第17章 17.5 | 高専の数学解説

(1)

\begin{vmatrix} x & 2 & -1 & 1 \\ y & 1 & 3 & -2 \\ z & 5 & 2 & 4 \\ 1 & 1 & 1 & 1 \end{vmatrix} = 0

\begin{vmatrix} x-1 & 1 & -2 & 0 \\ y+2 & 3 & 5 & 0 \\ z-4 & 1 & -2 & 0 \\ 1 & 1 & 1 & 1 \end{vmatrix} = \begin{vmatrix} x-1 & 1 & -2 \\ y+2 & 3 & 5 \\ z-4 & 1 & -2 \end{vmatrix} = -6(x-1) - 2(y+2) + 5(z-4) - 5(x-1) + 2(y+2) + 6(z-4)

$= -11x + 11z - 33 = 0 \quad x - z + 3 = 0$

\begin{vmatrix} x & 1 & -1 & 3 \\ y & 2 & 1 & 0 \\ z & 3 & -2 & 1 \\ 1 & 1 & 1 & 1 \end{vmatrix} = 0

\begin{vmatrix} x-3 & -2 & -4 & 0 \\ y & 2 & 1 & 0 \\ z-1 & 2 & -3 & 0 \\ 1 & 1 & 1 & 1 \end{vmatrix} = \begin{vmatrix} x-3 & -2 & -4 \\ y & 2 & 1 \\ z-1 & 2 & -3 \end{vmatrix} = -6(x-3) - 8y - 2(z-1) - 2(x-3) - 6y + 8(z-1)

$= -8x - 14y + 6z + 18 = 0 \quad 4x + 7y - 3z - 9 = 0$

\begin{vmatrix} x & 1 & 2 & 0 \\ y & 1 & 0 & -2 \\ z & 0 & -1 & 0 \\ 1 & 1 & 1 & 1 \end{vmatrix} = 0

\begin{vmatrix} x & 1 & 2 & 0 \\ y+2 & 3 & 2 & 0 \\ z & 0 & -1 & 0 \\ 1 & 1 & 1 & 1 \end{vmatrix} = \begin{vmatrix} x & 1 & 2 \\ y+2 & 3 & 2 \\ z & 0 & -1 \end{vmatrix} = -3x + 2z + (y+2) - 6z = 0

$3x - y + 4z - 2 = 0$