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I have been learning that the standard expression of a linear equation in two variables is of the form:
$a x + b y + c = 0$ while $a \neq 0$ and $b \neq 0$ . I want to understand the purpose of the constant 'c' in this equation and where it was derived from, whether it be graphically or algebraically.

horrupavinvo 2022-02-23

$A = (- 5, 1, 3), B = (- 2, 6, - 3), C = (- 4, 3, 4)$
I also have a line with points:
$D = (- 2, 7, 6), E = (- 1, 8, - 2)$
The vector equation form of the line:
$((- 2), (7), (6)) + r ((1), (1), (8)) : | r \in R R$
The vector equation form of the plane:
$((- 5), (1), (3)) + s ((3), (5), (- 6)) + t ((1), (2), (1)) : | s, t \in R R$
I am supposed to compute the intersection between these two.
What I know about intersection is that if I have a linear equation with no solution, then there is no intersection. If i have a Linear equation with one solution, then there is one intersection. If i have a linear equation with infinitely many solutions, then the line is contained within the plane.
I the answer is that the line is contained within the plane. I know the answer, but how is the solution squired is beyond me.

Tagiuraoob 2022-02-23

Write the following linear differential equations with constant coefficients in the form of the linear system $\dot{x} = A x$ and solve:
$(a) \ddot{x} + \dot{x} - 2 x = 0$
$(b) \ddot{x} + x = 0$
$(c) d \ddot{x} - 2 \ddot{x} - \dot{x} + 2 x = 0$
Hint: Let $x_{1} = x, x_{2} = \dot{x}$ , etc.
I have tried to do this in the following way but I do not know if I am doing well:
Let $x_{1} = x, x_{2} = \dot{x}, x_{3} = \ddot{x}, x_{4} = x^{\dots}$ , then the previous system becomes $(a) x_{3} + x_{2} - 2 x_{1} = 0, (b) x_{3} + x_{1} = 0, (c) x_{4} - x_{3} - x_{2} + 2 x_{1} = 0$
and thus $[\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}] = [\begin{matrix} \frac{x_{2}}{3} \\ x_{2} \\ \frac{- x_{2}}{3} \\ 0 \end{matrix}]$ . Is this what it should be or is it different? Thank you very much.

beuz89100g6c 2022-02-23

If we want to graph a horizontal line, we will do the following:
$y = 0 x + 3$
No matter the domain for x, the range for y will always be 3. Therefore, we have a horizontal line.
$y = 0 (0) + 3 = (0, 3)$
$y = 0 (1) + 3 = (1, 3)$
$y = 0 (2) + 3 = (2, 3)$
Now the formula to graph a vertical line looks like this:
$x = 3$
Well, wait a second. Where is the y? I would like to see the y in the equation. But it is missing. How can I write the equation for a vertical line that includes the y variable? This is all I can think of:
$x = 0 y + 3$
And with the following domain:
$x = 0 (0) + 3$
$x = 0 (1) + 3$
$x = 0 (2) + 3$ Is this correct? Is it ok to reverse the x and y, as I just did above? Or does this not make it a slope-intercept equation anymore? It should still be a linear equation, since the variables are raised to the first power, in my opinion. But the slope-intercept form looks like this: y = mx + b. So I am not sure if this is still a slope-intercept equation.

