JEE Main 26 Feb 2021 Shift 1 Solved Paper

Let A be the matrix as follows,
A=[

a	b
b	c

], since A is symmetric matrix.
Now, A2=[

a	b
b	c

][

a	b
b	c

]=[

a2+b2	ab+bc
ab+bc	b2+c2

]
Given that, diagonal entries of A2 is 1.
i.e. a2+b2+b2+c2=1 or a2+2b2+c2=1
Case 1. a=0
Then, 2b2+c2=1
(a) c=0 gives, b2=‌

1

2

or b=±‌

1

√2

∴a=0,b=1∕√2,c=0 (2 matrices) a=0,b=−1∕√2,c=0
(b) b=0, gives c2=1 or c=±1
∴a=0,b=0,c=1
and a=0,b=0,c=−1 (2 matrices)
Case 2. b=0, then a2+c2=1
(a) a=0, then c=±1
a=0,b=0,c=1 and a=0,b=0,c=−1
This is repeated case.
(b) c=0, then a=±1
a=1,b=0,c=0 and a=−1,b=0,c=0
Again 2 matrices.
Thus, only acceptable matrices are as follows
A=[

0	0
0	1

],[

0	0
0	−1

],[

1	0
0	0

],[

−1	0
0	0

]
Then possible number of such matrices are 4 .