Let A= bmatrix1 & 0 & 0\ 3 & 1 & 0\ 9 & 3 & 1 bmatrix and B= [bij ], 1 i, j 3. If B=A99-I, then the value of b31-b21b32 is :
- 99
- 199
- 149
- 159
Answer
(3) 149
Solution
Write A=I+N, where N= bmatrix0 & 0 & 0\ 3 & 0 & 0\ 9 & 3 & 0 bmatrix
N2= bmatrix0 & 0 & 0\ 0 & 0 & 0\ 9 & 0 & 0 bmatrix, N3=0 (zero matrix)
IN=NI, the binomial expansion can be used:
A99=I+99N+99C2N2=I+99N+4851N2
B=A99-I=99N+4851N2
b21=99 3=297, b32=99 3=297
b31=99 9+4851 9=9 4950=44550
b31-b21b32= 44550-297297=150-1=149
Answer: (3)