# Is my logic correct? - L. Algebra question

1. Jun 14, 2008

### JinM

Is my logic correct? -- L. Algebra question

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

$$\left[ \begin{array}{cccc} 1 & 2 & -3 & 2 \\ -1 & -2 & b & 3 \\ 0 & 0 & c & 2 \end{array} \right]$$

(1) For what values b and c is the system in the augmented matrix above consistent?

(2) For what values b and c does the system in the augmented matrix above have a unique solution?

3. The attempt at a solution

I tried to simplify, and after a few row operations, I end up with:

$$\left[ \begin{array}{cccc} 1 & 2 & -3 & 2 \\ 0 & 0 & c(b - 3) & 5c \\ 0 & 0 & 0 & 2b - 6 - 5c \end{array} \right]$$

Now I'm thinking this is it: the system is inconsistent for b = 3 and c != 0, and consistent for all other real numbers b and c.

Also, the system can never have a unique solution since there are free variables.

Is my reasoning faulty?

2. Jun 14, 2008

### rocomath

How did you eliminate your third row?

I have ...

$$\left[ \begin{array}{cccc} 1 & 2 & -3 & 2 \\ 0 & 0 & b-3 & 5 \\ 0 & 0 & 0 & \frac{2b-5c-6}{b-3} \end{array} \right]$$

Last edited: Jun 14, 2008
3. Jun 14, 2008

### JinM

I made a big mistake in calculating row equivalence! Never mind, I got what you have in your post now.

A few questions: how can you have unique solutions if you have a free variable? I'm inclined to think that we'll have infinite solutions if there is a free variable.

Also, how do you determine what c should be from the rightmost entry in the third row? I'm thinking we start from the second row, and say that b should not equal to 3 in order for the system to be consistent. I'm not sure how to deal with the constant 'c' afterwards.

Last edited: Jun 14, 2008
4. Jun 14, 2008

### rocomath

Sorry! I'm currently taking this class and we just started going over this stuff. I don't have my book with me so I couldn't tell you. Wait for Dick or Halls of Ivy to respond or someone at their level to answer your question.

5. Jun 14, 2008

### JinM

Oh I see. Thanks for your post.

$$\left[ \begin{array}{cccc} 1 & 2 & -3 & 2 \\ 0 & 0 & b - 3 & 5 \\ 0 & 0 & 0 & 2 - \frac{5c}{b-3} \end{array} \right]$$

For the system to be consistent, we can all be certain that b should NOT equal to 3. Following on the same logic,

$$\frac{5c}{b-3}$$

should equal to 2!

Let's simpify,

$$5c = 2b - 6$$

$$b = \frac{5c + 6}{2}$$

But we already know that b != 3, so

$$\frac{5c + 6}{2} \neq 3$$
$$\Rightarrow c \neq 0$$

I think I might be going in circles here.

Last edited: Jun 14, 2008
6. Jun 14, 2008

### Defennder

What do you mean by "free variables" ? Are you referring to b,c? In general, you would have infinite number of solutions if one or more rows of your final row-reduced augmented nxn matrix is nothing but zeros and the system is consistent. Another way to see it is that there are more unknowns than equations then you would have infinite solutions.

I don't understand your second question.

7. Jun 14, 2008

### Defennder

I assume you're addressing question 1. You have shown that it is necessary that $$2 = \frac{5c}{b-3}$$ holds. Since you only have one equation, you can't solve for specific values of b and c. That itself should be enough.