# $4 \times 4$ matrix and homogeneous system of equations.

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.

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

• 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. Dec 18 '18 at 7:44
• Never mind! It's because any scalar multiple of the solutions is also a valid solution. Dec 18 '18 at 7:56
• $(e)$ is actually untrue for $A=O$ Dec 18 '18 at 8:06
• @FutureMathperson Exactly! Dec 18 '18 at 16:02
• @ShubhamJohri Fair point, this probably should have been excluded in the question Dec 18 '18 at 16:03