BITSAT 2018 Paper

As A is an orthogonal matrix, AAT = I
⇒

1

3

[

1	2	2
2	1	−2
a	2	b

] .

1

3

[

1	2	a
2	1	2
2	−2	b

] = [

1	0	0
0	1	0
0	0	1

]
⇒

1

9

[

1	2	2
2	1	−2
a	2	b

] [

1	2	a
2	1	2
2	−2	b

] = [

1	0	0
0	1	0
0	0	1

]
⇒
= [

9	0	0
0	9	0
0	0	9

]
⇒ a + 4 + 2b = 0, 2a + 2 – 2b = 0 anda2 + 4 + b2 = 9
⇒ a + 2b + 4 = 0, a – b + 1 = 0 anda2+b2 = 5
⇒ a = – 2, b = – 1