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