Exercise 4.1.11

Answers

By property (ii) we have the fact

δ (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}) = 0 = δ (\begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix}) .

Since by property (i) and (ii) we know

0 = δ (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}) = δ (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}) + δ (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix})

= δ (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}) + δ (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) + δ (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) + δ (\begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix})

δ (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) + δ (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}),

we get that

δ (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) = - δ (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) = - 1

by property (iii). Finally, by property (i) and (iii) we can deduce the general formula for $δ$ below.

δ (\begin{matrix} a & b \\ c & d \end{matrix}) = aδ (\begin{matrix} 1 & 0 \\ c & d \end{matrix}) + bδ (\begin{matrix} 0 & 1 \\ c & d \end{matrix})

= acδ (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}) + adδ (\begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix}) + bcδ (\begin{matrix} 0 & 1 \\ c & d \end{matrix}) + bdδ (\begin{matrix} 0 & 1 \\ c & d \end{matrix})

= ad - bc = \det δ (\begin{matrix} a & b \\ c & d \end{matrix}) .

jephianlin

2011-06-27 00:00