Matrices and infinite solutions

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The discussion centers on determining the value of "h" in the system of equations: -8x - 7y = 7 and 16x + hy = 14, such that the system has infinitely many solutions. It is established that for the system to have infinitely many solutions, the two equations must be equivalent, which occurs when the lines they represent coincide. The analysis reveals that the only value of "h" that results in parallel lines is h = 14, which does not yield infinite solutions. Therefore, there is no real value of "h" that allows the system to have infinitely many solutions.

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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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