# Question about equations and a matrix

1. Jun 6, 2006

### finchie_88

I got a question today, that was a bit strange, it was this:
For what value(s) of p do the following equations have solutions?
$$4x + y + 5z = 12$$
$$3x + 2y + 5z = p$$
$$8x + 5y + 13z = 0$$

I said that it doesn't matter what value p is, there are never any solutions (regardless of what p is), since the matrix of the three equations is singular, was I right in thinking that? I also said that the equations give three planes which all intersect, but not at a unique solution, so they form a prism.

2. Jun 6, 2006

### HallsofIvy

Staff Emeritus
No, you are not. If the determinant of the matrix had been non-zero, then you would have been correct in saying that the system has solutions no matter what p is. However, if the determinant is zero, so that the matrix is singular, then there are two alterernatives (the "Fredholm alternative"): either the system has no solution or it has an infinite number of solutions. For what values of p does the augmented matrix reduce to something that has a solution?

Last edited: Jun 7, 2006
3. Jun 7, 2006

### finchie_88

Right, I can't see any values which p can take to give an infinite number of solutions, so I said that p can take any value (there would still be no solutions), that is why I said that there are no solutions, and that they form a prism since any pair of equations are consistent, however, are not consistent with the third. btw, the determinant is zero, so the matrix is singular.

4. Jun 7, 2006

### HallsofIvy

Staff Emeritus
Right?? You say "right" and then contradict what I just said! IF the determinant of the matrix is non-zero, then there is a unique solution for all p. If the determinant of the matrix is 0, then there exist no solution for some values of p and an infinite number of solutions for other values of p. For some values of p, the second equation coincides with one of the others so there are an infinite number of solutions. Doing a row-reduction, I get
$$\left(\begin{matrix}4&1&5&12\\0&\frac{5}{4}&\frac{5}{4}&p-9\\0&0&0&-\frac{12}{5}p-\frac{108}{5}\end{matrix}\right)$$
As long as that last term in the third row is not 0, there is no solution. If it is 0, then there are an infinite number of solutions. For what value of p is $-\frac{12}{5}p-\frac{108}{5}= 0$?

5. Jun 7, 2006

### finchie_88

That right was more of a "oh, you make it sound as though it's an obvious and simple concept" right, (it's a bit hard to get across such unwritten messages without speech! - lol, don't worry about it), anyway, that's not really important. Thank you for the help HOI. Help was much appreciated.

Last edited by a moderator: Jun 7, 2006