Homework Help: Questions on 3 simultaneous equations (3D vector planes)

1. Jan 31, 2010

Charismaztex

1. The problem statement, all variables and given/known data

1) Determine, by elimination, value of a (if any) such that the given system will have a unique solution

$x+2y+3z=2, 2x+2y+az=0, 3x+2y+z=0$

2) Determine, by elimination, values of a (if any) such that the given system:

a) Is consistent with and infinity of solutions;
b) has a unique solution;
c) is inconsistent, with no solution

$x+2y+3z=a, x+y+z=0, 3x+2y+z=0$

3) consider the following system of 3 equations in x,y, and z

$2x+2y+2z=9, x+3y+4z=5, Ax+5y+6z=B$

Give possibly values of A and B in the third equation which make this system:

a) inconsistent
b) consistent but with an infinite number of solutions

2. Relevant equations

N/A

3. The attempt at a solution

1) By a unique solution I'm presuming that all three planes meet at the same point. Would this be to solve and get a so that x,y, and z have a unique value?

2) Consistent with infinity of solutions, is that when we get a situation 0=0? Possibly with at least 2 of the plane equations the same, or intersecting like the "spine" of a book. So a value of a to get some sort of 0=0?

would inconsistent be a situation when we get 0=a number (a nonsense statement)?

To be honest, I have no idea how to approach the question which as to determine values of a or A and B. What sort of method would be suitable?

Charismaztex

2. Jan 31, 2010

HallsofIvy

Yes, just go ahead and solve the equations with the "a" in there. At some point, perhaps more than once, you will have to divide by an expression involving a. The system has a unique solution if, each time you have to divide by an expression involving a, that expression is NOT 0. a can be any number that does not make any of those expressions equal to 0.

Yes. Here, a will be such that, at some point, you have to "divide by 0" but the number you are dividing into is also 0.

Right. Again, a must be such that it makes some expression you divide by 0 but now the value you are dividing into is NOT 0.

The best way to do this would be to actually go ahead and solve for x, y, and z, equal to fractions with a in numerator and denominator. There will be a unique solution if a is such that none of those denominators is 0. There will be an infinite number of solutions if a is such that at least one of the denominators is 0 but so are the corresponding numerators. There will be no solution if a is such that there exist at least one denominator equal to 0 and the corresponding numerator is not equal to 0.

Last edited by a moderator: Jan 31, 2010
3. Jan 31, 2010

Charismaztex

Thanks for the reply. Using your method, I've successfully found for Q1) that a is anything except 2 (confirmed by inputting any number except 2 to get a unique solution on the calculator).

For Q2) I've found that no matter how I manipulate the equations, it always comes out to be something like, for example, $y+2z=a, y+2z=0$. Hence I can see that a must be 0 for infinite solutions, and a is any number other than 0 to be inconsistent with no solutions. I am stymied, however, at part b) where "a" should be a number where there is a unique solution. I can only see that a is either consistent with a=0 or inconsistent with a=any number, with not third option.

For Q3) I've worked out A=3 (in my equation, I got A=3 to make the denominator 0) and B=14 (I made the numerator =0 and solved for B) for infinite solutions and B=any number except 14 for inconsistent solution. I am wondering if this is the correct way to approach this.

Thanks,
Charismaztex