You need to find the two other rotation matrices using the exact same method.
The standard matrix for T is found by simply multiplying the three matrices together.
Let \lambda be the linear charge distribution in the wire.
Use Gauss's Law to express the electric field in terms of r and \lambda.
Then use this:
to determine the expression of \lambda as a function of V.
Your last determinant should be
\left|\begin{array}{cc} -18 & 1\\ -50 &9\end{array}\right|
not
\left|\begin{array}{cc} -18 & 0\\ -50 &9\end{array}\right|
In other words, we have \ker(A)\subset \ker(A^2).
Yes. You could also write : \operatorname{im}(A^2)\subset \operatorname{im}(A)
Now, how can you apply this to \ker(A^3), \ker(A^4)... and \operatorname{im}(A^3), \operatorname{im}(A^4)...?
a. If x \in \ker(A), what can you say about A^2 x? What does that tell you about \ker(A^2)?
b. For any x, A^2x is in \operatorname{im}(A^2). If you rewrite A^2x as A(Ax)=Ay, what can you say about \operatorname{im}(A)?
\int \frac{{\rm d}u}{\sqrt{1-u^2}}=\arcsin u
but here we have
\int \frac{{\rm d}x}{\sqrt{1-\left(\frac x 3 \right)^2}}
u= \frac x 3 \Rightarrow {\rm d}x= ?