- #1
stunner5000pt
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This is how the question appears in my textbook
Find the matrix of T corresponding to the bases B and D and use it to compute [itex] C_{D}[T(v)] [/itex] and hence T(v)
T; P2 - > R2
T(a + bx + cx^2) = (a+b,c)
B={1,x,x^2}
D={(1,-1),(1,1)}
v = a + bx + cx^2
ok i cna find Cd no problem it is
[tex] C_{D}[T(v)] = \frac{1}{2} \left(\begin{array}{ccc} 1&1&-1 \\ 1&1&1 \end{array}\right) [/tex]
now am i supposed to solve for X where
[tex] C_{D}[T(v)] X = T(v) [/tex]
but textbook doesn't do that...
it does [tex] C_{D}[T(v)] \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = X [/tex]
It doesn't make sense... doesn't the question ask to compute Cd[T(v)] and T(v) from it??
Next question
Verify this theorem for the given transformation and uses the standard basis in Rn
Theorem: [itex] M_{EB} (ST) = M_{ED} (S) \bullet M_{DB} (T) [/itex]
T;R3 -> R4
S: R4->R2
T(a,b,c) = (a+b,b+c,c+a,b-a)
S(a,b,c,d) = (a+b,c-d)
[tex] M_{EB} (S) = \left[C_{D}\left(\begin{array}{c} 1 \\ 0 \end{array}\right) \ C_{D}\left(\begin{array}{c} 1 \\ 0 \end{array}\right) \ C_{D}\left(\begin{array}{c} 0 \\ 1 \end{array}\right) \
C_{D}\left(\begin{array}{c} 0 \\ -1 \end{array}\right)\right] [/tex]
[tex] M_{EB} (S) = \left[\begin{array}{cc} 1&0 \\ 1&0 \\ 0&1 \\ 0&-1 \end{array}\right] [/tex]
[tex] M_{DB} (T) = \left[C_{D}\left(\begin{array}{c} 1 \\ 0 \\1 \\ -1 \end{array}\right) \ C_{D}\left(\begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \end{array}\right) \ C_{D}\left(\begin{array}{c} 0 \\ 1 \\ 1 \\ 0 \end{array}\right)\right] [/tex]
[tex] M_{DB} (T) = \left[\begin{array}{cccc} 1&0&1&-1 \\ 1&1&0&1 \\ 0&1&1&0 \end{array}\right] [/tex]
the dimensions of the matrices are not correct though..
is there something wrong in the way i am forming the matrices?
Find the matrix of T corresponding to the bases B and D and use it to compute [itex] C_{D}[T(v)] [/itex] and hence T(v)
T; P2 - > R2
T(a + bx + cx^2) = (a+b,c)
B={1,x,x^2}
D={(1,-1),(1,1)}
v = a + bx + cx^2
ok i cna find Cd no problem it is
[tex] C_{D}[T(v)] = \frac{1}{2} \left(\begin{array}{ccc} 1&1&-1 \\ 1&1&1 \end{array}\right) [/tex]
now am i supposed to solve for X where
[tex] C_{D}[T(v)] X = T(v) [/tex]
but textbook doesn't do that...
it does [tex] C_{D}[T(v)] \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = X [/tex]
It doesn't make sense... doesn't the question ask to compute Cd[T(v)] and T(v) from it??
Next question
Verify this theorem for the given transformation and uses the standard basis in Rn
Theorem: [itex] M_{EB} (ST) = M_{ED} (S) \bullet M_{DB} (T) [/itex]
T;R3 -> R4
S: R4->R2
T(a,b,c) = (a+b,b+c,c+a,b-a)
S(a,b,c,d) = (a+b,c-d)
[tex] M_{EB} (S) = \left[C_{D}\left(\begin{array}{c} 1 \\ 0 \end{array}\right) \ C_{D}\left(\begin{array}{c} 1 \\ 0 \end{array}\right) \ C_{D}\left(\begin{array}{c} 0 \\ 1 \end{array}\right) \
C_{D}\left(\begin{array}{c} 0 \\ -1 \end{array}\right)\right] [/tex]
[tex] M_{EB} (S) = \left[\begin{array}{cc} 1&0 \\ 1&0 \\ 0&1 \\ 0&-1 \end{array}\right] [/tex]
[tex] M_{DB} (T) = \left[C_{D}\left(\begin{array}{c} 1 \\ 0 \\1 \\ -1 \end{array}\right) \ C_{D}\left(\begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \end{array}\right) \ C_{D}\left(\begin{array}{c} 0 \\ 1 \\ 1 \\ 0 \end{array}\right)\right] [/tex]
[tex] M_{DB} (T) = \left[\begin{array}{cccc} 1&0&1&-1 \\ 1&1&0&1 \\ 0&1&1&0 \end{array}\right] [/tex]
the dimensions of the matrices are not correct though..
is there something wrong in the way i am forming the matrices?
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