Exercise 4.2.5

Answers

The determinant should be $- 12$ by following processes.

\det (\begin{matrix} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ 2 & 3 & 0 \end{matrix})

= 0 \det (\begin{matrix} 0 & - 3 \\ 3 & 0 \end{matrix}) - 1 \det (\begin{matrix} - 1 & - 3 \\ 2 & 0 \end{matrix}) + 2 \det (\begin{matrix} - 1 & 0 \\ 2 & 3 \end{matrix})

= 0 \times 9 - 1 \times 6 + 2 \times (- 3) = - 12

jephianlin

2011-06-27 00:00