I n b a s i s { 1 , t , t 2 } p = [ 5 − 1 3 ] T r a n s i t i o n m a t r i x T = [ 1 2 3 − 1 0 0 3 0 1 ] T h e c o o r d i n a t e s p ′ = T − 1 p = [ 1 2 3 − 1 0 0 3 0 1 ] − 1 [ 5 − 1 3 ] = [ 0 − 1 0 0.5 − 4 − 1.5 0 3 1 ] [ 5 − 1 3 ] = [ 1 2 0 ] In\,\,basis\,\,\left\{ 1,t,t^2 \right\} \\p=\left[ \begin{array}{c} 5\\ -1\\ 3\\\end{array} \right] \\Transition\,\,matrix\\T=\left[ \begin{matrix} 1& 2& 3\\ -1& 0& 0\\ 3& 0& 1\\\end{matrix} \right] \\The\,\,coordinates\\p'=T^{-1}p=\left[ \begin{matrix} 1& 2& 3\\ -1& 0& 0\\ 3& 0& 1\\\end{matrix} \right] ^{-1}\left[ \begin{array}{c} 5\\ -1\\ 3\\\end{array} \right] =\left[ \begin{matrix} 0& -1& 0\\ 0.5& -4& -1.5\\ 0& 3& 1\\\end{matrix} \right] \left[ \begin{array}{c} 5\\ -1\\ 3\\\end{array} \right] =\left[ \begin{array}{c} 1\\ 2\\ 0\\\end{array} \right] \\ I n ba s i s { 1 , t , t 2 } p = ⎣ ⎡ 5 − 1 3 ⎦ ⎤ T r an s i t i o n ma t r i x T = ⎣ ⎡ 1 − 1 3 2 0 0 3 0 1 ⎦ ⎤ T h e coor d ina t es p ′ = T − 1 p = ⎣ ⎡ 1 − 1 3 2 0 0 3 0 1 ⎦ ⎤ − 1 ⎣ ⎡ 5 − 1 3 ⎦ ⎤ = ⎣ ⎡ 0 0.5 0 − 1 − 4 3 0 − 1.5 1 ⎦ ⎤ ⎣ ⎡ 5 − 1 3 ⎦ ⎤ = ⎣ ⎡ 1 2 0 ⎦ ⎤
I find the inverse matrix with Gauss method:
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