This is the third part of a four-part problem.

If the given solutions

$$\boldsymbol{\vec{y}_1}(t) = \left\lbrack \begin{array}{c} t^{2}-2t \\ 2t \end{array} \right\rbrack, \ \ \ \boldsymbol{\vec{y}_2}(t) = \left\lbrack \begin{array}{c} t-1 \\ 1 \end{array} \right\rbrack.$$
form a fundamental set (i.e., linearly independent set) of solutions for the system

$$\boldsymbol{\vec{y}^{\,\prime}} = \left\lbrack \begin{array}{cc} 2t^{-2} & 1-2t^{-1}+2t^{-2} \\ -2t^{-2} & 2t^{-1}-2t^{-2} \end{array} \right\rbrack \boldsymbol{\vec{y}},$$
state the general solution of the linear homogeneous system, and if they do not, enter NONE in all of the answer blanks. Express the general solution as the product $\boldsymbol{\vec{y}}(t) = \Psi(t) \boldsymbol{\vec{c}}$, where $\Psi(t)$ is a square matrix whose columns are the solutions forming the fundamental set and $\boldsymbol{\vec{c}}$ is a column vector of arbitrary constants.