## mathcalculus 2 years ago 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 =

1. blurbendy

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?

2. mathcalculus

Yes, i transposed A^t= |dw:1374263685200:dw|

3. blurbendy

Looks good, so now I would multiply this matrix by the original a matrix.

4. mathcalculus

the problem i have is knowing which number to multiply with which and also last at c

5. blurbendy

A good rule of thumb is to multiply scalars at the end. The scalars in this case are -5 and 6

6. mathcalculus

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?

7. blurbendy

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

8. blurbendy

i swapped the negatives on the 2's. sorry about that

9. mathcalculus

oh okay. so basically I multiply.. row and columns?

10. blurbendy

Sort of. Matrix multiplication works like this: |dw:1374264402007:dw|

11. blurbendy

do you understand what you have to do?

12. blurbendy

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

13. mathcalculus

after? not really

14. blurbendy

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

15. blurbendy

Just do what you did to get 29, exception this time multiply the first row by the second column