设A-1=
(24
68),
则A=_____,|4A-1|=____,(AT)-1=______,|A|=_____.
【正确答案】:(-1 1/2
3/4 -1/4),
-128,
(2 6
4 8),
-1/8
本题主要考查的知识点为矩阵的运算及
性质.因为AA*=A*A=|A|E,所以A=|A|(A*)-1,又(A*)-1=(A-1)*,所以
A=|A|(A-1)*=-1/8
(8 -4
-6 2)
=
(-1 1/2
3/4 -1/4)
|4A-1|=42|A-1|=42×(-8)=-128,(AT)-1=(A-1)T=
(2 6
4 8)
设A-1= (24 68), 则A=_____,|4A-1|=____,(AT)-1=______,|A|=_____.
- 2024-11-07 03:12:44
- 线性代数(工)(13175)