JEE Mains 26-July-2022 Shift 2 Solved Paper

∵A=[

a	b
c	d

] then A2=[

a2+bc	b(a+d)
c(a+d)	bc+d2

]
For A−1 must exist ad−bc≠0...... (i)
and A=A−1⇒A2=I
∴a2+bc=d2+bc=1
and b(a+d)=c(a+d)=0
Case I : When a=d=0, then possible values of (b,c) are (1,1),(−1,1) and (1,−1) and (−1,1).
Total four matrices are possible.
Case II: When a=−d then (a,d) be (1,−1) or (−1,1).
Then total possible values of (b,c) are (12+11)×2=46.
∴ Total possible matrices =46+4=50.