Solution
= Solution
Introduce $z=Ux$. Solve $Lz=b$ by <forward substitution> and then $Ux=z$ by <back substitution>. For these factors, the recurrences are
$$
z_1=b_1,\quad z_2=b_2-z_1,\quad z_3=b_3-4z_2,
$$
followed by
$$
x_3=-z_3/4,\quad x_2=z_2-2x_3,\quad x_1=z_1-x_2-x_3.
$$
The nonzero diagonal entries make both triangular solves well-defined. No matrix inverse is calculated.