How does $\begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix} \cdot \vec{x} = \vec{0} \rightarrow \vec{x} = S\begin{bmatrix} 0\\1 \end{bmatrix}$?

Question

Where S is a constant in the title.

1. I understand that the statement in the title means, $\vec{x}$ can be any scalar of $\begin{bmatrix} 0\\1 \end{bmatrix}$, but where did $\begin{bmatrix} 0\\1 \end{bmatrix}$ come from?... Do I need to reduce $\begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix}$ to RREF...?

2. I also understand how to work out Linear combinations, which I verified on symbolab is correct.

3. Qn: How did Trefor Bazett(instructor) end up up with $S\begin{bmatrix} 0\\ 1 \end{bmatrix}$? Please can someone show me the steps?

For context, I'm trying to relearn foundation while coming across this part of the video:

Answer 1

Let's start at the beginning: $$ \begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix} \cdot \vec{x} = \vec{0} $$ This can also be written out in a more expanded form as: $$ \begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ Then, if one goes through all of the proper matrix operations on the left-hand-side, as you've done in your main post, we have: $$ \begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_1 \end{bmatrix} $$ and therefore we are looking for solutions where: $$ \begin{bmatrix} x_1 \\ x_1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ which means that $x_1=0$ is strictly defined. However, what does that say about our vector $\vec{x}$? Well, if we plug in the $x_1=0$ that we specifically know: $$ \vec{x} = \begin{bmatrix} 0 \\ x_2 \end{bmatrix} $$ then we are still left with a variable $x_2$ that is not strictly defined but yet will always satisfy the equation no matter what value we give it.

Matching to the form given in the problem then simply involves dividing the entire vector by a scalar to get it on the outside (and renaming it $x_2\to S$).

Answer 2

Thanks @mr_e_man & @1Rock. I believe the steps are:

$\begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix} \cdot \vec{x} = \vec{0} \rightarrow RREF: \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} \cdot \vec{x} = \vec{0} \to x_1 = 0$.

The key seem to be, that we must express $\vec{x}$ in terms of $x_2$. And since $x_2$ is a free column, the value of $x_2$ can be any real number.

thus, $\vec{x}=x_2 \begin{bmatrix} 0\\1 \end{bmatrix} \to \vec{x} = S\begin{bmatrix} 0\\1 \end{bmatrix}$ . This is achieved by replacing $x_2$ with S.