Let (0, ) and A= bmatrix1 & 2 & \ 1 & 0 & 1 \ 0 & 1 & 2 bmatrix. If det ( adj (2A-AT ) adj (A-2AT ) )=28, then ( det(A))2 is equal to:
- 16
- 36
- 49
- 1
Answer
(1) 16
Solution
Let P=A-2AT, Q=2A-AT QT=2AT-A=-P
|Q|=|-P|=(-1)3|P|=-|P|
| adjM|=|M|2 for order 3: |Q|2|P|2=28 |P|4=28 |P|= 4
P= bmatrix1&2& \\1&0&1\\0&1&2 bmatrix- bmatrix2&2&0\\4&0&2\\2 &2&4 bmatrix= bmatrix-1&0& \\-3&0&-1\\-2 &-1&-2 bmatrix
Expanding along C2: |P|=-(-1)(1+3 )=1+3
1+3 = 4 =1 ( >0)
|A|=1(0-1)-2(2-0)+1(1-0)=-4 ( detA)2=16
Answer: (1)