Which system of equations is not a linear system?

a)

b)

c)

d)

boitshupoO
2021-06-12
Answered

Which system of equations is not a linear system?

a)

b)

c)

d)

You can still ask an expert for help

yunitsiL

Answered 2021-06-13
Author has **108** answers

A system of linear equations has equations that can be written in the form

So, the correct answer is choice d.

asked 2022-07-24

Solve the following system of equations by using the inverse of the coefficient matrix A. (AX = B) x + 5y= - 10, -2x+7y=-31

asked 2022-10-31

Ratio between two numbers is 6:7 and the difference between them is 10. What are the two numbers?

asked 2022-07-03

Consider the system

${x}^{\prime}=\frac{-x}{2};{y}^{\prime}=2y+{x}^{2}$

Show that this system is topologically conjugate to the linear system $D{F}_{(0,0)}$

a) Solve both the linear and nonlinear systems and express your answers as a flows ${\varphi}_{t}^{L}(x,y)$ and ${\varphi}_{t}^{N}(x,y)$ respectively.

b) As I found ${\varphi}_{t}^{L}({x}_{0},{y}_{0})$ = $({x}_{0}{e}^{\frac{-1}{2}t},{y}_{0}{e}^{2t})$ and ${\varphi}_{t}^{N}({x}_{0},{y}_{0})$ = $({x}_{0}{e}^{\frac{-1}{2}t},({y}_{0}+\frac{1}{3}{x}_{0}^{2}){e}^{2t}-\frac{{x}_{0}^{2}}{3}{e}^{-t})$ Do they look right to you?

c) Find the topological conjugacy that maps the flow of the nonlinear system to that of the linear system.

${x}^{\prime}=\frac{-x}{2};{y}^{\prime}=2y+{x}^{2}$

Show that this system is topologically conjugate to the linear system $D{F}_{(0,0)}$

a) Solve both the linear and nonlinear systems and express your answers as a flows ${\varphi}_{t}^{L}(x,y)$ and ${\varphi}_{t}^{N}(x,y)$ respectively.

b) As I found ${\varphi}_{t}^{L}({x}_{0},{y}_{0})$ = $({x}_{0}{e}^{\frac{-1}{2}t},{y}_{0}{e}^{2t})$ and ${\varphi}_{t}^{N}({x}_{0},{y}_{0})$ = $({x}_{0}{e}^{\frac{-1}{2}t},({y}_{0}+\frac{1}{3}{x}_{0}^{2}){e}^{2t}-\frac{{x}_{0}^{2}}{3}{e}^{-t})$ Do they look right to you?

c) Find the topological conjugacy that maps the flow of the nonlinear system to that of the linear system.

asked 2022-11-04

An initial value problem ${\mathbf{x}}^{\prime}=A\mathbf{x}$

$$A=\left(\begin{array}{cccc}1& 1& 0& 0\\ 3& -1& 0& 0\\ 0& 0& 0& -2\\ 0& 0& 2& 0\end{array}\right)$$

$$A=\left(\begin{array}{cccc}1& 1& 0& 0\\ 3& -1& 0& 0\\ 0& 0& 0& -2\\ 0& 0& 2& 0\end{array}\right)$$

asked 2022-05-23

Suppose that $x$, $y$ and $z$ are three integers (positive,negative or zero) such that we get the following relationships simultaneously

1. $x+y=1-z$

and

2. ${x}^{3}+{y}^{3}=1-{z}^{2}$

Find all such $x$, $y$ and $z$.

1. $x+y=1-z$

and

2. ${x}^{3}+{y}^{3}=1-{z}^{2}$

Find all such $x$, $y$ and $z$.

asked 2022-07-02

Find the values of $k$ for which the simultaneous equations do not have a unique solution for $x,y$ and $z$.

