Question

# Find an LU factorization of A=\begin{bmatrix}h & -4 & -2 & 10 \\h & -9 & 4 & 2 \\ 0 & 0 & -4 & 2\\0 & 1 & 4 & 4 \\ 0 & 0 & 0 & \frac{h}{2} \end{bmatrix}. h=102

Polynomial factorization
Find an LU factorization of $$A=\begin{bmatrix}h & -4 & -2 & 10 \\h & -9 & 4 & 2 \\ 0 & 0 & -4 & 2\\0 & 1 & 4 & 4 \\ 0 & 0 & 0 & \frac{h}{2} \end{bmatrix}$$. h=102

2021-05-02
Step 1
$$A=\begin{bmatrix}102 & -4 & -2 & 10 \\102 & -9 & 4 & 2 \\ 0 & 0 & -4 & 2\\0 & 1 & 4 & 4 \\ 0 & 0 & 0 & 51 \end{bmatrix}$$
In this problem, we have to find matrices L(lower triangular) and U(upper triangular) for which
A=LU
Initial Matrix: $$\begin{bmatrix}1 & 0 & 0 & 0 & 0 \\0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0& 0\\0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}102 & -4 & -2 & 10 \\102 & -9 & 4 & 2 \\ 0 & 0 & -4 & 2\\0 & 1 & 4 & 4 \\ 0 & 0 & 0 & 51 \end{bmatrix}$$
Using some row transformations we convert this initial matrix into LU factorization,
where L will the matrix having all diagonal entries 1, and entries above diagonal 0.
Step 2
by some row transformations, we get the matrix L and matrix U
$$L=\begin{bmatrix}1 & 0 & 0 & 0 & 0 \\1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0& 0\\0 & -\frac{1}{5} & -\frac{13}{10} & 1 & 0 \\ 0 & 0 & 0 & \frac{51}{5} & 1 \end{bmatrix}$$
$$U=\begin{bmatrix}102 & -4 & 2- & 10\\0 & -5 & 6 & -8 \\ 0 & 0 & -4 & 2\\0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
If we multiply L and U, we will get the same matrix A.