D ( 1)t(1234)a11a22a33a44 ( 1)t(1243)a11a22a34a43 ( 1)t(2134)a12a21a33a44
( 1)
t(2143)
a12a21a34a43
a11a22a33a44 a11a22a34a43 a12a21a33a44 a12a21a34a43 从而
a11a21a12a33
a22a43
a34a44
(a11a22 a12a21)(a33a44 a34a43)
a11a22a33a44 a11a22a34a43 a12a21a33a44 a12a21a34a43 D 6.计算
a0
(1)
10
121102111
;
2 144 2 1110
123 n
xa a
110 0
ax a
; (4) Dn 101 0; (3) Dn
aa x
100 1
a00 111 10a0 0 11 1
(5) Dn 00a 0; (6) Dn 1 1 1.
100 a11 1a305
a30
按第4 按第3 0b02a34 4
解:(1) ( 1)d0b0( 1)3 3dc abcd.
12c3行展开0b列展开
12c
000d121112111211021110211102111
(2)
2 14403660366 2 11100331200 36
3b2000c052
; (2) 3d
·4·