I'm really stumped on this question, specifically regarding how to incorporate $T$ into the change of basis formula.
Let $$ \Gamma = \left( \begin{bmatrix} 1 \\\\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\\\ 1 \end{bmatrix} \right) \quad \text{and} \quad \Omega = \left( \begin{bmatrix} 1 \\\\ -1 \end{bmatrix}, \begin{bmatrix} 1 \\\\ 0 \end{bmatrix} \right)$$ be ordered bases for $\mathbb{R}^2$. Let $T$ be a linear transformation such that $$ T \left( \begin{bmatrix} x_1 \\\\ x_2 \end{bmatrix} \right) = \begin{bmatrix} -x_1 + 3x_2 \\\\ x_1 - x_2 \end{bmatrix}. $$ Then $$ [T]_\Gamma^\Omega = \begin{bmatrix} a & -1 \\\\ 2 & b \end{bmatrix}. $$ What is $a + b$?
We note that $$T(1,0) = (-1,1) = -1 \cdot (1,-1) + 0 \cdot (1,0) \text{ and }$$ $$T(0,1) = (3,-1) = 1 \cdot (1,-1) + 2 \cdot (1,0).$$ We have therefore that $$[T]_\Gamma^\Omega = \begin{pmatrix} -1 & 1 \\ \phantom{-} 0 & 2 \end{pmatrix}$$ so that $a + b = 1 + 0 = 1.$