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

Let

$$P = \left\lbrack \begin{array}{cc} 2t^{-2} & 1-2t^{-1}+2t^{-2} \\ -2t^{-2} & 2t^{-1}-2t^{-2} \end{array} \right\rbrack,$$ $$\boldsymbol{\vec{y}_1}(t) = \left\lbrack \begin{array}{c} t^{2}-2t \\ 2t \end{array} \right\rbrack, \ \ \ \boldsymbol{\vec{y}_2}(t) = \left\lbrack \begin{array}{c} t-1 \\ 1 \end{array} \right\rbrack.$$

1. Show that $\boldsymbol{\vec{y}_1}(t)$ is a solution to the system $\boldsymbol{\vec{y}^{\,\prime}} = P \boldsymbol{\vec{y}}$ by evaluating derivatives and the matrix product

   $$\boldsymbol{\vec{y}_1^{\,\prime}}(t) = \left\lbrack \begin{array}{cc} 2t^{-2} & 1-2t^{-1}+2t^{-2} \\ -2t^{-2} & 2t^{-1}-2t^{-2} \end{array} \right\rbrack \boldsymbol{\vec{y}_1}(t)$$
   Enter your answers in terms of the variable $t$.

   $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

   =

   $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

   
   
   
2. Show that $\boldsymbol{\vec{y}_2}(t)$ is a solution to the system $\boldsymbol{\vec{y}^{\,\prime}} = P \boldsymbol{\vec{y}}$ by evaluating derivatives and the matrix product

   $$\boldsymbol{\vec{y}_2^{\,\prime}}(t) = \left\lbrack \begin{array}{cc} 2t^{-2} & 1-2t^{-1}+2t^{-2} \\ -2t^{-2} & 2t^{-1}-2t^{-2} \end{array} \right\rbrack \boldsymbol{\vec{y}_2}(t)$$
   Enter your answers in terms of the variable $t$.

   $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

   =

   $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$