Exercise 2.4.33

Answers

Vector

b = [\begin{matrix} 3 \\ 5 \\ 8 \end{matrix}]

gives

x = 3 x_{1} + 5 x_{2} + 8 x_{3} = [\begin{matrix} 3 \\ 8 \\ 16 \end{matrix}];

The matrix

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

will have those $x_{1} = (1, 1, 1), x_{2} = (0, 1, 1), x_{3} = (0, 0, 1)$ as columns of its "inverse" $A^{- 1}$ .

mathchakra

2021-12-06 18:16