I've just started learning matrices, but I'm not sure how to go about solving this:
$\begin{pmatrix} 1 & 2 & 0 \\ 2 & 0 & 1\\1&0&2\end{pmatrix}$ $\begin{pmatrix} 0\\ x\\ y\end{pmatrix}$ $=$ $\begin{pmatrix} 6\\2\\4\end{pmatrix}$
I'm told to use the gaussian method, but I'm getting confused with the $\begin{pmatrix} 0\\ x\\ y\end{pmatrix}$ column, every example video online that shows how to solve matrices, does not have such a column, I'm not sure what to do, any help would be appreciated.
On
The most likely explanation for this is a typo, but let's assume that it's not.
That's certainly something I've never seen before but note that the system
$$\begin{pmatrix} 1 & 2 & 0 \\ 2 & 0 & 1\\1&0&2\end{pmatrix}\begin{pmatrix} 0\\ x\\ y\end{pmatrix} = \begin{pmatrix} 6\\ 2\\ 4\end{pmatrix}$$
is entirely equivalent to the system
$$\begin{pmatrix} 2 & 0 \\ 0 & 1\\0&2\end{pmatrix}\begin{pmatrix} x\\ y\end{pmatrix} = \begin{pmatrix} 6\\ 2\\ 4\end{pmatrix}$$
That $0$ essentially just negates the entire first column of matrix it's multiplying.
Can you solve this new system?
On
Multiply out the matrix to obtain a system of linear equations:
$\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1\\1&0&2\end{bmatrix}$ $\begin{bmatrix} 0\\ x\\ y\end{bmatrix}$ $=$ $\begin{bmatrix} 6\\2\\4\end{bmatrix}\equiv$$ \left\{ \begin{array}{c} 1\cdot 0+2x+0y=6 \\ 2\cdot 0+0x+1y=2 \\ 1\cdot 0+0x+2y=4 \end{array} \right. \equiv$$ \left\{ \begin{array}{c} 2x=6 \\ 1y=2 \\ 2y=4 \end{array} \right.$
$$\implies x=3, y=2$$ (assuming there was no typo)
Just proceed with Row Reduction as normal. If it helps, replace the zero with $z$. If the system is consistent you will conclude that $z=0$.
Can you reduce: $$\begin{pmatrix} 1 & 2 & 0 \\ 2 & 0 & 1\\1&0&2\end{pmatrix}\begin{pmatrix} z\\ x\\ y\end{pmatrix} = \begin{pmatrix} 6\\2\\4\end{pmatrix}$$
To