Exercise 2.5.3

Answers

We have

[\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} . 5 \\ - . 2 \end{matrix}] and [\begin{matrix} t \\ z \end{matrix}] = [\begin{matrix} - . 2 \\ . 1 \end{matrix}]

A^{- 1} = \frac{1}{10} [\begin{matrix} 5 & - 2 \\ - 2 & 1 \end{matrix}] .

This question solved $A A^{- 1} = I$ column by column, the main idea of Gauss-Jordan elimination. For a different matrix

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

you could find a first column for $A^{- 1}$ but not a second column-so $A$ would be singular (no inverse).

mathchakra

2021-12-19 16:56