Upper triangular matrix with big zero in lower triangular?

\documentclass[]{article}
\usepackage{mathtools}
\begin{document}

\[
    \left(
    \begin{array}{ccccc}
    1                                    \\
      & 1             &   & \text{\huge0}\\
      &               & 1                \\
      & \text{\huge0} &   & 1            \\
      &               &   &   & 1
    \end{array}
    \right)
\]  

\end{document}

or \makebox(0,0){\text{\huge0}} if you want to have the same line spacing.

Just in case : you might need repeated dots across the diagonal. Here is a ugly way to do so.

\newcount\dotcnt\newdimen\deltay
\def\Ddot#1#2(#3,#4,#5,#6){\deltay=#6\setbox1=\hbox to0pt{\smash{\dotcnt=1
\kern#3\loop\raise\dotcnt\deltay\hbox to0pt{\hss#2}\kern#5\ifnum\dotcnt<#1
\advance\dotcnt 1\repeat}\hss}\setbox2=\vtop{\box1}\ht2=#4\box2}

And an exemple (using amsmath, of course) :

\[\begin{pmatrix}
1\Ddot{12}.(6pt,-2pt,6pt,-5pt)&1\Ddot8.(9pt,2pt,6pt,0pt)&\quad&\quad&1\\
&&&&\\
&&&&\\
&&&&\\
&\mbox{\Huge 0}&&&\\
&&&&1\\
\end{pmatrix}\]