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derrick902

  • 2 years ago

How do you show that vector, x= (2,3,4)+ t1 (1,1,1) + t2 (1,2,3) is a subspace of R^n?

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  1. swissgirl
    • 2 years ago
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    Well for starters in order for a vector to be a subspace it must be linearly dependant

  2. joemath314159
    • 2 years ago
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    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)

  3. swissgirl
    • 2 years ago
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    hmmmm I would go about it just a drop differently

  4. swissgirl
    • 2 years ago
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    \[ \left( \begin{array}{ccc} 2 & 1 & 1 \\ 3 & 1 & 2 \\ 4 & 1 & 3 \end{array} \right)\] 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

  5. swissgirl
    • 2 years ago
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    \[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 3 & 1 & 2 \\ 4 & 1 & 3 \end{array} \right)\] 1/2R_1=R_1

  6. swissgirl
    • 2 years ago
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    \[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &- {1 \over 2} & - {1 \over 2} \\ 4 & 1 & 3 \end{array} \right)\] 3R_1+R_2=R_2

  7. swissgirl
    • 2 years ago
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    \[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 4 & 1 & 3 \end{array} \right)\] -2R_2=R_2

  8. joemath314159
    • 2 years ago
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    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).

  9. swissgirl
    • 2 years ago
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    \[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 & -2 & -2 \end{array} \right)\] -4R_1+R_3=R_3

  10. swissgirl
    • 2 years ago
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    \[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 & 1 & 1 \end{array} \right)\] -1/2R_3=R_3

  11. swissgirl
    • 2 years ago
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    \[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 &0 & 0 \end{array} \right)\] -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

  12. derrick902
    • 2 years ago
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    thanks :)

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