I am trying to figure out for what constant $a$ the following system has infinitely many solutions:
\begin{align*}
ax+y+z&=0 \\
x+ay+z&=0 \\
x+y+az&=0
\end{align*}
For $a=-2$ and $a = 1$ the determinant of the coefficient matrix is zero and since the rank of the coefficient matrix is the same as the rank of the augmented matrix, does this mean that for these two values there are infinitely many solutions?
More generally, what I want to ask is: does the determinant of the coefficient matrix being zero mean that there can't be unique solutions?
Best Answer
It's true that if a matrix is singular then any equation with at least one solution has infinitely many. Here is one way to see this.
Let $A$ be a singular matrix, so that $Ax=0$ has a nonzero solution $x_0$. Since $A$ is linear, any multiple of $x_0$ is also a solution, as mentioned in a comment. Thus we know at least there are infinitely many solutions to $Ax=0$. The set of such solutions is called the kernel of the matrix, and it is actually a vector subspace. This is easy to see; if $Ax=0$ and $Ay=0$ and $a,b$ are scalars, then $$A(ax+by)=aAx+bAy=0$$ so any linear combination of two elements of the kernel is still in the kernel.
Let $K$ be the kernel of $A$. Let's say we want to solve $$Ax=y$$ If a solution exists, we can find at least one by methods you're probably familiar with. Call some solution $x_0$. How do we get the rest of the solutions? Well, if we add an element $k$ of $K$ to $x_0$ we get $$A(x_0+k)=Ax_0+Ak=y+0=y$$ Hence $x_0+k$ is also a solution.
It turns out that this in fact gives you every solution. Why is that? Again this is because of linearity. Suppose $x_0$ and $x_1$ are both solutions. Then $$A(x_1-x_0)=Ax_1-Ax_0=y-y=0$$ Hence $x_1-x_0\in K$, and $x_1=x_0+(x_1-x_0)$.