system of equations

[martypanama]

I have the following system of equations in R²:

ax + y = a
x + ay = 1

and there's a constraint given, which is that a≠1, a≠-1.

Solving for x and y we get: x=1 and y=0. And so the system represents the point (1,0) in R².

In other words, the two original equations describe a family of pairs of lines through (1,0). (with the constraint that a≠1, a≠-1).

Here's my question: how can I find two perpendicular lines that include the point described by the system, (1,0), but that are different from the two given equations?

My hunch is to just let a=1 or -1 because then the lines can't be equal to the original equations (because of the constraint on a that we're given). Doing this I get the vertical line x = 1 from the 1st equation and the horizontal line y = 0 from the second equation, and these are perpendicular to each other. Is this the right way to solve this?

Thanks so much...

 

[martypanama]

Hi Denis,

Thanks for the response.

Let me try to clarify my original question.

The system of equations

ax + y = a
x + ay = 1 where a≠1, a≠-1 (i.e. a cannot be 1 or -1)

can be solved for x and y as (x,y)=(1,0). So the two equations represent pairs of lines (which can be many different pairs depending on the value of a) that intersect at the point (1,0).

The question is how to find a line (or really two lines perpendicular to each other) also through (1,0) but not included in the "family" of lines described by the original system of equations.

I believe that, since we're told that in the original system a≠1, a≠-1, if we let a=1 or -1 then we will get an equation for a line (or lines) not described by the original system. Does that make sense?

I asked the question because, although I'm confident I've done it correctly, I'm not certain.

Thanks a lot,

marty

 

[mmm4444bot]

the system represents the point (1,0)

No. The system represents two lines.

It's the system's solution that represents the point.

Is this the right way to solve this?

It's not good to use the same pronoun to represent distinct nouns in a single sentence.

I do not understand what you've been asked to do. Can you type the given exercise exactly as you received it?

if we let a=1 or -1 then we will get an equation for a line (or lines) not described by the original system.

If a = 1 or a = -1, then both equations represent the same line.

Write each equation in Slope-Intercept form. The slopes are -a and -1/a, so the lines will never be perpendicular.

Is the exercise simply to find a pair of perpendicular lines intersecting at the point (1,0) ?

The lines x = 1 and y = 0 are perpendicular and they intersect at (1, 0), but the vertical line has no slope. That's okay?

Let me try to clarify my original question.

If you're referring to a specific exercise, I would prefer that you post it verbatim -- as you received it.

I'm confused about what the posted system has to do with (what seems like) a request for two, arbitrary, perpendicular lines intersecting at (1,0).

Cheers :cool:

 

[martypanama]

Hi mmm4444bot,

Thanks a lot for responding. Sorry for the confusion. Below is the 2-part question exactly as it appears, and then below that are my answers to each part.
Also, looking at the two equations in slope intercept form, like you suggested, makes it pretty clear. Like you said, since the slopes are -a and -1/a, the system can't give us 2 perpendicular lines (regardless of the value of a). So that means that the lines x = 1 (it's ok there's no slope) and y = 0, since they are indeed perpendicular, can't possibly be described by the system. I think that basically resolves my doubt, but just in case here are the questions verbatim:

Part 1: Solve the following system of equations for x and y in R² by row-reduction:

ax + y = a , where a≠1, a≠-1 (Note that a may be 0; hence you may not divide by a.)
x + ay = 1

On the basis of your answer, it follows that the system of equations represents a __-flat (also called a ____) in R².

Part 2a: Now find, in non-parametric form, a 1-flat (i.e. a straight line in R²) that includes the flat described by the system of equations, but that is different from either of the two equations in Part 1.

Part 2b: Find, in vector-parametric form, a 1-flat (i.e. a straight line in R²) that includes the flat described by the system of equations, but that is different from either of the two equations in Part 1 and also perpendicular to the line you found in Part 2a.


And below are my answers to these questions.

Part 1:
After row reduction we get x = 1 and y = 0.
It follows that the system of equations represents a 0-flat (also called a point) in R².

Part 2a: In non-parametric form, a line that includes the point (1, 0) but that is different from either of the two equations in Part 1 is the line x = 1

Part 2b: y = 0

x = 1

y = 0, x = t can be expressed in vector-parametric form as [x, y] = [0, 0] + t [1, 0] **These are intended as column vectors but I don't know how to format those here

 

[HallsofIvy]

As mmm4444bot told you, no, those two equations do NOT "represent a point". Each equations represents a line and, since they are different lines, the two together represent the plane that those lines lie in. The two lines intersect at the point you determined.

 

[lookagain]

Not sure what you're asking; however:
ax + y = a [1]
ax + a^2y = a : [2] * a
Subtract above:
a^2y - y = 0
a^2y = y
a^2 = 1
> > a = 1 < <

\(\displaystyle a^2 \ = \ 1\)

\(\displaystyle a \ = \ \pm 1\)


And \(\displaystyle a^2y - y \ = \ 0\)


\(\displaystyle (a^2 - 1)y \ = \ 0\)


\(\displaystyle y \ = \dfrac{0}{a^2 - 1}\)


\(\displaystyle y \ = \dfrac{0}{(a - 1)(a + 1)}\)


\(\displaystyle y \ = \ 0, \ \ \ a \ \ne \ -1, \ 1\)