Probability Part 1

A=|

a	b
c	d

|‌‌‌‌|A|=ad−bc
Total case =64
For non-singular matrix |A|≠0⇒ad−bc≠0
⇒ad≠bc
And a,b,c,d are all different numbers in the set {1,2,3,4,5,6}
Now for ad=bc
(i) 6×1=2×3
⇒

a=6,b=2,c=3,d=1

or a=1,b=2,c=3,d=6

:

:

:

} 8 such cases
(ii) 6×2=3×4
⇒

a=6,b=3,c=4,d=2

or a=2,b=3,c=4,d=6

:

:

:

} 8 such cases
favourable cases
=‌6C4!−16
required probability
=

‌6C44!−16

64

=

43

162