Upper-triangular matrix is invertible iff its diagonal is invertible: C*-algebra case Exercise 1.14 of the book Rordam, Larsen and Laustsen "An introduction to K-theory for C*-algebras" asks to prove that an upper triangular matrix with elements from some C*-algebra $A$ is invertible in $M_n(A)$ iff all diagonal entries are invertible in $A$.
Trying to solve this I've found that if $a$ is invertible and $\delta$ is such that $(a^{-1}\delta)^n=0$ then $a+\delta$ is invertible too and its inverse is given by $(a+\delta)^{-1}=\sum_{k=0}^{n} (-a^{-1}\delta)^ka^{-1}$. Using this fact I can show that if diagonal is invertible, then upper-triangular matrix with this diagonal is invertible too, and also that if upper-triangular matrix has upper-triangular inverse, then its diagonal is invertible. So all I need to prove is that if upper-triangular matrix invertible, then its inverse is upper-triangular. I've failed to prove this.
Also there is a hint for this exercise: "Solve the equation $ab=1$ where $a$ is as above [i.e. upper-triangular matrix] and where $b$ is unknown upper triangular matrix". Solution of this equation follows from my reasoning above, but this doesn't help.
Update (counterexample attempt): I've made one more attempt and it looks for me like I have found a counterexample. However I think there is a mistake in it (because otherwise there is a mistake in the book). Here it is. Let $A=B(l^2(\mathbb{N}))$ --- algebra of bounded operators on sequences $x=\{x_i\}_ {i=1}^ \infty:\|x\|^2=\sum_{x=1}^{\infty}|x_i|^2<\infty$. Let $z\in A$ be defined by $(zx)_ {2n-1}=0$, $(zx)_{2n}=x_n$, and $t\in A$ be defined by $(tx)_{2n-1}=x_n$, $(tx)_ {2n}=0$. Then we have $t^*t=z^ * z=tt^* +zz^* =1$, $t^* z=z^* t=0$. From these we have that
$$\begin{pmatrix}z&tz^* \\\ 0&t^* \end{pmatrix}\begin{pmatrix}z^* &0\\\ zt^* &t\end{pmatrix}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$$ and
$$\begin{pmatrix}z^* &0\\\ zt^* &t\end{pmatrix}\begin{pmatrix}z&tz^* \\\ 0&t^* \end{pmatrix}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}.$$ 
So now my question should say "Where am I wrong?".
 A: So, the exercise is incorrect as stated, as the nice example in the question shows.  They probably meant to say that the matrix is invertible in the subalgebra of upper triangular matrices if and only if the diagonal entries are invertible.  This is the version given on page 16 in a set of lecture notes by Matthes and Szymański based primarily on the same book.  They also give a counterexample to the original statement.
A: Hi sorry to resurrect this post. I have a relevant example which I want to remember, and this seems like an appropriate place to put it.

Let $H$ be a separable Hilbert space with orthonormal basis $\{e_0,e_1,e_2,\ldots\}$. Let $S \in B(H)$ be the unilateral shift determined by $S(e_i) = e_{i+1}$ for all $i$. Let $P_0 \in B(H)$ be rank-1 projection onto the span of $e_0$. Then, $U = \begin{pmatrix} S & P_0 \\ 0 & S^* \end{pmatrix}$ is a unitary in $M_2(B(H))$, despite being nondiagonal, upper-triangular, and having both diagonal entries noninvertible. 

It's easy to check $U^* U = UU^* = 1$ directly, but the "real" reason this works is that, under the isomorphism $M_2(B(H)) \cong B(H^2)$, $U$ works on the natural basis  of $H^2$ by
$$ \cdots \mapsto (0,e_2) \mapsto (0,e_1) \mapsto (0,e_0) \mapsto (e_0,0) \mapsto (e_1,0) \mapsto (e_2,0) \mapsto (e_3,0) \mapsto \cdots $$
so $U$ is actually (up to a unitary conjugacy) the bilateral shift. 
