Exercise 2.5.31

Answers

We have

A^{- 1} = [\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{matrix}]

and

x = A^{- 1} [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix} 2 \\ 2 \\ 2 \\ 1 \end{matrix}] .

When the triangular $A$ alternates $1$ and $- 1$ on its diagonals, $A^{- 1}$ has $1^{'}$ s on the diagonal and first superdiagonal.

mathchakra

2021-12-19 17:38