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Exercise 4.2.4

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See the following process.

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

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

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

The first equality comes from adding one time the second row and one time the third row to the first column and the second equality comes from adding $- 1$ time the first row to the second and the third row. Finally we interchange the second and the third row and multiply the determinant by $- 1$ . Hence $k$ would be $2$ .

jephianlin

2011-06-27 00:00

Exercise 4.2.4

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