I have the following question here:
Let $A$ be a $4 \times 4$ matrix such that $x=\begin{matrix}[-4\ 0 \ 2 \ -8]^T \end{matrix}$ is a solution to the homogeneous system of linear equations $Ax=0$. Which of the following statements is false?
$(a)$ $det(A)=0$.
$(b)$ The linear system $Ax=b$ is consistent for every $ 4 \times 1$ column vector b.
$(c)$ The reduced row echelon form of $A$ has at least one row of zeroes.
$(d)$ $x=\begin{matrix}[6\ 0 \ -3 \ 12]\end{matrix}$ is also a solution to $Ax=0$
$(e)$ There exists a positive integer r so that the linear system $Ax=0$ has $4-$r free variables.
The answer is supposed to be $(b)$ but I'm just not seeing why. I thought that for a homogeneous system of equations, the solutions were always unique or there were infinitely many solutions.
Why is $(b)$ false? Does the statement not imply that we have a consistent system of equations for any b since it is a homogeneous system?
Also:
Why is $(a)$ true? What does the determinant have to do with this?
For $(c)$, how do we know the matrix has a row of zeroes? That would normally mean we have a free variable but I don't see how that's possible here? Why can we assume there are infinitely many solutions?
Why is $(d)$ true? I thought if the system is consistent, there is only one such solutions. How can $x$ have two sets of solutions?
For $(e)$ is there some sort of theorem I am missing?
My theory for linear algebra is fairly weak as you might think… I am decent at it but I can't just wrap my head around this.
Best Answer
The reason that (b) is false is because the only way for the system to be consistent for any column vector $b$ would be if the four columns of $A$ were linearly independent, which in turn can only be the case if $Ax = 0$ admits only the solution $x = 0$.