Phoebe Xiong 2022-02-23

This is my first question here, so hopefully I will be able to explain my problem in a coherent way. My ultimate question is: do you see any way I can simplify the following system so I can have a more intuitive solution to it? Let me explain in details what I mean:
I want to characterize x,y,z that solve the following system of non-linear equations:
$Δ x = \frac{θ Δ z Δ y}{Δ y - θ Δ z} (EQ1)$
$Δ x + Δ y + Δ z = \overset{―}{Y} (EQ2)$
$θ {(Δ z)}^{2} + {(Δ y)}^{2} = \overset{―}{U} (EQ3)$
where
$Δ x = x - \overset{―}{x}$
$Δ y = y - \overset{―}{y}$
$Δ z = z - \overset{―}{z}$
and $0 < θ < 1, \overset{―}{Y}, \overset{―}{U}, \overset{―}{x}, \overset{―}{y}$ and $\overset{―}{z}$ are parameters.
The way I've approached this was to parameterize (EQ2):
$Δ y + Δ z = α$
such that $α = \overset{―}{Y} - Δ x$
For $θ = 0.8, \overset{―}{x} = \overset{―}{y} = \overset{―}{z} = 0.5$ and $\overset{―}{Y} = - 0.5$ and $\overset{―}{U} = 0.2$ this parametric system would look like the following (I apologize I don't have reputation to embed the photo here):
The parametric system
I can characterize $Δ y$ and $Δ z$ as a function of $α$ :
$Δ y = \frac{θ α \pm \sqrt{\overset{―}{U} (1 + θ) - θ α^{2}}}{1 + θ}$
$Δ z = \frac{α \mp \sqrt{\overset{―}{U} (1 + θ) - θ α^{2}}}{1 + θ}$
My problem arrives in the characterization of α using (EQ1). Although we can show such an $α$ exists, it is far from an intuitive closed-form characterization, as $α$ solves:
$\pm \frac{(\overset{―}{Y} - α) (1 - θ) \sqrt{\overset{―}{U} (1 + θ) - θ α^{2}}}{θ α \pm \sqrt{\overset{―}{U} (1 + θ) - θ α^{2}}} = θ (α \mp \sqrt{\overset{―}{U} (1 + θ) - θ α^{2}})$
Am I using the correct approach to this problem? Or do you see anyway I could simplify this? Thank you!

Beverley Rahman 2022-02-23

My calculus book says the equation:
$y^{'} + x^{2} y = y^{2}$
is not linear.
Linear equations must be of the form:
$y^{'} + P (x) y = Q (x)$
Is the equation not in this form with Q(x) = 0?

ymhonnolfq 2022-02-23

I had one doubt in matrix form of linear equations
Say , we have a system of equations $A x = b$ such that A is an $n \times n$ matrix and b is a $n \times 1$ matrix and so is x then, if we are told that $rank (A) = n$ , then, do we need to check that $rank (A ∣ b) = n$ or can we say that since b is a linear combination of the component vectors in A then augmenting it in A won't increase the number of linearly independent column vectors and hence the $rank (A ∣ b) = n$ and can't ever be $n + 1$ .

ennuvolat4gv 2022-02-23

If a system of linear equations can be represented by the compact form,
$A x = b$
how will a system of nonlinear equations (system of quadratic equations and system of exponential equations) be represented in compact form?

Dachpunkt4cr 2022-02-23

In a previous task, I determined a polynomial interpolation using a system of linear equations.
The data points to be used were
$((0, f (0)), (\frac{1}{6}, f (\frac{1}{6})), (\frac{1}{4}, f (\frac{1}{4})))$
The linear equation used was of the form:
$((0^{2}, 0^{1}, 0^{0}), ({\frac{1}{6}}^{2}, {\frac{1}{6}}^{1}, {\frac{1}{6}}^{0}), ({\frac{1}{4}}^{2}, {\frac{1}{4}}^{1}, {\frac{1}{4}}^{0}))$
$((a_{2}), (a_{1}), (a_{0}))$
$((0), (\frac{1}{\sqrt{3}}), (1))$
and the polynomial of the form
$p_{2} (x) = a_{0} + a_{1} x + a_{2} x^{2}$
was determined to have the coefficients
$a_{0} = 0, a_{1} = - 8 + \frac{18}{\sqrt{3}}, a_{2} = 48 - \frac{72}{\sqrt{3}}$
I am not aware of how I would use a system of linear equations to determine Q(x). Is it perhaps possible to derive the coefficients from p2? Or is there some other method of interpolation I should pursue?

