# \displaystyle{b}{e}{g}\in{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}

Question
Alternate coordinate systems
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}

2021-02-28
Calculation:
The first equation is given below.
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{y}\right\rbrace}={\left\lbrace{1}\right\rbrace}\ldots{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}\right)}\right\rbrace}$$
The second equation is given below
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{4}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{2}\right\rbrace}{\left\lbrace{y}\right\rbrace}={\left\lbrace{3}\right\rbrace}\ldots{\left\lbrace{\left({\left\lbrace{2}\right\rbrace}\right)}\right\rbrace}$$
Step 1:
Write the equation (1) in slope-intercept standard form as follows:
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{y}\right\rbrace}={\left\lbrace{1}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{1}\right\rbrace}-{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}$$
Compare the equation $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}=-{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{1}\right\rbrace}$$ with the slope-intercept form $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{m}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{c}\right\rbrace}$$.
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{m}\right\rbrace}=-{\left\lbrace{2}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{c}\right\rbrace}={\left\lbrace{1}\right\rbrace}$$
Therefore, the value of slope m is -2 and c is 1.
Thus, the value of y-intercept (0, c) is (0, 1).
Subtitute 0 for y in equation $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}=-{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{1}\right\rbrace}$$ to obtain the value of x-intercept.
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{0}\right\rbrace}=-{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{1}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le-{\left\lbrace{1}\right\rbrace}=-{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}={\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{1}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace}$$
Therefore, the x-intercept is $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left({\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{1}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace},{\left\lbrace{0}\right\rbrace}\right)}\right\rbrace}.$$
Join the points (0, 1) and $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left({\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{1}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace},{\left\lbrace{0}\right\rbrace}\right)}\right\rbrace}$$ to make a line on the graph as shown below in Figure 1:

Step 2:
Write the equation (2) in slope-intercept standard form as follows:
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{4}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{2}\right\rbrace}{\left\lbrace{y}\right\rbrace}={\left\lbrace{3}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{2}\right\rbrace}{\left\lbrace{y}\right\rbrace}=-{\left\lbrace{4}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{3}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}=-{\left\lbrace{\frac{{{\left\lbrace-{\left\lbrace{4}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}=-{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\frac{{{3}}}{{{\left\lbrace{2}\right\rbrace}}}}$$
Compare the equation $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}=-{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace}$$ with the slope-intercept form $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{m}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{c}\right\rbrace}$$
. $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{m}\right\rbrace}=-{\left\lbrace{2}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{c}\right\rbrace}={\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace}$$
Therefore, the value of slope m is -2 and c is $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace}$$
Thus, the value of y-intercept (0, c) is $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\le{\left\lbrace{f}\right\rbrace}{\left\lbrace{t}\right\rbrace}{\left\lbrace{\left({\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{y}\right\rbrace}^{{{3}}}\right\rbrace}}}}\right\rbrace}\right)}\right\rbrace}$$
Substitute 0 for y in equation $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}=-{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace}$$ to obtain the value of x-intercept.
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{0}\right\rbrace}=-{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le-{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace}=-{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}={\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{4}\right\rbrace}\right\rbrace}}}}\right\rbrace}$$
Thus, the x-intercept is $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left({\left\lbrace{0}\right\rbrace},{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{4}\right\rbrace}\right\rbrace}}}}\right\rbrace}\right)}\right\rbrace}$$
Join the points $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left({\left\lbrace{0}\right\rbrace},{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}}\right\rbrace}\right)}\right\rbrace}{\left\lbrace\quad\text{and}\quad\right\rbrace}{\left\lbrace{\left({\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{4}\right\rbrace}\right\rbrace}}}}\right\rbrace},{\left\lbrace{0}\right\rbrace}\right)}\right\rbrace}$$ to make a line on the same graph of Figure 1 then the updated graph is shown below in Figure 2.

Step 3:
From Figure 2, it is observed that there is no point of intersection of lines.
Therefore, the lines never meet and are parallel to each so the system has no solution in the systems of linear equations.
Thus, the system is inconsistent and the system has no solution as shown below in Figure 2.

### Relevant Questions

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).

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. $$\displaystyle{\left({b}\right)}{\left[{S}\right]}_{{{E},{C}}}$$
$$\displaystyle{\left({c}\right)}{\left[{S}\right]}_{{{B},{C}}}$$

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

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.

All bases considered in these are assumed to be ordered bases. In Exercise, compute the coordinate vector of v with respect to the giving basis S for V. V is $$R^2, S = \left\{ \begin{bmatrix}1 \\ 0 \end{bmatrix}\begin{bmatrix} 0 \\1 \end{bmatrix} \right\}, v = \begin{bmatrix} 3 \\-2 \end{bmatrix}$$

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}

Interraption: To show that the system $$\displaystyle\dot{{r}}={r}{\left({1}-{r}^{2}\right)},\dot{\theta}={1}$$ is equivalent to $$\displaystyle\dot{{x}}={x}-{y}-{x}{\left({x}^{2}+{y}^{2}\right)},\dot{{y}}={x}+{y}-{y}{\left({x}^{2}+{y}^{2}\right)}$$ for polar to Cartesian coordinates.
$$\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}$$