Solving $L_2 = I - xe_2^T$ for $x$

Question

$$L_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} $$

$$e_2^T = \begin{bmatrix} 0 & 1 &0 \\ \end{bmatrix} $$

I need to find the value of $x$ given this equation

$$L_2 = I - xe_2^T$$

I figured out that its a $3 \times 1$ matrix since we need to create a $3 \times 3$ matrix but rest I'm not sure how to find

Answer 1

$$L_2-I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & -1 & 0 \end{bmatrix}$$

Thus you are looking for $$x=\begin{bmatrix}a\\b\\c\end{bmatrix}$$ such that $$xe_2^T = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$

Note that $$xe_2^T= \begin{bmatrix}a\\b\\c\end{bmatrix}[0,1,0]= \begin{bmatrix} 0 & a & 0 \\ 0 & b & 0 \\ 0 & c & 0 \end{bmatrix}$$

Therefore we have $$x=\begin{bmatrix}0\\0\\1\end{bmatrix}=e_3$$