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I was wondering if you guys could help me out with this. I honestly have no idea how to solve it, since I've been going through different examples but they all use determinants, which we're not supposed to know yet, so we have to solve it with another method. (Gauss, Gauss-Jordan)

Given the system: $$x+by+az=1$$ $$ax+by+z=a$$ $$x+aby+z=b$$

Find the values of "a" and "b" so that the system has an unique solution, infinite solutions and no solution.

I turned it into a matrix and tried to solved it, but I got nowhere useful. I got this:

\begin{matrix} 1 & b & a & 1 \\ 0 & 1 & {a^2\over(ab-b)} & 0 \\ 0 & 0 & a^2+a-1 & 1-b \\ \end{matrix}

I would be happy if you could give me some sort of hint or useful way so I can solve this.

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Hint: What happens if $a^2+a-1=0$? Does it depend on what happens to $1-b$? –  Gerry Myerson Feb 20 '13 at 2:13
    
Let me try again. What does the last row of the matrix tell you about the system of equations, if $a^2+a-1=0$? and does the answer to that question depend on the value of $1-b$? –  Gerry Myerson Feb 20 '13 at 2:29
    
I'm really sorry if I'm being stupid here. I guess it does depend on the value of $1-b$, to make $a^2+a-1=0$ $b=1$. And having $a^2+a-1=0$ means I could have two or one answers; right? –  ChairOTP Feb 20 '13 at 2:35
    
If $a^2+a-1=0$, and $1-b=17$, then the last row of your matrix corresponds to the equation, $0=17$. How many solutions does that equation have? If $a^2+a-1=0$, and $b=1$, then the last row of your matrix corresponds to the equation, $0=0$, which is true no matter what $x,y,z$ are. No matter what $z$ is, the second row then gives you a value of $y$, and the first row then gives you a value of $x$. So, how many solutions then? –  Gerry Myerson Feb 20 '13 at 4:33
    
1) None. 2) Infinite solutions. Well I've been trying to find a way to solve it, but I haven't really found anything consistent, I have come to some conclusions though. I have found that when $a=0$,$b=1$ there's a unique solution. When $a=1$,$b=1$ there are infinite solutions. And when $a=1$,$b=0$ there are no solutions. I got to these answers by analysing the process of the Gaussian method. And they seem to be right, but I'm not sure if that's the correct way to solve these kind of exercises. –  ChairOTP Feb 20 '13 at 4:44
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2 Answers

Turning the system into a matrix is a good first approach.

Hint: When a system has a unique solution, what does the echelon form of its augmented matrix look like?

Also, is that supposed to be a z in the third equation and not another x?

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I'm guessing each unknown is equal to some value. And when it has no solution is because some unknown is cancelled and it turns $0=$another real number and infinite solutions is when two or more equations are the same. Am I right? –  ChairOTP Feb 20 '13 at 2:21
    
Yes, if the system is inconsistent (no solution at all) the its augmented matrix has a row with all zeroes in the coefficient side and something nonzero in the rightmost column. Under what conditions does this happen with the system you have? –  Robert Talbert Feb 20 '13 at 2:26
    
Could it be when a and b are equal to $1$? In that case we'd have a zero in the dominator, so y wouldn't exist? I'm sorry if this is wrong, it's a little bit late here and I'm not really understanding much of this. –  ChairOTP Feb 20 '13 at 2:30
    
Try setting $a=1, b=1$ and see what that gives you. Also try getting the augmented matrix to echelon, not reduced echelon, form. The form you have above can only be obtained if you assume $a,b \neq 1$. –  Robert Talbert Feb 20 '13 at 2:32
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First problem is that your row reduction went wrong somewhere. You should get to the matrix $$\pmatrix{1&b&a&|&1\cr0&b(1-a)&1-a^2&|&0\cr0&0&-a^2-a&|&b-1\cr}\tag1$$ I think. Now we have to look at what happens when $b(1-a)=0$, and what happens when $-a^2-a=0$.

When $a=0$, the matrix becomes $$\pmatrix{1&b&0&|&1\cr0&b&1&|&0\cr0&0&0&|&b-1\cr}$$ Looking at the third row we conclude that if $a=0$ and $b\ne1$ then the number of solutions is ... ?While if $a=0$ and $b=1$ then the number of solutions is ... ?

When $a=-1$, the matrix becomes $$\pmatrix{1&b&-1&|&1\cr0&2b&0&|&0\cr0&0&0&|&b-1\cr}$$ Again, when $a=-1$, you get two cases, depending on whether or not $b=1$.

When $a=1$, the matrix becomes $$\pmatrix{1&b&1&|&1\cr0&0&0&|&0\cr0&0&-2&|&b-1\cr}$$ How many solutions when $a=1$?

When $b=0$, the matrix becomes $$\pmatrix{1&0&a&|&1\cr0&0&1-a^2&|&0\cr0&0&-a^2-a&|&-1\cr}$$ We can assume $a\ne\pm1$, since we have already done those cases, and further reduce this matrix to $$\pmatrix{1&0&0&|&1\cr0&0&1&|&0\cr0&0&0&|&-1\cr}$$ So, when $b=0$, and $a\ne\pm1$, how many solutions?

From here on, we can assume $b(1-a)\ne0$ and $-a^2-a\ne0$, since we have dealt with those cases. So, when you look at matrix (1) under these assumptions, how many solutions?

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I must have done something wrong, you're right. But in the third row I keep getting $2-a^2-a$ –  ChairOTP Feb 20 '13 at 6:31
    
I'll check my computations (but probably not until tomorrow). –  Gerry Myerson Feb 20 '13 at 6:38
    
Supposing I got the third row thing right. I got this: $a=1$,$b=1$ then there are infinite solutions. If $a=1$ and $b/=1$ then there's no solution, and if $b=0$,$a/=+-1$ then there are also no solutions. But if I'm right, there's no case when $a=0$ which I need it so I can get unique solutions. –  ChairOTP Feb 20 '13 at 6:45
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