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I know how to apply that on a 3 by 2 matrix. However, when it comes to a 3 by 3 matrix, I am kind of lost.

For example A is $$ \begin{matrix} 4 & 8 & 1 \\ 0 & 2 & -2 \\ 3 & 6 & 7 \\ \end{matrix} $$ And I know v, w and x. I used the formula to find H_1, which is $$ \begin{matrix} 4/5 & 0 & 3/5 \\ 0 & 4/5 & 0 \\ 3/5 & 0 & -4/5 \\ \end{matrix} $$ And $H_1A$ is $$ \begin{matrix} 5 & 10 & 5 \\ 0 & 8/5 & -8/5 \\ 0 & 0 & -5 \\ \end{matrix} $$ Despite the first column, there are still two columns left. How am I supposed to find the new x and w and how to proceed from here? Thanks!

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Now you forget about the first row and column and compute the reflector for the remaining $2×2$ matrix. This gets extended to a $3×3$ matrix as block-diagonal matrix with the unit matrix in the upper diagonal block.

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    Sorry, can you elaborate it? – 2017-02-09
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    This answer says the same as the more detailed answer to your other question on the topic. In your specific problem, you are done as $H_1A$ already is in upper triangular form. If you need to find a second reflector, it would be the one that flips the sign on the second entry in each column. Please post a source or the formulas for $v$, $w$ and $x$ as these variable names by themselves to not explain much. – 2017-02-09