Let B = left{ begin{bmatrix} 1 -2 end{bmatrix}

Let B = left{ begin{bmatrix} 1 -2 end{bmatrix}

Question
Alternate coordinate systems
asked 2021-01-19

Let \(B = \left\{ \begin{bmatrix} 1 \\ -2 \end{bmatrix}, \begin{bmatrix}2 \\1 \end{bmatrix} \right\} \ and\ C = \left\{ \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix}1 \\3 \end{bmatrix} \right\}\)
be bases for\(\displaystyle{R}^{{2}}.\) Change-of-coordinate matrix from C to B.

Answers (1)

2021-01-20

Given that \(B = \left\{ \begin{bmatrix} 1 \\ -2 \end{bmatrix}, \begin{bmatrix}2 \\1 \end{bmatrix} \right\} and C = \left\{ \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix}1 \\3 \end{bmatrix} \right\}\) Let x, y are scalare such that \(\begin{bmatrix} 2 \\ 1 \end{bmatrix} = x \begin{bmatrix} 1 \\ -2 \end{bmatrix} + y \begin{bmatrix} 2 \\ 1 \end{bmatrix}\) clearly \(\displaystyle{x}={1},{y}={0}\)
\(\begin{bmatrix} 2 \\ 1 \end{bmatrix} = 0 \begin{bmatrix} 1 \\ -2 \end{bmatrix} + 1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} \rightarrow (c)\) and \(\begin{bmatrix} 1 \\ 3 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ -2 \end{bmatrix} + y \begin{bmatrix} 2 \\ 1 \end{bmatrix}\)
\(= 1 \begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} x + 2y \\ -2x + y \end{bmatrix}\)
\(\displaystyle{x}+{2}{y}={1},-{2}{x}+{y}={3}\) Solving above equation \(\begin{array}{c} 2\not{x} + 4y = 2\\ -\not{2}x + y = 3 \\ \hline 5y = 5 \Rightarrow y = 1\end{array}\) pul \(\displaystyle-{y}={1}\in{x}+{2}{y}={1}\)
\(\displaystyle\Rightarrow{x}+{2}={1}\)
\(\displaystyle\Rightarrow{x}=-{1}\)
\(\therefore \begin{bmatrix} 1 \\ 3 \end{bmatrix} = (-1)\begin{bmatrix} 1 \\ -2 \end{bmatrix} + 1\begin{bmatrix} 2 \\ 1\end{bmatrix} \rightarrow (d)\) from c, d Transition matrix of C over D \([T]_{C, B} = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}\)

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