Lily Deleon 2022-02-23

I have just started auditing Linear Algebra, and the first thing we learned was how to solve a system of linear equations by reducing it to row echelon form (or reduced row echelon form). I know how to solve a system of linear equations using determinants, and so the using row echelon form seems to be very inefficient and an easy way to make mistakes.
The lecturer seemed to indicate that this will be useful in the future for something else, but a friend who has a PhD in mathematics said it is totally useless.
Is this true? Is this technique totally useless or is there some "higher mathematics" field/ technique where row echelon form will be useful?

Milla Werner 2022-02-23

Show that the solutions of a homogeneous linear differential equation y''+a(x)y'+b(x)y=0 form a vector space. What is its dimension?
I understand that the dimension is 2 and that 0 is a solution to the differential equation
$(0^{″} + a (x) ⋆ 0^{'} + b (x) ⋆ 0 = 0)$ .
How does one go about proving the other two properties of a vector space: closed under addition and closed under multiplication?

Akbar Witt 2022-02-23

Can anyone tell me how would I answer these type of questions? I already know how to answer the normal way which doesn't include any variables such as Alpha in the question.
a. $[\begin{matrix} 1 & α & 4 \\ 3 & 6 & 8 \end{matrix}]$
b. $[\begin{matrix} 1 & α & - 3 \\ - 2 & 4 & 6 \end{matrix}]$

Aman Wyatt 2022-02-23

I was reading Linear Algebra by Hoffman and Kunze and I encountered some concepts which were arguably not explained or expounded upon thoroughly. I will present my questions below, and I am looking for answers that do not invoke concepts like determinant, vector spaces etc. Note that this is an introductory chapter that assumes no knowledge of the above concepts.
Question 1: Consider a linear system A with k equations. If we form a new equation by taking a linear combination of these k equations, then any solution of A is also a solution of this new equation. Why is this so?
Question 2a: Consider two linear systems A and B. If each equation of A is a linear combination of the equations of B, then any solution of B is also a solution of A. Why is this so?
Question 2b: If each equation of A is a linear combination of the equations of B, then it is not necessary that any solution of A is also a solution of B. Why is this so?

Fearne Castro 2022-02-23

I am stucked on the following challenge: "If the line determined by two distinct points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is not vertical, and therefore has slope $\frac{(y_{2} - y_{1})}{(x_{2} - x_{1})}$ , show that the point-slope form of its equation is the same regardless of which point is used as the given point." Okay, we can separate $(x_{0}, y_{0})$ from the form to get:
$y (x_{2} - x_{1}) - x (y_{2} - y_{1}) = y_{0} (x_{2} - x_{1}) - x_{0} (y_{2} - y_{1})$
But how exclude this point $(x_{0}, y_{0})$ and leave only $x, y, x_{1}, y_{1}, x_{2}, y_{2}$ in the equation? UPDATE: There is a solution for this challenge:
$(y_{1} - y_{2}) x + (x_{2} - x_{1}) y = x_{2} y_{1} - x_{1} y_{2}$
From the answer I found that
$y_{2} (x - x_{1}) - y_{1} (x - x_{2}) = y (x_{2} - x_{1})$
but why this is true?

