1. ## Evaluate

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}

2. Originally Posted by gretchen
Another one I'm not sure on is this one:

Evaluate
{1 5}{1 2}
{2 3}{0 1}
first row first column (1,5)(1,0)' = 1.1+5.0=1
first row second column = (1,5)(2,1)'=1.2+5.1=7

second row first column = (2,3)(1,0)'=2.1+3.0=2
second row second colum = (2,3)(2,1)'=2.2+3.1=7

so the matrix product is:

[ 1 7 ]
[ 2 7 ]

RonL

3. 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}

4. Originally Posted by gretchen
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}
First element is [3, 2, 1][2, 2, 2]' = 3.2 + 2.2 + 1.2 = 12
Second element is [3, 1, 0][2, 2, 2]' = 3.2 + 1.2 + 0.2 = 8
Third element is [0, -2, 1][2, 2, 2]' = 0.2 + (-2).2 + 1.2 = -2

[12]
[ 8]
[-2]

RonL

5. Originally Posted by gretchen
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}
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.

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