\(\alpha = \left\{ \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ -1 \end{bmatrix} \right\} \beta = \left\{ \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} \right\}\)
\([x]_{\alpha} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}_{\alpha} \rightarrow x = 1. \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} + 2 \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} - 1\begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -3 \\ 0 \\ 6 \end{bmatrix}\) 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 \([x]_{\beta} = \begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}\)
Consider the following two bases for R^3
Answers (1)
Relevant Questions
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}
Let \(\displaystyle\gamma={\left\lbrace{t}^{{2}}-{t}+{1},{t}+{1},{t}^{{2}}+{1}\right\rbrace}{\quad\text{and}\quad}\beta={\left\lbrace{t}^{{2}}+{t}+{4},{4}{t}^{{2}}-{3}{t}+{2},{2}{t}^{{2}}+{3}\right\rbrace}{b}{e}{\quad\text{or}\quad}{d}{e}{r}{e}{d}{b}{a}{s}{e}{s}{f}{\quad\text{or}\quad}{P}_{{2}}{\left({R}\right)}.\) Find the change of coordinate matrix Q that changes \(\beta \text{ coordinates into } \gamma-\text{ coordinates}\)
(10%) In \(R^2\), there are two sets of coordinate systems, represented by two distinct bases: \((x_1, y_1)\) and \((x_2, y_2)\). If the equations of the same ellipse represented by the two distinct bases are described as follows, respectively: \(2(x_1)^2 -4(x_1)(y_1) + 5(y_1)^2 - 36 = 0\) and \((x_2)^2 + 6(y_2)^2 - 36 = 0.\) Find the transformation matrix between these two coordinate systems: \((x_1, y_1)\) and \((x_2, y_2)\).
Given the elow bases for \(R^2\) and the point at the specified coordinate in the standard basis as below, (40 points)
\((B1 = \left\{ (1, 0), (0, 1) \right\} \)&
\( B2 = (1, 2), (2, -1) \}\)(1, 7) = \(3^* (1, 2) - (2, 1)\)
\(B2 = (1, 1), (-1, 1) (3, 7 = 5^* (1, 1) + 2^* (-1,1)\)
\(B2 = (1, 2), (2, 1) \ \ \ (0, 3) = 2^* (1, 2) -2^* (2, 1)\)
\((8,10) = 4^* (1, 2) + 2^* (2, 1)\)
B2 = (1, 2), (-2, 1) (0, 5) =
(1, 7) =
a. Use graph technique to find the coordinate in the second basis. (10 points) b. Show that each basis is orthogonal. (5 points) c. Determine if each basis is normal. (5 points) d. Find the transition matrix from the standard basis to the alternate basis. (15 points)
