Assume $\hat{M}_1, \hat{M}_2, \hat{T}_{11}, \hat{T}_{12}, \hat{T}_{21}, \hat{T}_{22}$ are $2\times 2$ matrix. And $a, b, A, B, C, D$ are all numbers, satisfying the following relation:
\begin{align} \left( \begin{array}{cc} \hat{M}_1 \\ \hat{M}_2 \end{array} \right) \left( \begin{array}{cc} a \\ b \end{array} \right) = \left( \begin{array}{cc} \hat{T}_{11} & \hat{T}_{12} \\ \hat{T}_{21} & \hat{T}_{22} \end{array} \right) \left( \begin{array}{cc} A \\ B\\ C \\ D \end{array} \right). \end{align}
If we define $\begin{align} \left( \begin{array}{cc} A \\ B \end{array} \right) = \hat{S} \left( \begin{array}{cc} a \\ b \end{array} \right) \end{align}$, where S is a $2\times2$ matrix. There is definitely one equation between $\hat{S}$ and $\hat{M}_i$, $\hat{T}_{ij}$.
But my question here is: does the expression of S have any relation with Det($\check{M}$). \begin{align} \check{M}\equiv\left( \begin{array}{cc} \hat{T}_{11} & \hat{T}_{12} \\ \hat{T}_{21} & \hat{T}_{22} \end{array} \right). \end{align}
Thanks anyway, I've solved the problem.
If we redefine $\check{M}\equiv[\phi_1,\phi_2,\phi_3,\phi_4]$, where $\phi_i$ is $4\times1$ matrix. And $\left( \begin{array}{cc} \hat{M}_1 \\ \hat{M}_2 \end{array} \right) \equiv [\psi_1,\psi_2]$.
According to Cramer's rule, \begin{align*} A&=\frac{Det[(a\psi_1+b\psi_2,\phi_2,\phi_3,\phi_4)]}{Det[\check{M}]}\\ &= a\frac{Det[(\psi_1,\phi_2,\phi_3,\phi_4)]}{Det[\check{M}]} +b\frac{Det[(\psi_2,\phi_2,\phi_3,\phi_4)]}{Det[\check{M}]}. \end{align*}
So $\hat{S}_{11} = \frac{Det[(\psi_1,\phi_2,\phi_3,\phi_4)]}{Det[\check{M}]}$. Same way, we can derive out the expression of $\hat{S}$.