Exercise 2.5.25

Answers

We have

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

and

B [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}],

so $B^{- 1}$ does not exist.

mathchakra

2021-12-19 17:31