Answer
$$A^{-1}= \left[ \begin {array}{cccccc} 1&-1&0\\ 0
&1&-1\\ 0&0&1\end {array} \right]
.$$
Work Step by Step
Let $A$ be a matrix given by
$$A=\left[ \begin {array}{cccccc} 1&1&1 \\ 0&1&1 \\ 0&0&1 \end {array} \right].$$
To find $A^{-1}$, we have
$$\left[ A \ \ I \right]=\left[ \begin {array}{cccccc} 1&1&1&1&0&0\\ 0&1&1&0
&1&0\\ 0&0&1&0&0&1\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[I \ \ A^{-1} \right]= \left[ \begin {array}{cccccc} 1&0&0&1&-1&0\\ 0&1&0&0
&1&-1\\ 0&0&1&0&0&1\end {array} \right]
.
$$
Then $A^{-1}$ is given by
$$A^{-1}= \left[ \begin {array}{cccccc} 1&-1&0\\ 0
&1&-1\\ 0&0&1\end {array} \right]
.$$