$LDL^t$ Factorization Algorithm to find a factorization of the form A

Question

For $$ \begin{pmatrix} 4 & 2 & 2 \\ 2 & 6 & 2 \\ 2 & 2 & 5 \\ \end{pmatrix} $$ I found that $$ L=\begin{pmatrix} 1 & 0 & 0 \\ 1/2 & 1 & 0 \\ 1/2 & 1/5 & 1 \\ \end{pmatrix} $$ and $$ D=\begin{pmatrix} 4 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 18/5 \\ \end{pmatrix} $$ However, when I calculate $LDL^t$, I always get $$ \begin{pmatrix} 4 & 2 & 2 \\ 2 & 6 & 2 \\ 2 & 2 & 24/5 \\ \end{pmatrix}, $$ which does not equal to Matrix A. Did I make a mistake? but 24/5 and 5 are pretty close. Can someone please help? Thanks!

Answer 1

Using the algorithm located at this site, you should get:

$$L = \left( \begin{array}{ccc} 1 & 0 & 0 \\ \dfrac{1}{2} & 1 & 0 \\ \dfrac{1}{2} & \dfrac{1}{5} & 1 \\ \end{array} \right)$$

$$D = \left( \begin{array}{ccc} 4 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & \dfrac{19}{5} \\ \end{array} \right)$$