Exercise 4.2.2

Answers

Determinant is linear when we fixed all but one row. So we have that

\det (\begin{matrix} 3 a_{1} & 3 a_{2} & 3 a_{3} \\ 3 b_{1} & 3 b_{2} & 3 b_{3} \\ 3 c_{1} & 3 c_{2} & 3 c_{3} \end{matrix}) = 3 \det (\begin{matrix} a_{1} & a_{2} & a_{3} \\ 3 b_{1} & 3 b_{2} & 3 b_{3} \\ 3 c_{1} & 3 c_{2} & 3 c_{3} \end{matrix})

= 9 \det (\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ 3 c_{1} & 3 c_{2} & 3 c_{3} \end{matrix}) = 27 \det (\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}) .

Hene we know that $k = 27$ .

jephianlin

2011-06-27 00:00