1. **State the problem:** We need to find the dimension of the space of all upper triangular matrices of order $n$. These are square matrices $A = (a_{ij})$ where $a_{ij} = 0$ for $i > j$, and the entries $a_{ij}$ with $i \leq j$ are arbitrary real numbers. 2. **Understanding the structure:** Every entry above or on the main diagonal can be any real number, and every entry below the diagonal is zero. 3. **Count the number of free entries:** - The matrix is $n \times n$, so total entries are $n^2$. - The zero entries are those where $i > j$, which are below the diagonal. - The free entries correspond to all entries on the main diagonal and above, i.e., where $i \leq j$. 4. **Calculate the number of such entries:** The entries on or above the diagonal are: $$ 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} $$ This formula counts the number of entries in the upper triangle including the diagonal. 5. **Conclusion:** The dimension of the space of all upper triangular $n \times n$ matrices is therefore: $$ \boxed{\frac{n(n+1)}{2}} $$ This matches the option $(n(n+1))/2$.