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

[jshayhsei]

Homework Statement

Consider the real system:

<br /> \begin{pmatrix}<br /> 1 &amp; 2 &amp; 3 &amp; 4\\<br /> 4 &amp; 3 &amp; 2 &amp; 1\\<br /> -5 &amp; -5 &amp; -5 &amp; -5<br /> \end{pmatrix}x=\begin{pmatrix}a\\b\\c\end{pmatrix}<br />

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.

Homework Equations

No relevant equations

The Attempt at a Solution

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

<br /> \begin{pmatrix}<br /> 1 &amp; 0 &amp; -1 &amp; -2\\<br /> 0 &amp; 1 &amp; 2 &amp; 3\\<br /> 0 &amp; 0 &amp; 0 &amp; 0<br /> \end{pmatrix}x=\begin{pmatrix}(-c/5)+b+(4/5)c\\-b-(4/5)c\\a+b+c\end{pmatrix}<br />

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

<br /> \begin{pmatrix}1\\1\\1\end{pmatrix}<br />

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.

 

[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?