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

(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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Ancessitere
(a) Begin by forming the augmented matrix (B’B):
$(BB)=\left(\begin{array}{c}-12 &-4 &1 &-2 \\ 0 &4 &3 &-2\end{array}\right)$
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
(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:
$(P)=\left(\begin{array}{c}6 &4\\ 9 &4\end{array}\right)$
user_27qwe

(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

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