(a) find the transition matrix from B to B', (b) find the transitio

hrostentsp6 2021-11-20 Answered
(a) find the transition matrix from B to \(\displaystyle{B}'\),
(b) find the transition matrix from \(\displaystyle{B}'\) to B,
(c) verify that the two transition matriced are inverses of each other, and
(d) find the coordinate matrix \(\displaystyle{\left[{x}\right]}_{{{B}}}\), given the coordinate matrix \(\displaystyle{\left[{x}\right]}_{{{B}}}\). \(\displaystyle{B}={\left\lbrace{\left({1},{3}\right)},{\left(-{2},-{2}\right)}\right\rbrace},{B}`={\left\lbrace{\left(-{12},{0}\right)},{\left(-{4},{4}\right)}\right\rbrace}\)
\[\left[x\right]_{B'}=\begin{bmatrix}-1\\3\end{bmatrix}\]

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Expert Answer

Ancessitere
Answered 2021-11-21 Author has 492 answers
(a) Begin by forming the augmented matrix (B’B):
\[(B`B)=\left(\begin{array}{c}-12 &-4 &1 &-2 \\ 0 &4 &3 &-2\end{array}\right)\]
Using Gauss-Jordan elimination we obtain the above in the form \(\displaystyle{\left({I}{P}^{{-{1}}}\right)}\)
\[(IP^{-1})=\left(\begin{array}{c}1 &0 &-1/3 &1/3 \\0 &1 &3/4 &-1/2\end{array}\right)\]
Then, the transition matrix form B to B’ is:
\[P^{-1}=\left(\begin{array}{c}-1/3 &1/3\\ 3/4 &-1/2\end{array}\right)\]
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Jeffrey Parrish
Answered 2021-11-22 Author has 151 answers
(b) First form the matrix (BB’)
\[(BB`)=\left(\begin{array}{c}1 &-2 &-12 &-4\\ 3 &-2 &0 &4\end{array}\right)\]
By Gauss-lordan process we wrotte:
\[(IP)=\left(\begin{array}{c}1 &0 &6 &4\\ 0 &1 &9 &4\end{array}\right)\]
The transition matrix from B’ to B is then:
\[(P)=\left(\begin{array}{c}6 &4\\ 9 &4\end{array}\right)\]
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user_27qwe
Answered 2021-11-24 Author has 1923 answers

(c) Calculating \(P^{-1}P\) and \(PP^{-1}\) we obtain:

\(P^{-1}P=\left(\begin{array}{c}-1/3 &1/3\\ 3/4 &-1/2\end{array}\right)\left(\begin{array}{c}6 &4\\ 9 &4\end{array}\right)=\left(\begin{array}{c}1 &0\\ 0 &1\end{array}\right)=I\)

\(P^{-1}=\left(\begin{array}{c}6 &4\\ 9 &4\end{array}\right)\left(\begin{array}{c}-1/3 &1/3\\ 3/4 &-1/2\end{array}\right)=\left(\begin{array}{c}1 &0\\ 0 &1\end{array}\right)\)

Then \(P^{-1}\) the inverse of P.

(d) We have in the basis B

\((x)_{B}=P(x)_{B`}=\left(\begin{array}{c}6 &4\\ 9 &4\end{array}\right)\left(\begin{array}{c}-1\\ 3\end{array}\right)=\left(\begin{array}{c}6\\ 3\end{array}\right)\)

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