# Consider the real system of matrix A, determine all solutions.

1. Sep 17, 2014

### jshayhsei

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

Consider the real system:

$$\begin{pmatrix} 1 & 2 & 3 & 4\\ 4 & 3 & 2 & 1\\ -5 & -5 & -5 & -5 \end{pmatrix}x=\begin{pmatrix}a\\b\\c\end{pmatrix}$$

and denote the system by matrix A.

1. What is the rank of A? By inspection, determine a non-zero vector in the null space of A-transpose.
2. Under what conditions on the numbers a, b, c is the system solvable? (Use your solution of part a).
3. Determine all solutions of the system for a = b = 1, c = -2.

2. Relevant equations

No relevant equations

3. The attempt at a solution

For part 1, I reduced to row echelon and got:

$$\begin{pmatrix} 1 & 0 & -1 & -2\\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 0 \end{pmatrix}x=\begin{pmatrix}(-c/5)+b+(4/5)c\\-b-(4/5)c\\a+b+c\end{pmatrix}$$

I found that rank = 2. I also found that the vector for the null space of A-transpose is:

$$\begin{pmatrix}1\\1\\1\end{pmatrix}$$

For part 2, I said that the solution is solvable under the condition that a+b+c = 0 because there is only one row with all zeros.

I am stumped on part 3. I thought that I should just plug a = b = 1, c = -2 into my reduced row echelon, giving me infinitely many solutions. I don't think this is correct though. Any advice would be helpful. Thanks.

2. Sep 17, 2014

### LCKurtz

Why don't you think there should be infinitely many solutions? You have more unknowns than equations, so it wouldn't be surprising if there were some free variables, would it?