$$\begin{bmatrix} 4 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ \frac{3}{2} \end{bmatrix}$$
Show that the matrix above is positive definite and solve the system using the decomposition $A = \mathbf L\mathbf D\mathbf L^\top$ with $\mathbf L$ unit triangular and $\mathbf D$ diagonal.
I think I can use the decomposition below:
So we have
$$\begin{bmatrix} 4 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 3 \end{bmatrix} = \begin{bmatrix} D_1 \\ L_{21}D_1 & L_{21}^2D_1+D_2 \\ L_{31}D_1 & L_{31}L_{21}D_1+L_{32}D_2 & L_{31}^2D_1+L_{32}^2D_2 + D_3 \end{bmatrix}$$
But what do I do with the other coefficients of the matrix which are not equal to the right side? I mean, what do I do with $a_{12}, a_{13}$ for example?
And how do I proceed to solve the system?
If you have $LDL^Tx=b$, multiply on the left by $L^{-1}$ to get $$ DL^Tx=L^{-1}b. $$ So you have a system $Dy=L^{-1}b$, with $D$ diagonal; then you simply have $y_j=(L^{-1}b)_j/D_{jj}$ for each $j$. Since $y=L^Tx$, you have $x=(L^{T})^{-1}y$, which you can calculate directly.