设 $dps{A=sex{a{cccc} 1&2&&\ 1&3&&\ &&0&2\ &&-1&0 ea}}$, 且 $dps{sez{sex{frac{1}{2}A}^*}^{-1}BA=6AB+12E}$. 求 $B$.
解答:
(1) 先给出几个性质: $$eex ea detsex{a{cc} X&\ &Y ea} &=det Xcdot det Y,\ sex{a{cc} X&\ &Y ea}sex{a{cc} U&\ &V ea} &=sex{a{cc} XU&\ &YV ea},\ sex{a{cc} U&\ &V ea}^{-1}&=sex{a{cc} U^{-1}&\ &V^{-1} ea}. eea eeex$$
(2) 再做题目. $$eex ea &quad sez{sex{frac{1}{2}A}^*}^{-1}BA=6AB+12E\ & a BA=frac{1}{2}A^*cdot 6AB+frac{1}{2}A^*cdot 12 E =3|A|B+6A^*=6B+6A^*\ & a B(A-6E)=6A^*\ & a B=6A^*(A-6E)^{-1} =6|A|A^{-1}(A-6E)^{-1} =12[(A-6E)cdot A]^{-1}\ &quadquad =12sex{a{cccc} sex{a{cc} -5&2\ 1&-3 ea}sex{a{cc} 1&2\ 1&3 ea}&\ &sex{a{cc} -6&2\ -1&-6 ea}sex{a{cc} 0&2\ -1&0 ea} ea}^{-1}\ &quadquad =12sex{a{cccc} -3&-4&&\ -2&-7&&\ &&-2&-12\ &&6&-2 ea}^{-1}=12sex{a{cccc} -cfrac{7}{13}&cfrac{4}{13}&&\ cfrac{2}{13}&-cfrac{3}{13}&&\ &&-cfrac{1}{38}&cfrac{3}{19}\ &&-cfrac{3}{38}&-cfrac{1}{38} ea}. eea eeex$$