anonymous
  • anonymous
How do you show that vector, x= (2,3,4)+ t1 (1,1,1) + t2 (1,2,3) is a subspace of R^n?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
swissgirl
  • swissgirl
Well for starters in order for a vector to be a subspace it must be linearly dependant
anonymous
  • anonymous
Since ( 2,3,4)=(1,1,1)+(1,2,3), it follows that (2,3,4)+t1(1,1,1)+t2(2,3,4)=(t1+1)(1,1,1)+(t2+1)(2,3,4)
swissgirl
  • swissgirl
hmmmm I would go about it just a drop differently

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

swissgirl
  • swissgirl
\[ \left( \begin{array}{ccc} 2 & 1 & 1 \\ 3 & 1 & 2 \\ 4 & 1 & 3 \end{array} \right)\] I forgot the terms but basically x=(2+1t_1+1t_2)+(3+1t_1+2t_2)+(4+1t_1+3t_2) Now if you would row reduce this then you will see that this matrix is linearly dependant
swissgirl
  • swissgirl
\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 3 & 1 & 2 \\ 4 & 1 & 3 \end{array} \right)\] 1/2R_1=R_1
swissgirl
  • swissgirl
\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &- {1 \over 2} & - {1 \over 2} \\ 4 & 1 & 3 \end{array} \right)\] 3R_1+R_2=R_2
swissgirl
  • swissgirl
\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 4 & 1 & 3 \end{array} \right)\] -2R_2=R_2
anonymous
  • anonymous
you are going to get all 0s in the first column, with one 1 in each of the second and third columns. Thats because (2,3,4)=1(1,1,1)+1(1,2,3).
swissgirl
  • swissgirl
\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 & -2 & -2 \end{array} \right)\] -4R_1+R_3=R_3
swissgirl
  • swissgirl
\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 & 1 & 1 \end{array} \right)\] -1/2R_3=R_3
swissgirl
  • swissgirl
\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 &0 & 0 \end{array} \right)\] -1R_2+R_3=R_3 As you see the last row is just zeroes this means that this matrix is linearly dependant meaning that it is a subspace
anonymous
  • anonymous
thanks :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.