Question

Consider the following two bases for R^3

Alternate coordinate systems
ANSWERED
asked 2020-11-02

Consider the following two bases for \(\displaystyle{R}^{{3}}\) :
\(\alpha := \left\{ \begin{bmatrix} 2 \\ 1\\ 3 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix}3 \\ 1 \\ -1 \end{bmatrix} \right\} and\ \beta := \left\{ \begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix}-2 \\ 3\\ 1 \end{bmatrix}, \begin{bmatrix}2 \\ 3\\ -1 \end{bmatrix} \right\}\) If \([x]_{\alpha} = \begin{bmatrix}1 \\ 2 \\-1 \end{bmatrix}_{\alpha} then\ find\ [x]_{\beta}\)
(that is, express x in the \(\displaystyle\beta\) coordinates).

Answers (1)

2020-11-03

\(\alpha = \left\{ \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ -1 \end{bmatrix} \right\} \beta = \left\{ \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} \right\}\)
\([x]_{\alpha} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}_{\alpha} \rightarrow x = 1. \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} + 2 \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} - 1\begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -3 \\ 0 \\ 6 \end{bmatrix}\) Let \(x = C_1 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + C_2 \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix} + C_3 \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} \Rightarrow \begin{array}{c}C_1 - 2C_2 + 2C_3 = -3 - (1)\\ C_1 + 3C+2 + 3C_3 = 0 - (2) \\ C_1 + C_2 - C_3 = 6 - (3)\end{array}\) From (3) \(\displaystyle{C}_{{3}}={C}_{{1}}+{C}_{{2}}-{6}\) From (1) \(\displaystyle{C}_{{1}}-{2}{C}_{{2}}+{2}{C}_{{1}}+{2}{C}_{{2}}-{12}=-{3}\Rightarrow{3}{C}_{{1}}=+{9}\Rightarrow{C}_{{1}}={3}\) From (2) \(\displaystyle{C}_{{1}}+{3}{C}_{{2}}+{3}{C}_{{1}}+{3}{C}_{{2}}-{18}={0}\Rightarrow{4}{C}_{{1}}+{6}{C}_{{2}}={18}\Rightarrow{6}{C}_{{2}}={C}_{{2}}\)
\(\displaystyle{C}_{{3}}={3}+{1}-{6}=-{2}\) So \(\begin{bmatrix} -3 \\ 0 \\ 6 \end{bmatrix} = + 3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + 1 \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix} - 2 \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}\) So \([x]_{\beta} = \begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}\)

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