Proof involving two linear equations

In summary: When is this not valid?3)## \frac{a}{b} = ...## as that certainly isn't valid either?We can rewrite the equations as a matrix equation, where the matrix is in the form of a determinant. This allows us to use the property that the determinant of a matrix multiplied by its inverse is equal to the identity matrix. From this, we can see that the determinant of the matrix is not equal to zero, as that would mean the inverse does not exist. Therefore, the two sets of equations must have the same solution, proving that ## (x_
  • #1
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TL;DR Summary
Given ## a,b,c,d,e,f \in \mathbb {R}, ad - bc \neq 0 ##, if ##(x_1,y_1)## and ##(x_2,y_2)## are pairs of real numbers satisfying:

## ax_1 + by_1 = e, cx_1 + dy_1 =f ##
## ax_2 + by_2 = e, cx_2 + dy_2 = f ##

then ## (x_1,y_1) = (x_2,y_2). ##
Given ## a,b,c,d,e,f \in \mathbb {R}, ad - bc \neq 0 ##, if ##(x_1,y_1)## and ##(x_2,y_2)## are pairs of real numbers satisfying:

## ax_1 + by_1 = e, cx_1 + dy_1 =f ##
## ax_2 + by_2 = e, cx_2 + dy_2 = f ##

then ## (x_1,y_1) = (x_2,y_2). ##

Here is my attempt at a proof, I have gotten stuck:
1)
## ax_1 + by_1 = ax_2 + by_2 ##
## cx_1 + dy_1 = cx_2 + dy_2 ##

2)
## a(x_1 - x_2) = b(y_2 - y_1) ##
## c(x_1 - x_2) = d(y_2 - y_1) ##

3)
## \frac{a}{b} = \frac{y_2 - y_1}{x_1 - x_2} ##
## \frac{c}{d} = \frac{y_2 - y_1}{x_1 - x_2} ##

4)
## \frac{a}{b} = \frac{c}{d} ##

5)
## ad = bc ##

6)
## ad - bc = 0 ## but this is false as the hypothesis states that ## ad - bc \neq 0. ##

I am stuck now and am not sure how to proceed.
 
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  • #2
Isn't that essentially a proof? What assumptions did you make in order to get ##ad = bc##?

Note: have you studied matrices yet?
 
  • #3
I don't think it is a proof as I need to prove that ## (x_1,y_1) ## and ## (x_2,y_2) ## are equal, or in other words ## x_1 - x_2 = 0 ## and ## y_1 - y_2 = 0. ## I cannot see how to get there from the result I have ended with?

I have done some linear algebra a long time ago but I am very rusty on everything and am just trying to learn some proofs as that's something I haven't done before. I can't remember all the details of the determinant at this stage...
 
  • #4
RM86Z said:
3)
## \frac{a}{b} = \frac{y_2 - y_1}{x_1 - x_2} ##
## \frac{c}{d} = \frac{y_2 - y_1}{x_1 - x_2} ##

4)
## \frac{a}{b} = \frac{c}{d} ##
What are you assuming here?
 
  • #5
Sorry I know there is a connection here I'm just not getting it at this stage. By the properties of the determinant the result ## ad - bc = 0 ## tells me that the two rows of the linear system are the same so ## a = b ## and ## c = d. ##
 
  • #6
RM86Z said:
Sorry I know there is a connection here I'm just not getting it at this stage. By the properties of the determinant the result ## ad - bc = 0 ## tells me that the two rows of the linear system are the same so ## a = b ## and ## c = d. ##
When is this not valid?

RM86Z said:
3)
## \frac{a}{b} = \frac{y_2 - y_1}{x_1 - x_2} ##
## \frac{c}{d} = \frac{y_2 - y_1}{x_1 - x_2} ##
 
  • #7
When ## x_1 - x_2 = 0 ## and actually now that I think about it it wasn't defined in the hypothesis that ## b,d \neq 0 ## as that certainly isn't valid either?

EDIT: I suppose the last bit about ## a,b \neq 0 ## doesn't make sense as it depends on the ordering of the algebra. In other words instead of ## \frac{a}{b} ## we could have ## \frac{b}{a} ##...
 
  • #8
RM86Z said:
When ## x_1 - x_2 = 0 ## and actually now that I think about it it wasn't defined in the hypothesis that ## b,d \neq 0 ## as that certainly isn't valid either?
You tacitly assumed that ##x_1 \ne x_2## and reached a contradiction - in terms of ##ad = bc##. You've effectively proved, therefore, that ##x_1 = x_2##, as that is the only way around the contradiction.

You always need to be careful that the denominator on any expression in not zero. Technically, you should have done that in going from step 2) to step 3)

RM86Z said:
2)
## a(x_1 - x_2) = b(y_2 - y_1) ##
## c(x_1 - x_2) = d(y_2 - y_1) ##

3)
## \frac{a}{b} = \frac{y_2 - y_1}{x_1 - x_2} ##
## \frac{c}{d} = \frac{y_2 - y_1}{x_1 - x_2} ##
You should have indicated here that you were assuming that ##x_1 - x_2 \ne 0##. Otherwise, that is not valid mathematically.
 
