Checkerboard inverse of a totally nonnegative matrix ($(-1{)}^{i+j}(A^{-1}{)}_{ij}\ge 0$)

If an invertible square totally nonnegative matrix $A$ has size $n$, then $\det A>0$ andIndeed the adjugate identity gives $(A^{-1}{)}_{ij}=(-1{)}^{i+j}\det A_{\hat{j},\hat{i}}/\det A$, and the deleted-row/deleted-column minor is nonnegative. Consequently an alternating vector $v_j=(-1{)}^j$ gives $(A^{-1}v{)}_i=(-1{)}^i{\sum }_j|(A^{-1}{)}_{ij}|$.