Krdytkyu

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.

One of the many equivalent conditions for a matrix $A$ to be invertible is that the only solution to $Ax=0$ is the trivial solution i.e. if $Ax=0$ then $x=0$. In this problem, you are given a nontrivial solution to $Ax=0$ so we know $A$ must not be invertible.

This immediately implies that $detA = 0$, so $(a)$ is true. Statements $(c)$ and $(e)$ are essentially saying the same thing: each row of zeroes in the reduced row echelon form of $A$ will correspond to a free variable in the equation $Ax=0$.

This leaves $(b)$. "I thought that for a homogeneous system of equations, the solutions were always unique or there were infinitely many solutions." Yes, this is true if there are solutions. However, there will be some vectors $b$ such that $Ax=b$ does not have a solution. The reason for that is because $A$ will have at most $3$ pivots, not the maximum $4$. Since the columns of $A$ are not linearly independent, there will be vectors $b$ that cannot be expressed as a linear combination of the columns. Hence $(b)$ is false.

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 $x = 0$.

(b) is false because $Ax=[-4;0;2;-8]^T$ is not consistent.