What does $A = L + D + U$ look like? Exercise: 
Consider the problem
\begin{split}
10u_1 + u_2 &= 1\\
u_1 + 10u_2 &= 10
\end{split}
with the solution $(u_1,u_2)^T = (0,1)^T$. For a general system of equations
$$Au = b$$
with an $n\times n$ matrix $A$, which consists of a lower triangular submatrix $L$, the diagonal $D$ and the uppertriangular submatrix $U\,(A=L+D+U)$, a Gauss–Seidel iteration is defined by $$u^{i+1} = D^{-1}(b - Lu^{i+1} - Uu^i)$$
and a Jacobi iteration is defined by $$u^{i+1} = D^{-1}(b - Lu^i - Uu^i)$$
Perform:
a) Four Gauss-Seidel iterations.
b) Four Jacobi iterations.
Question:


*

*What does $A = L + D + U$ mean? I suppose that I doesn't mean that $a_{ij} = l_{ij} + d_{ij} + u_{ij}$? If that is what it's supposed to mean, do we get $$L = \begin{bmatrix}0 & 0\\
1 &0 \end{bmatrix}, \,D = \begin{bmatrix}10 & 0\\
0&10\end{bmatrix},\,U = \begin{bmatrix} 0&1\\
0&0\end{bmatrix} $$


Thanks in advance!
 A: Your $L$, $D$, and $U$ are correct. Note that $a_{ij} = l_{ij} + d_{ij} + u_{ij}$  is true, however, two of the three terms are always zero; the upper right entry would be $a_{ij} = 0 + 0 + 1$.
In general, $\color{red}A = \color{blue}L + \color{orange}D + \color{green}U$ looks like
$$
\color{red}{\begin{pmatrix}
a_{1,1} & a_{1,2} & \ldots & a_{1,n-1} & a_{1,n}
\\ a_{2,1} & a_{2,2} & \ldots & a_{2,n-1} & a_{2,n}
\\ \vdots & \vdots & \ddots & \vdots & \vdots
\\ a_{n-1,1} & a_{n-1,2} & \ldots & a_{n-1,n-1} & a_{n-1,n}
\\ a_{n,1} & a_{n,2} & \ldots & a_{n,n-1} & a_{n,n}
\end{pmatrix}}
=\\
\color{blue}{\begin{pmatrix}
0 & 0 & \ldots & 0 & 0
\\ a_{2,1} & 0 & \ldots & 0 & 0
\\ \vdots & \vdots & \ddots & \vdots & \vdots
\\ a_{n-1,1} & a_{n-1,2} & \ldots & 0 & 0
\\ a_{n,1} & a_{n,2} & \ldots & a_{n,n-1} & 0
\end{pmatrix}}
+\\
\color{orange}{\begin{pmatrix}
a_{1,1} & 0 & \ldots & 0 & 0
\\ 0 & a_{2,2} & \ldots & 0 & 0
\\ \vdots & \vdots & \ddots & \vdots & \vdots
\\ 0 & 0 & \ldots & a_{n-1,n-1} & 0
\\ 0 & 0 & \ldots & 0 & a_{n,n}
\end{pmatrix}}
+\\
\color{green}{\begin{pmatrix}
0 & a_{1,2} & \ldots & a_{1,n-1} & a_{1,n}
\\ 0 & 0 & \ldots & a_{2,n-1} & a_{2,n}
\\ \vdots & \vdots & \ddots & \vdots & \vdots
\\ 0 & 0 & \ldots & 0 & a_{n-1,n}
\\ 0 & 0 & \ldots & 0 & 0
\end{pmatrix}}
$$
Note that in this case $L$ and $U$ are strictly triangular matrices. A triangular matrix may have non-zero entries in its diagonal, e.g.: $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ is an upper triangular matrix (but not a strictly upper triangular matrix).
