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}
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