In this problem, we will see that if we use a different basis, the resulting matrix A for the transformation T in the standard basis is still the same, even though the P, M and P-1 may be different. This is what we would expect and want! It's like if one friend translated a math problem to Spanish to solve it and another friend translated it to Chinese. When they both translated the solution back to English, the solution should be the same. Here's what we'll do. Use the a different basis B* = {w, w2} 31 = { " 2 -6 }. Follow the same steps above, but with basis B*. Show that the matrix A is the same as what we determined in class (which was A = [-4/5 3/5] = 3/5 4/5 )