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If A = [(1,2,2),(2,1,-2),(a,2,b)] is a m...

If A = `[(1,2,2),(2,1,-2),(a,2,b)]` is a matrix satisfying the equation `"AA"^(T) = 9I`, where `I is 3xx3` identity matrix, then the ordered pair (a,b) is equal to

A

(2, -1)

B

(-2, 1)

C

(2, 1)

D

(-2, -1)

Text Solution

Verified by Experts

The correct Answer is:
D

Given, `A = [{:(1, 2, 2), (2, 1, -2), (a,2, b):}], A^(T) = [{:(1, 2, a), (2, 1, 2), (2, -2, b):}]` and
`"AA"^(T) = [{:(1, 2, 2), (2, 1, -2), (a,2, b):}] [{:(1, 2, a), (2, 1, 2), (2, -2, b):}]`
`= [{:(9, 0, a+ 4 +2b), (0, 9, 2a+2-2b), (a+4+2b, 2a+2-2b, a^(2) + 4 +b^(2)):}]`
It is given that, `"AA"^(T) = 9I`
`rArr [{:(9, 0, a+ 4 +2b), (0, 9, 2a+2-2b), (a+4+2b, 2a+2-2b, a^(2) + 4 +b^(2)):}] =9[{:(1, 0, 0), (0, 1, 0), (0, 0, 1):}]`
`rArr [{:(9, 0, a+ 4 +2b), (0, 9, 2a+2-2b), (a+4+2b, 2a+2-2b, a^(2) + 4 +b^(2)):}] =[{:(9, 0, 0), (0, 9, 0), (0, 0, 9):}]`
On comparing, we get
`a +4 +2b = 0 rArr a + 2b = -4 " " ...(i)`
`2a + 2 -2b =0 rArr a-b =-1 " " .... (ii)`
and `a^(2) + 4 + b^(2) = 9 " " ...(iii)`
On solving Eqs. (i) and (ii), we get
a= -2, b =-1
This satisfies Eq. (iii)
Hence, `(a, b) -= (-2, -1)`
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