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beketso Group Title

linear combinations on vectors.help

  • 2 years ago
  • 2 years ago

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  1. beketso Group Title
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    write the matrix\[E= \left[\begin{matrix}3 & 1 \\ 1 & -1\end{matrix}\right]\]

    • 2 years ago
  2. beketso Group Title
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    as a linear combination of the matrices \[A=\left[\begin{matrix}1 & 1 \\ 1& 0\end{matrix}\right]\]

    • 2 years ago
  3. beketso Group Title
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    \[B=\left[\begin{matrix}0 & 0 \\ 1 & 1\end{matrix}\right]\]

    • 2 years ago
  4. beketso Group Title
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    \[C=\left[\begin{matrix}0 & 2 \\ 0 & -1\end{matrix}\right]\]

    • 2 years ago
  5. richyw Group Title
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    are there just three?

    • 2 years ago
  6. beketso Group Title
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    basically ,you write the matrix E as a linear combination of the matrices A,B, and C

    • 2 years ago
  7. beketso Group Title
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    @richyw ,yes there only three

    • 2 years ago
  8. richyw Group Title
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    ok so the concepts you need here are matrix addition and scalar multiplication.

    • 2 years ago
  9. richyw Group Title
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    you can solve this one in different ways but it's simple enough to do by inspection

    • 2 years ago
  10. richyw Group Title
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    so looking at E, the first thing I notice is that \(e_{11}=3\) now looking at matrices A, B, and C. The only one that has anything but 0 in that position is matrix A, so we know already that it must be 3 times matrix A. which gives. \[\left[\begin{matrix}3 & 3 \\ 3 & 0\end{matrix}\right]\] right?

    • 2 years ago
  11. beketso Group Title
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    ok i got that part

    • 2 years ago
  12. richyw Group Title
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    alright so now look at matrix E and notice that position \(e_{12}\) is 2, so we need to subtract a certain amount from matrix A to produce a 2 there.

    • 2 years ago
  13. richyw Group Title
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    well, matrix B has a zero in that position, so it is no good, so subtract one times matrix C from matrix A and we get \[3A-C=\left[\begin{matrix}3&1\\3&1\end{matrix}\right]\]

    • 2 years ago
  14. richyw Group Title
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    sorry I should have said "subtract one times matrix C from three times matrix A". Now I hope you can see what multiple of matrix B you bust subtract to get E

    • 2 years ago
  15. richyw Group Title
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    are you following?

    • 2 years ago
  16. beketso Group Title
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    yeah

    • 2 years ago
  17. richyw Group Title
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    alright cool well then you have the answer!

    • 2 years ago
  18. beketso Group Title
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    so it is 3A-2B-C=E?

    • 2 years ago
  19. richyw Group Title
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    yes!

    • 2 years ago
  20. beketso Group Title
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    thanx a lot!!!!!!!

    • 2 years ago
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