Another one I'm not sure on is this one:

Evaluate

{1 5}{1 2}

{2 3}{0 1}

Possible solutions:

{1 10}

{0 3 }

-------

{1 10}

{0 7}

-------

{1 7}

{5 7}

-------

{1 7}

{2 7}

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- Jan 24th 2007, 09:20 PMgretchenEvaluate
Another one I'm not sure on is this one:

Evaluate

{1 5}{1 2}

{2 3}{0 1}

Possible solutions:

{1 10}

{0 3 }

-------

{1 10}

{0 7}

-------

{1 7}

{5 7}

-------

{1 7}

{2 7} - Jan 24th 2007, 11:08 PMCaptainBlack
- Jan 25th 2007, 02:24 AMgretchen
Thanks Ron, you have helped me so much, if only you knew. Because you show me how you get the answer, I am getting the other problems like them correct.

I have two other types of evaluations if you could help me out:

First one is similar to the one before but it takes a 3x3 matrix:

Q: Evaluate

{3 2 1}{2}

{3 1 0}{2}

{0 -2 1}{2}

Solutions:

{8}

{6}

{1}

-----

{12}

{8}

{-2}

-----

{12}

{8}

{1}

-----

{12}

{6}

{-2}

Second one is finding the inverse:

Q: Find A[-1 power] where A=

{2 4}

{2 5}

Solutions:

{5 -2}

{-4 2}

-----

{5 -4}

{-2 2}

-----

{1 -2}

{-1 2.5}

-----

{2.5 -2}

{-1 1} - Jan 25th 2007, 07:39 AMCaptainBlack
- Jan 25th 2007, 07:45 AMCaptainBlack
Given that you are given a list of options the easiest way to do this is

to multiply each of the candidates by the original matrix.

Start with the first row first column.

Case 1:

[2, 4][5,-4]' = 2.5-4.4 = -6 != 1, so not this one.

Case 2:

[2, 4][5,-2]' = 2.5-4.2 = -2 != 1, so not this one.

Case 3:

[2, 4][1,-1]' = 2.1-4.1 = -2 != 1, so not this one.

Case 4:

[2, 4][2.5,-1]' = 2.(2.5)-4.1 = 1 != 1, so this is the one, now just

check the other elements of the product are the elements of the

identity matrix.

RonL