I have to do the Cholesky factorization of the following matrix : $A = \begin{pmatrix} 14 & 23 & 18\\ 23 & 41 & 30\\ 18 & 30 & 36 \end{pmatrix}$.
I found $A = LL^T = \begin{pmatrix} \sqrt{14} & 0& 0\\ 23/\sqrt{14} & 3\sqrt{5}/\sqrt{14}& 0\\ 18/\sqrt{14} & \sqrt{14}/7\sqrt{5} & 8/\sqrt{5} \end{pmatrix} \begin{pmatrix} \sqrt{14} & 23/\sqrt{14}& 18/\sqrt{14}\\ 0 & 3\sqrt{5}/\sqrt{14}& \sqrt{14}/7\sqrt{5}\\ 0 & 0 & 8/\sqrt{5} \end{pmatrix}$.
And on a correction, they found : $A = LL^T = \begin{pmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0\\ 2 & 4 & 0\\ 3 & 5 & 6 \end{pmatrix}$.
I really don't understand their answer... Someone could explain to me ? Thank you in advance...
Let $$\begin{pmatrix} 14 & 23 & 18\\ 23 & 41 & 30\\ 18 & 30 & 36 \end{pmatrix}=\begin{pmatrix} a_{11} & a_{12} & a_{13}\\ 0 & a_{22} & a_{23}\\ 0 & 0 & a_{33} \end{pmatrix} \begin{pmatrix} a_{11} & 0 & 0\\ a_{12} & a_{22} & 0\\ a_{13} & a_{23} & a_{33} \end{pmatrix}=\begin{pmatrix} a_{11}^2+a_{12}^2+a_{13}^2 & a_{12}a_{22}+a_{13}a_{23} &a_{13}a_{33}\\ a_{12}a_{22}+a_{13}a_{23} &a_{22}^2+a_{33}^2 & a_{23}a_{33}\\ a_{33}a_{13} & a_{33}a_{23} & a_{33}^2\end{pmatrix}$$
Now start comparing the matrices elementwise beginning from the last column.