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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 =

Calculus1
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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?
Yes, i transposed A^t= |dw:1374263685200:dw|
Looks good, so now I would multiply this matrix by the original a matrix.

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the problem i have is knowing which number to multiply with which and also last at c
A good rule of thumb is to multiply scalars at the end. The scalars in this case are -5 and 6
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?
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
i swapped the negatives on the 2's. sorry about that
oh okay. so basically I multiply.. row and columns?
Sort of. Matrix multiplication works like this: |dw:1374264402007:dw|
do you understand what you have to do?
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
after? not really
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
Just do what you did to get 29, exception this time multiply the first row by the second column

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