Exercise 2.7.20

Answers

We have

[\begin{matrix} 1 & 3 \\ 3 & 2 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & - 7 \end{matrix}] [\begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix}];

[\begin{matrix} 1 & b \\ b & c \end{matrix}] = [\begin{matrix} 1 & 0 \\ b & 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & c - b^{2} \end{matrix}] [\begin{matrix} 1 & b \\ 0 & 1 \end{matrix}];

[\begin{matrix} 2 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{matrix}] = [\begin{matrix} 1 \\ - \frac{1}{2} & 1 \\ 0 & - \frac{2}{3} & 1 \end{matrix}] [\begin{matrix} 2 \\ \frac{3}{2} \\ \frac{4}{3} \end{matrix}] [\begin{matrix} 1 & - \frac{1}{2} & 0 \\ 1 & - \frac{2}{3} \\ 1 \end{matrix}] = {LDL}^{T}

mathchakra

2022-01-16 12:28