Exercise 4.2.24

Answers

Let $z$ be the zero row vector. Thus we have that

\det (\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{r - 1} \\ z \\ a_{r + 1} \\ ⋮ \\ a_{n} \end{matrix}) = \det (\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{r - 1} \\ 0 z \\ a_{r + 1} \\ ⋮ \\ a_{n} \end{matrix}) = 0 \det (\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{r - 1} \\ z \\ a_{r + 1} \\ ⋮ \\ a_{n} \end{matrix}) = 0

jephianlin

2011-06-27 00:00