Assuming $x,y,z \neq 0$, find the inverse of the following matrix using Gauss-Jordan elimination.
$$\begin{bmatrix} 1&1&1&1\\ 1&1+x&1&1\\ 1&1&1+y&1\\ 1&1&1&1+z \end{bmatrix}$$
I started with this to find the inverse, but I actually don't get which row operations I should do.
$$\left[\begin{array}{cccc|cccc} 1&1&1&1& 1&0&0&0\\ 1&1+x&1&1&0&1&0&0\\ 1&1&1+y&1&0&0&1&0\\ 1&1&1&1+z&0&0&0&1\\ \end{array}\right]$$
As a frst step, you replace the second, the third, and the fourth rows by themselves minus the first row, thereby getting$$\begin{bmatrix}1 & 1 & 1 & 1 & | & 1 & 0 & 0 & 0 \\ 0 & x & 0 & 0 & | & -1 & 1 & 0 & 0 \\ 0 & 0 & y & 0 & | & -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & z & | & -1 & 0 & 0 & 1\end{bmatrix}.$$Then, you divide the second row by $x$, the third row by $y$ and the fourth row by $z$, which will give you:$$\begin{bmatrix}1 & 1 & 1 & 1 & | & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & | & -\frac1x & \frac1x & 0 & 0 \\ 0 & 0 & 1 & 0 & | & -\frac1y & 0 & \frac1y & 0 \\ 0 & 0 & 0 & 1 & | & -\frac1z & 0 & 0 & \frac1z\end{bmatrix}.$$All that remains is to use the second, the third and the fourth rows to eliminate those unwanted $1$'s from the first row. Can you do that?