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?


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.

  • $\begingroup$ You have not mentioned the option ($e$) in your question. $\endgroup$ Dec 18 '18 at 7:07
  • $\begingroup$ Thanks. I edited. $\endgroup$ Dec 18 '18 at 7:15
  • 1
    $\begingroup$ $Ax=b$ is not homogeneous. $\endgroup$
    – amd
    Dec 18 '18 at 7:49

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 $x = 0$.


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.

  • $\begingroup$ Hey so I edited my question and added the correct choice $(d)$. In your edit, it should say "Statements $(c)$ and $(e)$ are essentially saying the same thing: each...." I get everything you said now but why is choice $(d)$ (The one I just put in) true? I thought that if a system of equations has one solutions, it can't have another. $\endgroup$ Dec 18 '18 at 7:44
  • $\begingroup$ Never mind! It's because any scalar multiple of the solutions is also a valid solution. $\endgroup$ Dec 18 '18 at 7:56
  • $\begingroup$ $(e)$ is actually untrue for $A=O$ $\endgroup$ Dec 18 '18 at 8:06
  • $\begingroup$ @FutureMathperson Exactly! $\endgroup$
    – pwerth
    Dec 18 '18 at 16:02
  • $\begingroup$ @ShubhamJohri Fair point, this probably should have been excluded in the question $\endgroup$
    – pwerth
    Dec 18 '18 at 16:03

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