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

Question
Alternate coordinate systems

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.

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 $$[T]_{C, B} = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$$

### Relevant Questions

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 R^2. Find the change-of-coordinate matrix from B to C.

Consider the bases $$B = \left(\begin{array}{c}\begin{bmatrix}2 \\ 3 \end{bmatrix}, \begin{bmatrix}3 \\ 5 \end{bmatrix}\end{array}\right) of R^2 \ and\ C = \left(\begin{array}{c}\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix}1\\0 \\ 1 \end{bmatrix}\end{array}, \begin{bmatrix}0 \\ 1\\1 \end{bmatrix}\right) of R^3$$.
and the linear maps $$S \in L (R^2, R^3) \ and\ T \in L(R^3, R^2)$$ given given (with respect to the standard bases) by $$[S]_{E, E} = \begin{bmatrix}2 & -1 \\ 5 & -3\\ -3 & 2 \end{bmatrix} \ and\ [T]_{E, E} = \begin{bmatrix}1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}$$ Find each of the following coordinate representations. $$\displaystyle{\left({b}\right)}{\left[{S}\right]}_{{{E},{C}}}$$
$$\displaystyle{\left({c}\right)}{\left[{S}\right]}_{{{B},{C}}}$$

Consider the bases $$B=(\begin{bmatrix}2 \\3 \end{bmatrix},\begin{bmatrix}3\\5 \end{bmatrix}) \ of \ r^{2} \ and \ E \ (\begin{bmatrix}1 \\1\\0 \end{bmatrix}, \begin{bmatrix}1 \\0\\1 \end{bmatrix}, \begin{bmatrix}0 \\0\\1 \end{bmatrix}) of \ R^{3}$$.
and the linear maps $$S\in \alpha (R^{2},R^{3}) \ and \ T \in\alpha (R^{3},R^{2})$$ given given (with respect to the standard bases) by $$[S]_{\epsilon,\epsilon}\begin{bmatrix}2 & -1 \\5 & -3\\-3 & 2 \end{bmatrix} \ and \ [T]_{\epsilon,\epsilon}\begin{bmatrix}1 & -1 & 1 \\1 & 1 & -1\end{bmatrix}.$$ Find each of the following coordinate representations. $$a) [S]_{\beta,\epsilon}, b) [S]_{\epsilon, E}, c) [S]_{\beta, E}$$

To solve:
$$\displaystyle{\left(\begin{matrix}{x}-{2}{y}={2}\\{2}{x}+{3}{y}={11}\\{y}-{4}{z}=-{7}\end{matrix}\right)}$$

The reason ehy the point $$(-1, \frac{3\pi}{2})$$ lies on the polar graph $$r=1+\cos \theta$$ even though it does not satisfy the equation.

The quadratic function $$\displaystyle{y}={a}{x}^{2}+{b}{x}+{c}$$ whose graph passes through the points (1, 4), (2, 1) and (3, 4).

The system of equation \begin{cases}2x + y = 1\\4x +2y = 3\end{cases} by graphing method and if the system has no solution then the solution is inconsistent. Given: The linear equations is \begin{cases}2x + y = 1\\4x +2y = 3\end{cases}

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}

The coordinate vector of $$\displaystyle{\left[{\mathbf{{{p}}}}\ {\left({t}\right)}\ =\ {6}\ +\ {3}{t}\ -{t}^{{{2}}}\right]}$$ relative to the basis $$\displaystyle \mathscr{\left\lbrace{B}\right\rbrace}\ ={\left\lbrace{1}\ +\ {t},\ {1}\ +\ {t}^{{{2}}},\ {t}\ +\ {t}^{{{2}}}\right\rbrace}$$

To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}$$

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