Exercise 1.14 of the book Rordam, Larsen and Laustsen "An introduction to K-theory for C*-algebras" asks to prove, that 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?".