\(\text{I'm going to assume you're not at the point where you are diagonalizing matrices yet}\\ \text{So what they probably want you to do is notice that}\\ A^7 x = A^6 A x = A^6 2x = 2 A^5 Ax = 4A^5x ~\text{ etc.}\\ \text{see if you can work the whole thing through and find $Y$}\)
\((AB)^{-1}<1,2> =\\ B^{-1}A^{-1}<1,2>\)
\(A<2,-1> = <1,2>\\ <2,-1>=A^{-1}<1,2>\)
\(B<1,3>=<2,-1>\\ <1,3> = B^{-1}<2,-1>\)
\(B^{-1}A^{-1}<1,2> = \\ B^{-1}<2,-1> = \\ <1,3>\)
Encinitas here and I napped right through it.
\(\text{We're talking about the set of numbers}\\ \Large \frac{1}{5},\frac{1}{4},\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac{2}{3},\frac{3}{4},\frac{4}{5},1\)
\(\text{There are ten of them so we have to take the average of the middle two}\\ \text{This is the average of $\dfrac 1 2$ and $\dfrac 3 5$ which is $\dfrac{11}{20}$}\)
since they have to be different
1x1 = 1
1x2 = 2
1x3 = 3
1x4=4
2x2 = 4
1x5=5
1x6=6
2x3=6
1x7=7
1x8=8
46 tiles
\(\text{No dividing by zero. So }\\ x=64, ~x = \pm 8,~x = 4\\ \text{are all excluded from the domain}\)
\(\text{Rational roots will occur when the discriminant is a non-negative perfect square}\\ D=b^2 - 4ac = 25-(4)(2)(b) = 25-8b\\ \text{$D$ is a perfect square for $b=0, 2, 3$}\\ \sum b = 5\)
\(\text{Each digit can be any of 1,3,5,7, 9}\\ N=5^4 = 625\)
\(A = x y = 2x + 4y\)
Now we just have to solve this for integer values of x and y.
There are 4 integer solutions
(x,y) = (5,10), (6,6), (8,4), (12,3)
Of these (5,10) yields an area of 50 which is the maximum
did you graph it? The answers are pretty obvious if you do.