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mathcalculus Group Title

HELP MATRICES: If a= [-4,-3, 2; 0, -5,-2] (2x3 dimension matrix) c= [ 7,-1; -2,-7] (2x2 dimension matrix) then -5AA^t+6c =

  • one year ago
  • one year ago

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  1. blurbendy Group Title
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    I'm assuming A is referring to a? If so, the first thing I would do is take the transpose of matrix a, since you have A^t in your equation. Are you familiar with how to do that?

    • one year ago
  2. mathcalculus Group Title
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    Yes, i transposed A^t= |dw:1374263685200:dw|

    • one year ago
  3. blurbendy Group Title
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    Looks good, so now I would multiply this matrix by the original a matrix.

    • one year ago
  4. mathcalculus Group Title
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    the problem i have is knowing which number to multiply with which and also last at c

    • one year ago
  5. blurbendy Group Title
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    A good rule of thumb is to multiply scalars at the end. The scalars in this case are -5 and 6

    • one year ago
  6. mathcalculus Group Title
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    i did this: -5(-4) (-4)+6c -5(-3)(-3)+6c -5(2) (2)+6c -5(0)(0)+6c -5(-5)(-5)+6c -5(-2)(-2)+6c and also, how do i know what dimension the new matrix will look like if a is a 2x3 dimensions and c is a 2x2 dimension? doesn't that say not allowed to be multiplied? unless the middle numbers are the same?

    • one year ago
  7. blurbendy Group Title
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    Hold on, you have to actually multiply matrix a by a^t, so you'd have: |dw:1374264066231:dw| this is a 2 x 3 matrix * 3 x 2 matrix, which will give you a 2 x 2 matrix. Then you would multiply each value in your 2 x 2 matrix by -5

    • one year ago
  8. blurbendy Group Title
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    i swapped the negatives on the 2's. sorry about that

    • one year ago
  9. mathcalculus Group Title
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    oh okay. so basically I multiply.. row and columns?

    • one year ago
  10. blurbendy Group Title
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    Sort of. Matrix multiplication works like this: |dw:1374264402007:dw|

    • one year ago
  11. blurbendy Group Title
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    do you understand what you have to do?

    • one year ago
  12. blurbendy Group Title
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    To calculate the position next to 29 in the new matrix, you would multiply the first row in a by the SECOND column in a^t

    • one year ago
  13. mathcalculus Group Title
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    after? not really

    • one year ago
  14. blurbendy Group Title
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    you need the next position in the matrix (next to 29), so to calculate the position next to 29 in the new matrix, you would multiply the first row in a by the SECOND column in a^t

    • one year ago
  15. blurbendy Group Title
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    Just do what you did to get 29, exception this time multiply the first row by the second column

    • one year ago
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