線形代数学1 No.11 2005. 7.13
4.3
行列式の余因子展開（解答）
担当：市原問題 18 行列A=
1 −1 0
2 0 4
0 3 −3
の余因子ea13,ea21,ea32 を計算しなさい.
e a13=
¯¯
¯¯
¯ Ã2 0
0 3
!¯¯
¯¯
¯×(+1) = (2×3−0×0)×(+1) = 6
e a21=
¯¯
¯¯
¯ Ã
−1 0 3 −3
!¯¯
¯¯
¯×(−1) = ((−1)×(−3)−0×(3))×(−1) = −3
e a32=
¯¯
¯¯
¯ Ã1 0
2 4
!¯¯
¯¯
¯×(−1) = (1×4−0×2)×(−1) =−4
問題 19 次の行列の行列式を余因子展開を使って計算しなさい.
(1) W =
1 0 0 4 6 0 3 5 7
|W| = a1×af11
= 1×
¯¯
¯¯
¯ Ã
6 0 5 7
!¯¯
¯¯
¯
= 1×(6×7−0×5)) = 42
(2) X =
1 0 2 0 3 0 5 0 1
|X| = a22×af22
= 3×
¯¯
¯¯
¯ Ã1 2
5 1
!¯¯
¯¯
¯
= 3×(1×1−2×5)) =−27
(3) Y =
0 −1 −4
−1 0 −2
0 3 0
|Y| = a32×af32
= 3× −
¯¯
¯¯
¯ Ã
0 −4
−1 −2
!¯¯
¯¯
¯
= 3× −(0×(−2)−(−4)×(−1))) = 12
(4) Z =
0 1 0 3 1 0 2 0 0 −1 0 0
−2 0 3 1
|z| = a32×af32
= (−1)× −
¯¯
¯¯
¯¯
¯
0 0 3 1 2 0
−2 3 1
¯¯
¯¯
¯¯
¯
= (−1)× − Ã
3×
¯¯
¯¯
¯
à 1 2
−2 3
!¯¯
¯¯
¯
!
= (−1)× −(3×(1×3−2×(−2))) = 21