Kelvin Gregory 2022-02-23

Here is the example I encountered :
A matrix $M (5 \times 5)$ is given and its minimal polynomial is determined to be ${(x - 2)}^{3}$ . So considering the two possible sets of elementary divisors
${{(x - 2)}^{3}, {(x - 2)}^{2}} and {{(x - 2)}^{3}, (x - 2), (x - 2)}$
we get two possible Jordan Canonical forms of the matrix , namely $J_{1}$ and $J_{2}$ respectively. So $J_{1}$ has 2 and $J_{2}$ has 3 Jordan Blocks.
Now we are to determine the exact one from these two. From the original matrix M, we determined the Eigen vectors and 2 eigen vectors were linearly independent. So the result is that $J_{1}$ is the one .So, to determine the exact one out of all possibilities , we needed two information -
1) the minimal polynomial ,together with 2) the number of linearly independent eigen vectors.
Now this was a question-answer book so not much theoretical explanations are given . From the given result , I assume the number of linearly independent eigen vectors -which is 2 in this case - decided $J_{1}$ to be the exact one because it has 2 Jordan Blocks. So the equation
"Number of linearly independent eigen vectors=Number of Jordan Blocks"
must be true for this selection to be correct .
Now this equation is not proved in this book or the text book I have read says nothing of this sort
So, that is my question here : How to prove the equation "Number of linearly independent eigen vectors=Number of Jordan Blocks"?

Corbin Walton 2022-02-23

$T : V \to W$ is the linear transformation of the vector spaces V and W. Let's take $v_{0}$ which is an element of V as $T (v_{0}) = ω_{0}$ .
How can I show that all the solutions of the equation $T (v) = ω_{0}$ are of the form of $v = v_{0} + u$ ? u is an element of N(T).

Margaux Cole 2022-02-22

How would you solve the following system of linear equations:
$2 x_{1} + (3 + a) x_{2} + 2 x_{3} = 2 + a$
$x_{1} + a x_{2} + 2 x_{3} = a$
$a x_{1} + 2 x_{2} + 2 a x_{3} = 0$
assuming that $a \neq \pm \sqrt{2}$ ? I feel confident in solving linear equation systems with just constants as the coefficients but the variable coefficient a is what gives me problems.
Here is the solution I find: solution as augmented matrix but how would the assumption about a make the reduced echelon form any different compared to an assumption about e.g. $a \neq \pm \sqrt{2}$ ?

Kris Patton 2022-02-22

How do I craft a linear equation so that it is in the form of $a x + b x + c = 0$ where $a^{2} + b^{2} = 1$ if I have two points? I know how to get it into the form ax+bx+c=0 but I can't figure out the algorithm for satisfying the second condition.

erycletrefeebr 2022-02-22

In describing the solution of a system of linear equations with many solutions, why do we use a free variable as a parameter to describe the other variables in the solution? Why do we not we use a leading variable? Since by the commutative property of addition we can swap between the free and leading variables, e.g. x + y + z = x + z + y; The solution set will be essentially identical (albeit having different orders).
Definitions:
1. A parameter that is not a leading variable is referred to as a free variable.
2. A leading variable is the first variable in reduced form with a non-zero coefficient.
3. These definitions are clearest when applied to the Echelon form of a system of linear equations expressed as a Matrix.
For example:
Let S be the solution set of the system
$x + y + z = 3$
$y - z = 4$
Using the free variable z as the parameter
$S = {(- 2 z - 1, z + 4, z) ∣ z \in R}$ .
Using the leading variable y as the parameter
$S = {(- 2 y + 7, y, y - 4) ∣ y \in R}$ .

starwolf3wx 2022-02-22

Suppose the 4-vector c gives the coefficients of a cubic polynomial $p (x) = c_{1} + c_{2} x + c_{3} c^{2} + c_{4} c^{3}$ . Express the conditions
$p (1) = p (2), p^{'} (1) = p^{'} (2)$
as a set of linear equations of the form Ac=b. Give the sizes of A and b, as well as their entries.
So far, what I've done was sub in the values and differentiate the polynomials and let it equal to each other after subbing their values.
I get
$c_{1} + c_{2} + c_{3} + c_{4} = c_{1} + 2 c_{2} + 4 c_{3} + 8 c_{4}$
then,
$0 = c_{2} + 3 c_{3} + 7 c_{4}$
as well as
$c_{2} + 2 c_{3} + 3 c_{4} = c_{2} + 4 c_{3} + 12 c_{4}$
then,
$0 = 2 c_{3} + 9 c_{4}$
What should I do now?

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