1)=0$$?
– gi2302
Nov 22 at 19:54
@gi2302 because $p=ap_1+bp_2+cp_3$.
– hamam_Abdallah
Nov 22 at 19:55
@gi2302 because $p=ap_1+bp_2+cp_3$.
– hamam_Abdallah
Nov 22 at 19:55
This is what I have on letter d). What I'm missing.. $$p(t)=xp_1+yp_2+zp_3$$ $$2t^2-t-1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$ $$2t^2-t-1=xt^2-xt+yt^2-2ty+y-t^2z+z$$ $$2t^2-t-1=xt^2+yt^2-t^2z-xt-2ty+y+z$$ $$2t^2-t-1=(x+y-z)t^2-(x+2y)t-(-y-z)$$ Then I get and this equation system: $$x+y-z=2$$ $$x+2y=1$$ $$-y-z=1$$ I don't get a solution for this system.. :-(
– gi2302
Nov 22 at 20:10
This is what I have on letter d). What I'm missing.. $$p(t)=xp_1+yp_2+zp_3$$ $$2t^2-t-1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$ $$2t^2-t-1=xt^2-xt+yt^2-2ty+y-t^2z+z$$ $$2t^2-t-1=xt^2+yt^2-t^2z-xt-2ty+y+z$$ $$2t^2-t-1=(x+y-z)t^2-(x+2y)t-(-y-z)$$ Then I get and this equation system: $$x+y-z=2$$ $$x+2y=1$$ $$-y-z=1$$ I don't get a solution for this system.. :-(
– gi2302
Nov 22 at 20:10
@gi2302 $(1,0,-1)$ is a solution.
– hamam_Abdallah
Nov 22 at 20:29
@gi2302 $(1,0,-1)$ is a solution.
– hamam_Abdallah
Nov 22 at 20:29
you're right... I probably missed something while solving the system. Thank you very much!
– gi2302
Nov 22 at 20:32
you're right... I probably missed something while solving the system. Thank you very much!
– gi2302
Nov 22 at 20:32
add a comment |
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You approach is correct, but there is a quicker way of eliminating (a), (b) and (c) (see the answer from hamam_Abdallah) and you must have made a slip in your calculations in (d): $2t^2 - t - 1 = (t^2 - t) - (-t^2 + 1) = p_1 - p_3$
– Rob Arthan
Nov 22 at 19:58