Exercise 4.3.13

Answers

1.

For the case

n = 1

we have

\det (\bar{M}) = \bar{M_{11}} = \bar{\det (M)}

. By induction, suppose that $\det (\bar{M}) = \bar{\det (M)}$ for $M$ is a $k \times k$ matrix. For a $(n + 1) \times (n + 1)$ matrix $M$ , we have

\det (\bar{M}) = \sum {j = 1}^{n} {(- 1)}^{i + j} {\bar{M}}_{1 j} \det ({\tilde{\bar{M}}}_{i} j)

= \sum {j = 1}^{n} {(- 1)}^{i + j} {\bar{M}}_{1 j} \det ({\bar{\tilde{M}}}_{i} j) = \bar{\det (M)} .

2.

This is an instant result by

1 = | \det (I) | = | \det (Q Q^{∗}) | = | \det (Q) | | \det (Q^{∗}) | = | \det (Q) |^{2} .

jephianlin

2011-06-27 00:00