A row $[b_1,\dots,b_{d+1}]$ is unimodular if there is $[a_1,\dots,a_{d+1}]$ such that $a_1b_1+\cdots+a_{d+1}b_{d+1}=1$.
Let $A$ be a Noetherian ring of dimension $d$ and $I$ be an ideal generated by $(b_1,\dots ,b_{d+1})$ elements. Now if $\mathrm{height}(I)>d$, then how do we get that the row $[b_1,\dots,b_{d+1}]$ is unimodular?
I have been told that this is a trivial thing but I am not able to understand how. Can someone explain how this is the case?
As we discussed in the comments, the condition $\operatorname{height}(I)>\dim A$ implies $I=A$, as the height of $I$ is the infimum of the heights of all prime ideals in $A$ containing $I$. Thus, we have $$1\in I = (b_1,\dots,b_{d+1})$$ and this means there exist $a_1,\dots,a_{d+1}\in A$ such that $$1 = a_1b_1+\dots+a_{d+1}b_{d+1}$$