Linear Algebra: Help understanding why three lines have common point of intersection

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Nemanjavuk69

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I have the following homework

1662923279592.png

I know the answer is YES and I know the matrix can be written as [imath]\begin{bmatrix} 1 & 0 & -\frac{13}{7}\\ 0 & 1 & -\frac{5}{7}\\ 0 & 0 & 0 \end{bmatrix}[/imath]

which is in reduced Echelon form

However these are my three questions I am curious about...

1. WHY is the answer YES
2. HOW can we see that they have a common point of intersection
3. WHAT does row 3 tell us (since it is all 0's)?
 
Last edited:
Nemanjavuk69 said:
I have the following homework

View attachment 34019

I know the answer is YES and I know the matrix can be written as [imath]\begin{bmatrix} 1 & 0 & -\frac{13}{7}\\ 0 & 1 & -\frac{5}{7}\\ 0 & 0 & 0 \end{bmatrix}[/imath]

which is in reduced Echelon form (sorry, I don't know how to do matrix via LaTex, so I used Maple and took a screenshot).

However these are my three questions I am curious about...

1. WHY is the answer YES
2. HOW can we see that they have a common point of intersection
3. WHAT does row 3 tell us (since it is all 0's)?
Click to expand...
I don't know what is the meaning of your matrix, but to answer your question #2 : find the intersection point -- if it exists -- of the first two lines and verify that it belongs to the third line.
 
blamocur said:
I don't know what is the meaning of your matrix, but to answer your question #2 : find the intersection point -- if it exists -- of the first two lines and verify that it belongs to the third line.
Click to expand...
Of course, that is one way of doing it, however, I need to understand WHY the answer is YES, HOW we can see that it has a common point of intersection (ONLY looking at the matrix) and WHAT the last row tells us.

When you stated you don't know what the meaning of my matrix is, is it because you don't know how I got the matrix (calculated it) or is it because you are not familiar with a linear system on the Echelon/ reduced Echelon form?
 
blamocur said:
I don't know how the matrix is related to the line equations.
Click to expand...
I am in a class called "Linear Algebra" and we are reading "linear algebra and its applications sixth global edition". The book shows you how to solve linear equations by using a linear system. Usually, the system is used when dealing with more than two equations. In my example, we have three equations. The book also tells you, that if you write the equations as an augmented matrix and rewrite it on Echelon form (sometimes reduced echelon form) you can find the values for x, y, z... and that is what I am using my matrix for.
 
Nemanjavuk69 said:
I am in a class called "Linear Algebra" and we are reading "linear algebra and its applications sixth global edition". The book shows you how to solve linear equations by using a linear system. Usually, the system is used when dealing with more than two equations. In my example, we have three equations. The book also tells you, that if you write the equations as an augmented matrix and rewrite it on Echelon form (sometimes reduced echelon form) you can find the values for x, y, z... and that is what I am using my matrix for.
Click to expand...
I see -- thank you.

  1. So initially the matrix contains all three equations including their right hand sides.
  2. The fact that you zeroed one row by echeloning means that at least one of the equations can be represented as a linear combination of another two.
  3. But this means that any pair of values (a.k.a. point) which turns those two equations to 0 also turns to zero the third equation. Or, to put this differently, any point belonging to both of the two lines belongs to the third one (the one whose equation can be represented as a linear combination of the other two).
 
blamocur said:
I see -- thank you.

  1. So initially the matrix contains all three equations including their right hand sides.
  2. The fact that you zeroed one row by echeloning means that at least one of the equations can be represented as a linear combination of another two.
  3. But this means that any pair of values (a.k.a. point) which turns those two equations to 0 also turns to zero the third equation. Or, to put this differently, any point belonging to both of the two lines belongs to the third one (the one whose equation can be represented as a linear combination of the other two).
Click to expand...
So would this also be correct to say "Since the last row (row 3) is zeroed out, that means that all three lines share a common point intersection"?

Also, your your answer to my second point (HOW can we see that they have a common point of intersection), would this also be correct to say "Since we can write at least one row on echelon form, that must mean, that at least two lines intersect on a common point"?

Lastly, I am not quite sure what you mean with "So initially the matrix contains all three equations including their right hand sides.".

However, thank you for your help so far!
 
Nemanjavuk69 said:
I am in a class called "Linear Algebra" and we are reading "linear algebra and its applications sixth global edition". The book shows you how to solve linear equations by using a linear system. Usually, the system is used when dealing with more than two equations. In my example, we have three equations. The book also tells you, that if you write the equations as an augmented matrix and rewrite it on Echelon form (sometimes reduced echelon form) you can find the values for x, y, z... and that is what I am using my matrix for.
Click to expand...
The determinant of "that" matrix is the area defined by the points of intersection of those lines. The elements of the last row being 0, implied that the area of the triangle is 0 - implieng the lines are concurrent.
 
