高専の数学解説

第16章 16.12

(1)

∣1+a1111+b1111+c∣\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix}

=(1+a)(1+b)(1+c)+1+1−(1+a)−(1+b)−(1+c)= (1+a)(1+b)(1+c) + 1 + 1 - (1+a) - (1+b) - (1+c)

=1+ab+bc+ca+a+b+c+abc+2−3−a−b−c= 1 + ab + bc + ca + a + b + c + abc + 2 - 3 - a - b - c

=ab+bc+ca+abc= ab + bc + ca + abc

=abc(1+1a+1b+1c)= abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)

(2)

∣1+x2xyxzyx1+y2yzzxzy1+z2∣\begin{vmatrix} 1+x^2 & xy & xz \\ yx & 1+y^2 & yz \\ zx & zy & 1+z^2 \end{vmatrix}

=(1+x2)(1+y2)(1+z2)+x2y2z2+x2y2z2−y2z2(1+x2)−x2z2(1+y2)−x2y2(1+z2)= (1+x^2)(1+y^2)(1+z^2) + x^2y^2z^2 + x^2y^2z^2 - y^2z^2(1+x^2) - x^2z^2(1+y^2) - x^2y^2(1+z^2)

=1+x2+y2+z2+x2y2+y2z2+z2x2+x2y2z2+2x2y2z2−y2z2−x2y2z2−x2z2−x2y2z2−x2y2−x2y2z2−x2y2−x2y2z2= 1 + x^2 + y^2 + z^2 + x^2y^2 + y^2z^2 + z^2x^2 + x^2y^2z^2 + 2x^2y^2z^2 - y^2z^2 - x^2y^2z^2 - x^2z^2 - x^2y^2z^2 - x^2y^2 - x^2y^2z^2 - x^2y^2 - x^2y^2z^2

=1+x2+y2+z2= 1 + x^2 + y^2 + z^2

解説: r31bn1z