Let $K$ be an algebraically closed field of characteristic different from $2$ or $3$, and let $E: y^2 = x^3 +Ax+B$ and $E′: y^2 = x^3 +A′x+B′$ be elliptic curves over $K$.
Assume that $E$ and $E′$ are isomorphic via an invertible linear map $M : \mathbb{P}^2 \rightarrow \mathbb{P}^2$ that maps the infinite point $\mathcal O$ of $E$ to the infinite point $\mathcal O$ of $E'$. I want to show that that $j(E) = j(E′)$.
So what I have tho show is that $j(E) = 1728\cdot\frac{4A^3}{4A^3+27B^2} = 1728\cdot\frac{4A'^3}{4A'^3+27B'^2} = J(E')$.
My idea was to represent the linear map $M$ as follows:
$M =\begin{bmatrix}a & b & c\\d&e&f\\ g&h&i\end{bmatrix}$
Since $M * \mathcal O = M[0:1:0] = [b:e:h] = [0:1:0] = \mathcal O$ it should follow than $ b = 0, e = 1$ and $h = 0$.
However, this doesn't seem to lead anywhere, since I don't learn anything useful about the coefficients $A'$ & $B'$ which are used to calculate $j(E')$.
Am I somewhat on the right track, or is this done completely different?