Given the matrix
\begin{bmatrix} 1 & 0 & 1 &\bigm| & 0 \\ 0 & 1 & -1 &\bigm| & 0 \\ 0 & -4 & 4 &\bigm| & 0 \end{bmatrix}
Attempt: Using some matrix operations (changing equation 3 by the sum of (4 times equation 2) and (equation 3)), I transformed the given matrix to \begin{bmatrix} 1 & 0 & 1 &\bigm| & 0 \\ 0 & 1 & -1 &\bigm| & 0 \\ 0 & 0 & 0 &\bigm| & 0 \end{bmatrix}
Therefore, $x= -z$ and $y=z$, so the solution is $(-z, z, z).$
Are my solution and answer valid?
Yes, actually if you transform a linear equation $Ax=0$ to $Rx=0$ where \begin{equation} R =\left[\begin{array}{ll} I & F \\ 0 & 0 \end{array}\right] \end{equation} and denote $x$ as \begin{equation} x = \left[\begin{array}{l} x_i \\ x_j \end{array}\right] \end{equation} here $I$ is an identity matrix, $F$ is a block matrix. From above equations, we obtain \begin{align} I x_i + F x_j = 0 \\ x_i = -F x_j \end{align} In this example, \begin{equation} \left[\begin{array}{l} x_1 \\ x_2 \end{array}\right]=-\left[\begin{array}{c} 1 \\ -1 \end{array}\right] x_3=\left[\begin{array}{c} -1 \\ 1 \end{array}\right] x_3 \end{equation} Hence, your solution is valid.
Remark: the above statement is not religious since we omit the dimensions of $I$ and $F$.