Matrices and infinite solutions

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To find the value of h that allows the system of equations to have infinitely many solutions, the equations must be equivalent, meaning one is a multiple of the other. The discussion reveals that when h = 14, the system represents two parallel lines that do not intersect, indicating no solutions. It is concluded that for any other value of h, the lines will intersect at a single point, resulting in a unique solution. The analysis shows that there is no value of h that leads to coinciding lines, which would be necessary for infinite solutions. Thus, the system cannot have infinitely many solutions for any real value of h.
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Homework Statement



Find h so that:

-8x + -7y = 7
16x + hy = 14

has infinitely many solutions (solve this exercise with matrices).

Homework Equations



-

The Attempt at a Solution



I converted the system to matrix form, but when I try to convert it to echelon form, I get the following result:

-8x + -7y = 7
0x + (h - 14)y = 28

How can you proceed from that point? I know that the y coefficient shouldn't be 0, because then it would mean that the system has no solutions. However, the problem asks for a value of "h" in which the system has infinitely many solutions.
 
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Ipos Manger said:

Homework Statement



Find h so that:

-8x + -7y = 7
16x + hy = 14

has infinitely many solutions (solve this exercise with matrices).

Homework Equations



-

The Attempt at a Solution



I converted the system to matrix form, but when I try to convert it to echelon form, I get the following result:

-8x + -7y = 7
0x + (h - 14)y = 28

How can you proceed from that point? I know that the y coefficient shouldn't be 0, because then it would mean that the system has no solutions. However, the problem asks for a value of "h" in which the system has infinitely many solutions.

Are you sure you copied the problem correctly? Geometrically, your system of equations represents two straight lines. If h = 14 in the original system, the system represents two parallel lines that don't intersect. For there to be an infinite number of solutions, the two equations have to be equivalent. In this case, each would have to be a nonzero multiple of the other. As you show the equations, this can't happen.
 
You need a fixed value of H in which case the 2 lines will always intersect.
Since it's a 2d world I would go about it showing a value of H in which case the 2 functions' ascension angle or whatever it's called is not the same.

I already know the tan of the ascension angle is the coefficient of the argument.

I would express both functions as Y and get:
y = (8/-7)x -1
y = (-16/H)x + 14/H

now all that's left is to show that (8/-7) =/= -16/H - I get that they are only equal if and only if H = 14. No matter what ever else H, except H=0, value will result in intersection or in other words will mean the system has a specific solution. Right now it seems to me the system is solveable unless H=14 or H=0.

For the system to have Infinite amount of solutions for 1 specific value of H means that the 2 lines are coinciding? (is that the word I'm looking for?) Coinciding is a special case of parallel, but I just showed the lines are parallel only if H=14. I drew the 2 graphs when H=14, they are not coinciding and they never will, which means there is no real value of H in which case the system has infinite number of solutions.

Disecting this with Cramer's method end up in a brickwall aswell, same story, coinciding is impossible and only parallel when H=14, but no real solutions for the system.
 
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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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