Subhotosh Khan said:
The determinant of "that" matrix is the area defined by the points of intersection of those lines. The elements of the last row being 0, implied that the area of the triangle is 0 - implieng the lines are concurrent.
Click to expand...
Ohhh. So, let me get this straight. The last row being zeroed out is telling us that all three lines intersect each other at a common point? (because you can only get the last row zeroed out by using one of the other rows, but both the other lines intersect). What would happen if the last row was [imath]\begin{matrix} 0 & 0 & 1 \end{matrix}[/imath] what would that tell us than? What if the last row was [imath]\begin{matrix} 1 & 0 & 0 \end{matrix}[/imath], what would that then tell us?
 
Nemanjavuk69 said:
I have the following homework

View attachment 34019

I know the answer is YES and I know the matrix can be written as [imath]\begin{bmatrix} 1 & 0 & -\frac{13}{7}\\ 0 & 1 & -\frac{5}{7}\\ 0 & 0 & 0 \end{bmatrix}[/imath]

which is in reduced Echelon form

However these are my three questions I am curious about...

1. WHY is the answer YES
2. HOW can we see that they have a common point of intersection
3. WHAT does row 3 tell us (since it is all 0's)?
Click to expand...
The 1st row of your matrix says that 1*x1 +0*x2 = -13/7
The 2nd row of your matrix says that 0*x1 +1*x2 =-5/7
What does this tell you??
 
Using determinants as Dr Khan suggested is a good idea, but that method does not tell you the point of intersection.
 
For the record linear algebra and its applications sixth global edition is not a good description of your book as there are many many linear algebra books titled linear algebra and its applications and many of those are 6th edition. You really need to include the author as well.
I mention this as someone here might have the same book and might be able to help you better than other helpers.
 
Steven G said:
The 1st row of your matrix says that 1*x1 +0*x2 = -13/7
The 2nd row of your matrix says that 0*x1 +1*x2 =-5/7
What does this tell you??
Click to expand...
Those two equations tell me, that the two lines have a common point of intersection, namely ([imath]-\frac{13}{7}, -\frac{5}{7}[/imath]), with other words, we KNOW the first two lines have a common point of intersection. However, is this the reason why the third row is zeroed out? Is that always the case? What if we had 4 lines or n'th number of lines, would they also all be zeroed out if they all shared a common point of intersection?

Thank you for your help

-Nemanja
 
Steven G said:
For the record linear algebra and its applications sixth global edition is not a good description of your book as there are many many linear algebra books titled linear algebra and its applications and many of those are 6th edition. You really need to include the author as well.
I mention this as someone here might have the same book and might be able to help you better than other helpers.
Click to expand...
The book linear algebra and its applications sixth global edition is written by David Lay (Author), Steven Lay (Author), Judi McDonald (Author)
 
Nemanjavuk69 said:
Those two equations tell me, that the two lines have a common point of intersection, namely ([imath]-\frac{13}{7}, -\frac{5}{7}[/imath]), with other words, we KNOW the first two lines have a common point of intersection. However, is this the reason why the third row is zeroed out? Is that always the case? What if we had 4 lines or n'th number of lines, would they also all be zeroed out if they all shared a common point of intersection?

Thank you for your help

-Nemanja
Click to expand...
Read the response #8 - carefully. Write it down on a piece of paper, with pencil (or pen) - and read again.
 
Nemanjavuk69 said:
Those two equations tell me, that the two lines have a common point of intersection, namely ([imath]-\frac{13}{7}, -\frac{5}{7}[/imath]), with other words, we KNOW the first two lines have a common point of intersection. However, is this the reason why the third row is zeroed out? Is that always the case? What if we had 4 lines or n'th number of lines, would they also all be zeroed out if they all shared a common point of intersection?

Thank you for your help

-Nemanja
Click to expand...
No, the 3rd row is all zeros since the 3rd equation is a linear combination of the other 2 rows.
 
Nemanjavuk69 said:
Lastly, I am not quite sure what you mean with "So initially the matrix contains all three equations including their right hand sides.".
Click to expand...
How did you get the reduced echelon form? I.e., which matrix did you reduce it from?
 
Nemanjavuk69 said:
What if we had 4 lines or n'th number of lines, would they also all be zeroed out if they all shared a common point of intersection?
Click to expand...
If n lines all had a common intersection point then the rank of the matrix would be 2. I.e., the reduced echelon form would have all-zero rows except the first two.
 
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