  • #9
PS In fact, it's not valid if ##b = 0## or ##d = 0## either!
 
  • #10
Ah brilliant that makes sense now thank you so much PeroK! That was a very slight detail that I wasn't seeing. I appreciate the help.
 
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  • #11
Your equations can be written as
$$
\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot \begin{bmatrix}x_1\\y_1\end{bmatrix}=\begin{bmatrix}e\\f \end{bmatrix}\text{ and }\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot \begin{bmatrix}x_2\\y_2\end{bmatrix}=\begin{bmatrix}e\\f \end{bmatrix}
$$
From there you can use
$$
\begin{bmatrix}d&-b\\-c&a \end{bmatrix}\cdot\begin{bmatrix}a&b\\c&d\end{bmatrix}=(ad-bc)\cdot\begin{bmatrix}1&0\\0&1\end{bmatrix}
$$
 
  • #12
RM86Z said:
Summary:: Given ## a,b,c,d,e,f \in \mathbb {R}, ad - bc \neq 0 ##, if ##(x_1,y_1)## and ##(x_2,y_2)## are pairs of real numbers satisfying:

## ax_1 + by_1 = e, cx_1 + dy_1 =f ##
## ax_2 + by_2 = e, cx_2 + dy_2 = f ##

then ## (x_1,y_1) = (x_2,y_2). ##

Given ## a,b,c,d,e,f \in \mathbb {R}, ad - bc \neq 0 ##, if ##(x_1,y_1)## and ##(x_2,y_2)## are pairs of real numbers satisfying:

## ax_1 + by_1 = e, cx_1 + dy_1 =f ##
## ax_2 + by_2 = e, cx_2 + dy_2 = f ##

then ## (x_1,y_1) = (x_2,y_2). ##

Here is my attempt at a proof, I have gotten stuck:
1)
## ax_1 + by_1 = ax_2 + by_2 ##
## cx_1 + dy_1 = cx_2 + dy_2 ##

2)
## a(x_1 - x_2) = b(y_2 - y_1) ##
## c(x_1 - x_2) = d(y_2 - y_1) ##

3)
## \frac{a}{b} = \frac{y_2 - y_1}{x_1 - x_2} ##
## \frac{c}{d} = \frac{y_2 - y_1}{x_1 - x_2} ##

4)
## \frac{a}{b} = \frac{c}{d} ##

5)
## ad = bc ##

6)
## ad - bc = 0 ## but this is false as the hypothesis states that ## ad - bc \neq 0. ##

I am stuck now and am not sure how to proceed.
Each ##(x_k,y_k)## pair will have a unique solution. Since all constants are the same for both, the pairs must be equal.
 
  • #13
mathman said:
Each ##(x_k,y_k)## pair will have a unique solution.
This is the statement that has to be shown. It is not automatically obvious.
mathman said:
Since all constants are the same for both, the pairs must be equal.
This is wrong. E.g. ##a=b=c=d=e=f=1.##
 
  • #14
a=a, b=b, etc. is what I meant.
I didn't think it was necessary to prove what is well known. Two linear equations in two unknowns has either a unique solution or none. Basic point: both pairs of equations use the same (a,b, etc.) they will have the same solution. The unknowns are dummies, so it doesn't matter what pair of letters are used, the answer will always be the same for a given set of knowns.
 
  • #15
mathman said:
Two linear equations in two unknowns has either a unique solution or none.
Or infinitely many.
 
  • #16
mathman said:
I didn't think it was necessary to prove what is well known.
It's not well known to someone just learning the subject!
 
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  • #17
RM86Z said:
Summary:: Given ## a,b,c,d,e,f \in \mathbb {R}, ad - bc \neq 0 ##, if ##(x_1,y_1)## and ##(x_2,y_2)## are pairs of real numbers satisfying:

## ax_1 + by_1 = e, cx_1 + dy_1 =f ##
## ax_2 + by_2 = e, cx_2 + dy_2 = f ##

then ## (x_1,y_1) = (x_2,y_2). ##

Given ## a,b,c,d,e,f \in \mathbb {R}, ad - bc \neq 0 ##, if ##(x_1,y_1)## and ##(x_2,y_2)## are pairs of real numbers satisfying:

## ax_1 + by_1 = e, cx_1 + dy_1 =f ##
## ax_2 + by_2 = e, cx_2 + dy_2 = f ##

then ## (x_1,y_1) = (x_2,y_2). ##

You can consider ## ax + by = e## as the equation for a straight and ##cx + dy =f ## as the equation of a second straight. Both straights are in the ##(x,y)## coordinate plane. Now, either those two lines are a) equal, b) parallel, or c) intersect in one point.

##ad-bc\neq 0## rules out the first two cases, but this needs some considerations to see.
 
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  • #18
fresh_42 said:
You can consider ## ax + by = e## as the equation for a straight and ##cx + dy =f ## as the equation of a second straight. Both straights are in the ##(x,y)## coordinate plane. Now, either those two lines are a) equal, b) parallel, or c) intersect in one point.

##ad-bc\neq 0## rules out the first two cases, but this needs some considerations to see.

Somehow I've never thought of it this way. Genius.
 

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