Question

Let B and C be the following ordered bases of R^3

Alternate coordinate systems
ANSWERED
asked 2020-10-25

Let B and C be the following ordered bases of \(\displaystyle{R}^{{3}}:\)
\(B = (\begin{bmatrix}1 \\ 4 \\ -\frac{4}{3} \end{bmatrix},\begin{bmatrix}0 \\ 1 \\ 8 \end{bmatrix},\begin{bmatrix}1 \\ 1 \\ -2 \end{bmatrix})\)
\(C = (\begin{bmatrix}1 \\ 1 \\ -2 \end{bmatrix}, \begin{bmatrix}1 \\ 4 \\ -\frac{4}{3} \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 8 \end{bmatrix})\) Find the change of coordinate matrix I_{CB}

Answers (1)

2020-10-26

\(B = (\begin{bmatrix}1 \\ 4 \\ -\frac{4}{3} \end{bmatrix},\begin{bmatrix}0 \\ 1 \\ 8 \end{bmatrix},\begin{bmatrix}1 \\ 1 \\ -2 \end{bmatrix})\)
\(C = (\begin{bmatrix}1 \\ 1 \\ -2 \end{bmatrix}, \begin{bmatrix}1 \\ 4 \\ -\frac{4}{3} \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 8 \end{bmatrix})\)
\(\begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 1 \\ 1 & 4 & 1 & 4 & 1 & 1 \\ -2 & - \frac{4}{3} & 8 & - \frac{4}{3} & 8 & -2 \end{bmatrix}\)
\(\begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & \frac{1}{3} & 1 & \frac{1}{3} & 0 \\ 0 & - \frac{2}{3} & 8 & - \frac{2}{3} & 8 & 0 \end{bmatrix} \begin{array}{}-R_{1}+R_{2}\rightarrow R_{2}\\ 2R_{1}+R_{3}\rightarrow R_{3}\\ R_{2}\times\frac{1}{3}\rightarrow\frac{R_{2}}{3}\end{array}\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & \frac{1}{3} & 1 & \frac{1}{3} & 0 \\ 0 &0 & 1 & 0 & 1 & 0 \end{bmatrix} \begin{array}{} -\frac{2}{3} R_{2}+R_{3}\rightarrow R_{3}\\ R_{3} \times \frac{9}{70}\rightarrow\frac{R_{9 R_3}}{70}\end{array}\)
\(\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \end{bmatrix} \begin{array}{} -\frac{1}{3} R_{3}+R_{2}\rightarrow R_{2}\\ - R_{2} + R_1 \rightarrow R_1\end{array}\)
\(I_{CB} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\)

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