Inverse of an invertible triangular matrix (either upper or lower) is triangular of the same kind How can we prove that the inverse of an upper (lower) triangular matrix is upper (lower) triangular?

Just to be add a simple answer, I will add another one. Note that, by the $QR$ decomposition, for any invertible upper-triangular matrix $U$, there exist an upper-triangular matrix $R$ and an orthogonal matrix $Q$ so that $U^{-1} = Q R.$ Hence, ${(R U)^{-1}} = Q.$ This implies that, inverting both sides, $$R U = Q^{T}.$$ Since the product of two upper-triangular matrices are still upper triangular,  we have found an  orthogonal upper-triangular matrix, which is impossible unless $Q=D$, for some diagonal Matrix $D$, see this link.. Thus the result follows by noting that $$U^{-1}=Q R = DR,$$ which is upper-triangular.
