Let b = {(1, 3), (−2, −2)} and b' = {(−12, 0), (−4, 4)} be bases for r2, and let a = 0 2 3 4 be the matrix for t: r2 → r2 relative to b.

(a) find the transition matrix p from b' to b. p =

(b) use the matrices p and a to find [v]b and [t(v)]b, where [v]b' = [−2 4]t. [v]b = [t(v)]b =

(c) find p−1 and a' (the matrix for t relative to b'). p−1 = a' = (

(d) find [t(v)]b' two ways. [t(v)]b' = p−1[t(v)]b = [t(v)]b' = a'[v]b' =

Step-by-step explanation:

a) Let M =

1          -2       -12      -4

3         -2         0       4

The RREF of M is

1       0        6        4

0       1        9        4

Hence, the transition matrix P from B' to B is P =

6       4

9        4

(b).

Since [v]B’ = (2  -1)T, hence [v]B = P[v]B’ = (8,14)T.

( c).

Let N = [P|I2] =

6       4        1        0

9       4        0        1

The RREF of N is

1        0        -1/3            1/3

0        1         ¾             -1/2

Hence, P^-1 =

-1/3            1/3

¾             -1/2

Also, A’ = PA =

12          28

12          34

(d). [T(v)]B’ = A’[v]B’ = (-4,10)T

answer: all the values are within because it has a pattern that repeats from -6 to 4.

good luck!