Given the matrix A=$ \left( \begin{array}{ccc} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8\\ 1 & -1 & 2 & 3 \\ 2 & 1 & 1 &2\end{array} \right) $, write it in the $L_{4 \times 4}U_{4 \times 4}$, where L is the lower triangular matrix and U is the upper triangular matrix.
To be honest, I don't even understand what the question is asking of me, however I do know what upper and lower triangular matrices are.
Thanks in advance for your help.
As @Julien pointed out, every square matrix admits a $PLU$ decomposition, where $P$ is a permutation matrix. We have: $A = P \cdot L \cdot U$, such that:
$A=\begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8\\1 & -1 & 2 & 3 \\2 & 1 & 1 &2 \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\\0 & 1 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 & 0\\1 & 1 & 0 & 0\\2 & 1 & 1 & 0\\5 & \dfrac{4}{3} & \dfrac{5}{3} & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 & 3 & 4\\0 & -3 & -1 & -1\\0 & 0 & -4 & -5\\0 & 0 & 0 & -\dfrac{7}{3} \end{bmatrix}$
You could try manually cranking this one to find its $LU$ factorization. We want:
$L \cdot U = \begin{bmatrix} 1 & 0 & 0 & 0\\l_{21} & 1 & 0 & 0\\l_{31} & l_{32} & 1 & 0\\l_{41} & l_{42} & l_{43} & 1 \end{bmatrix} \cdot \begin{bmatrix} u_{11} & u_{12} & u_{13} & u_{14}\\0 & u_{22} &u_{23} & u_{24}\\0 & 0 & u_{33} & u_{34}\\0 & 0 & 0 & u_{44} \end{bmatrix} = \begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8\\1 & -1 & 2 & 3 \\2 & 1 & 1 &2 \end{bmatrix}$
We start off by solving the first row, so we get:
$$u_{11} = 1, u_{12} = 2, u_{13} = 3, u_{14} = 4$$
The portion of the multiplication that determines the remaining entries in the first column of $A$ yields:
$$l_{21}u_{11} = 5 \rightarrow l_{21} = 5$$
$$l_{31}u_{11} = 1 \rightarrow l_{31} = 1$$
$$l_{11}u_{11} = 2 \rightarrow l_{41} = 2$$
At this point rewrite all the variables you solved for and then continue the process and see if you can solve the remaining variables. Of course it is easy to check the result if you can solve all of the equations.
So, we currently have:
$L \cdot U = \begin{bmatrix} 1 & 0 & 0 & 0\\5 & 1 & 0 & 0\\1 & l_{32} & 1 & 0\\2 & l_{42} & l_{43} & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 & 3 & 4\\0 & u_{22} &u_{23} & u_{24}\\0 & 0 & u_{33} & u_{34}\\0 & 0 & 0 & u_{44} \end{bmatrix} = \begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8\\1 & -1 & 2 & 3 \\2 & 1 & 1 &2 \end{bmatrix}$
Try solving for $u_{22},u_{23}, u_{24}$, and then $l_{32}, l_{42}$ and continue this process.
Let's go step by step. We want an equation of the following form:
$$\begin{pmatrix}1&2&3&4\\5&6&7&8\\1&-1&2&3\\2&1&1&2\end{pmatrix}=\begin{pmatrix}\star&0&0&0\\\star&\star&0&0\\\star&\star&\star&0\\\star&\star&\star&\star\end{pmatrix}\begin{pmatrix}\star&\star&\star&\star\\0&\star&\star&\star\\0&0&\star&\star\\0&0&0&\star\end{pmatrix}$$
From the first column and first row of our known matrix, it's not too hard to see that we can start with this:
$$\begin{pmatrix}1&2&3&4\\5&6&7&8\\1&-1&2&3\\2&1&1&2\end{pmatrix}=\begin{pmatrix}1&0&0&0\\5&\star&0&0\\1&\star&\star&0\\2&\star&\star&\star\end{pmatrix}\begin{pmatrix}1&2&3&4\\0&\star&\star&\star\\0&0&\star&\star\\0&0&0&\star\end{pmatrix}$$
Next, we can choose the diagonal elements of our upper triangular matrix to be $1$, and fill in the lower triangular matrix column by column:
$$\begin{pmatrix}1&2&3&4\\5&6&7&8\\1&-1&2&3\\2&1&1&2\end{pmatrix}=\begin{pmatrix}1&0&0&0\\5&-4&0&0\\1&-3&\star&0\\2&-3&\star&\star\end{pmatrix}\begin{pmatrix}1&2&3&4\\0&1&\star&\star\\0&0&1&\star\\0&0&0&1\end{pmatrix}$$
$$\begin{pmatrix}1&2&3&4\\5&6&7&8\\1&-1&2&3\\2&1&1&2\end{pmatrix}=\begin{pmatrix}1&0&0&0\\5&-4&0&0\\1&-3&5&0\\2&-3&1&\star\end{pmatrix}\begin{pmatrix}1&2&3&4\\0&1&2&\star\\0&0&1&\star\\0&0&0&1\end{pmatrix}$$
$$\begin{pmatrix}1&2&3&4\\5&6&7&8\\1&-1&2&3\\2&1&1&2\end{pmatrix}=\begin{pmatrix}1&0&0&0\\5&-4&0&0\\1&-3&5&0\\2&-3&1&7/5\end{pmatrix}\begin{pmatrix}1&2&3&4\\0&1&2&3\\0&0&1&8/5\\0&0&0&1\end{pmatrix}$$