I would not use matrices at all. The third equation, \(\displaystyle y_3'= 3y_3\), involves only \(\displaystyle y_3\) and can be integrated directly: \(\displaystyle y_3(t)= c_3e^{3t}\).

The second equation is \(\displaystyle y_2'= 3y_2+ y_3= 3y_2+ c_3e^{3t}\). That is a "non-homogeneous equation with constant equations". We can first solve the "associated homogeneous equation", \(\displaystyle y_2'= 3y_2\). The general solution to that is \(\displaystyle y= c_2e^{3t}\) just as before. We also seek a single solution to the entire equation. Normally, since the "non-homogeneous" part is \(\displaystyle c_3e^{3t}\) we would try a multiple of that but that is already a solution to the homogeneous equation so we try \(\displaystyle y= Ac_3te^{3t}\). Then \(\displaystyle y'= Ac_3e^{3t}+ 3Ac_3te^{3t}\) so the equation becomes \(\displaystyle Ac_3e^{3t}+ 3Ac_3te^{3t}= Ac_3te^{3t}+ c_3e^{3t}\). That reduces to \(\displaystyle Ac_3e^{3t}= c_3e^{3t}\) so A= 1.

\(\displaystyle y_2= c_2e^{3t}+ c_3te^{3t}\).

Now, the first equation is \(\displaystyle y_1'= 2y_1+ y_2- y_3= 2y_1+ c_2e^{3t}+ 2c_3te^{3t}- c_3e^{3t}\). The associated homogeneous equation is \(\displaystyle y_1'= 2y_1\) which has general solution \(\displaystyle y= c_1e^{2t}\). To find a solution to the entire equation try \(\displaystyle y= (Ax+ B)e^{3t}\) and find values for A and B that make that true.