## System with positive definite matrix

Use Cholesky factorization of $$\mathbf A$$ to solve system $$\mathbf A\mathbf x = (10, 21, -32, 26, 23)^T$$, where

$$\mathbf A=\begin{pmatrix} 1& 2& -3& 2& 1\\ 2& 5& -6& 3& 2\\ -3& -6& 10& -5& -3\\ 2& 3& -5& 15& 11\\ 1& 2& -3& 11& 14 \end{pmatrix}$$

• #### Hint

For $$\mathbf A=\mathbf U^H\mathbf U\mathbf x$$ use the substitution $$\mathbf U\mathbf x=\mathbf y$$.

• #### Resolution

The decomposition is $$\mathbf U= \begin{pmatrix} 1& 2&-3& 2& 1\\ 0& 1& 0&-1& 0\\ 0& 0& 1& 1& 0\\ 0& 0& 0& 3& 3\\ 0& 0& 0& 0& 2 \end{pmatrix}$$.

The system $$\mathbf U^H\mathbf y=\mathbf b$$ yields $$\mathbf y=(10, 1,-2, 3, 2)^T$$ and $$\mathbf U\mathbf x=\mathbf y$$ yields $$\mathbf x=(1, 1, -2, 0, 1)^T$$.

Since both matrices are in the echelon form, it suffices to use twice the backward subtitution.

• #### Result

The system has a solution $$\mathbf x=(1, 1, -2, 0, 1)^T$$.