Also show that when $k=-2$ the equations are inconsistent

$kx+2y+z=0$

$3x+0y-2z=4$

$3x-6ky-4z=14$

Using a determinant and setting to zero, then solving the quadratic gives

$\left|\begin{array}{ccc}k& 2& 1\\ 3& 0& -2\\ 3& -6k& -4\end{array}\right|=0$

$k=-2,k=\frac{1}{2}$

So far so good, but when subbing $-2$ for $k$

$-2x+2y+z=0\phantom{\rule{thickmathspace}{0ex}}...\phantom{\rule{thinmathspace}{0ex}}eqn\phantom{\rule{thinmathspace}{0ex}}1$

$3x+0y-2z=4\phantom{\rule{thickmathspace}{0ex}}...\phantom{\rule{thinmathspace}{0ex}}eqn\phantom{\rule{thinmathspace}{0ex}}2$

$3x+12y-4z=14\phantom{\rule{thickmathspace}{0ex}}...\phantom{\rule{thinmathspace}{0ex}}eqn\phantom{\rule{thinmathspace}{0ex}}3$

subbing eqn $2$ from eqn $3$ gives

$12y-2z=10$

$\Rightarrow y=\frac{5+z}{6}$

Subbing for $y$ into eqn $1$

$-2x+\frac{5+z}{3}-z=0$

$\Rightarrow x=\frac{5-2z}{6}$

Subbing for $y$ and $x$ into eqn $3$

$\frac{5-2z}{2}+10-2z=14$

$\Rightarrow z=\frac{-1}{2},\phantom{\rule{thickmathspace}{0ex}}x=1\phantom{\rule{thickmathspace}{0ex}}and\phantom{\rule{thickmathspace}{0ex}}y=\frac{3}{4}$

These values for $x,y$ and $z$ seem to prove unique solutions for these equations, yet from the determinant and also the question in the text book should they not be inconsistent?

Also show that when $k=-2$ the equations are inconsistent

$kx+2y+z=0$

$3x+0y-2z=4$

$3x-6ky-4z=14$

Using a determinant and setting to zero, then solving the quadratic gives

$\left|\begin{array}{ccc}k& 2& 1\\ 3& 0& -2\\ 3& -6k& -4\end{array}\right|=0$

$k=-2,k=\frac{1}{2}$

So far so good, but when subbing $-2$ for $k$

$-2x+2y+z=0\phantom{\rule{thickmathspace}{0ex}}...\phantom{\rule{thinmathspace}{0ex}}eqn\phantom{\rule{thinmathspace}{0ex}}1$

$3x+0y-2z=4\phantom{\rule{thickmathspace}{0ex}}...\phantom{\rule{thinmathspace}{0ex}}eqn\phantom{\rule{thinmathspace}{0ex}}2$

$3x+12y-4z=14\phantom{\rule{thickmathspace}{0ex}}...\phantom{\rule{thinmathspace}{0ex}}eqn\phantom{\rule{thinmathspace}{0ex}}3$

subbing eqn $2$ from eqn $3$ gives

$12y-2z=10$

$\Rightarrow y=\frac{5+z}{6}$

Subbing for $y$ into eqn $1$

$-2x+\frac{5+z}{3}-z=0$

$\Rightarrow x=\frac{5-2z}{6}$

Subbing for $y$ and $x$ into eqn $3$

$\frac{5-2z}{2}+10-2z=14$

$\Rightarrow z=\frac{-1}{2},\phantom{\rule{thickmathspace}{0ex}}x=1\phantom{\rule{thickmathspace}{0ex}}and\phantom{\rule{thickmathspace}{0ex}}y=\frac{3}{4}$

These values for $x,y$ and $z$ seem to prove unique solutions for these equations, yet from the determinant and also the question in the text book should they not be inconsistent?

asked 2022-09-23

For which real values of t does the following system of linear equations:

$$\{\begin{array}{c}t{x}_{1}+{x}_{2}+{x}_{3}=1\\ {x}_{1}+t{x}_{2}+{x}_{3}=1\\ {x}_{1}+{x}_{2}+t{x}_{3}=1\end{array}$$

Have:

a) a unique solution?

b) infinitely many solutions?

c) no solutions?

$$\{\begin{array}{c}t{x}_{1}+{x}_{2}+{x}_{3}=1\\ {x}_{1}+t{x}_{2}+{x}_{3}=1\\ {x}_{1}+{x}_{2}+t{x}_{3}=1\end{array}$$

Have:

a) a unique solution?

b) infinitely many solutions?

c) no solutions?