Four dice are thrown simultaneously and the numbers shown on these dice are recorded in 2 $\times$ 2 matrices. The probability that such formed matrix have all different entries and are non-singular, is :

$A = \left| {\matrix{ a &amp; b \cr c &amp; d \cr } } \right|$<br><br>| A | = ad $-$ bc<br><br>Total case = 6⁴<br><br>For non-singular matrix | A | $\ne$ 0 $\Rightarrow$ ad $-$ bc $\ne$ 0<br><br>$\Rightarrow$ ad $\ne$ bc<br><br>And a, b, c, d are all different numbers in the set {1, 2, 3, 4, 5, 6}<br><br>Now for ad = bc<br><br>(i) 6 $\times$ 1 = 2 $\times$ 3<br><br>$\Rightarrow$ $$\left. \matrix{ a = 6,b = 2,c = 3,d = 1 \hfill \cr or\,a = 1,b = 2,c = 3,d = 6 \hfill \cr : \hfill \cr : \hfill \cr} \right\}$$ 8 each cases<br><br>(ii) 6 $\times$ 2 = 3 $\times$ 4<br><br>$\Rightarrow$ $$\left. \matrix{ a = 6,b = 3,c = 4,d = 2 \hfill \cr or\,a = 2,b = 3,c = 4,d = 6 \hfill \cr : \hfill \cr : \hfill \cr} \right\}$$ 8 such cases<br><br>favourable cases<br><br>= $^6C_4 \times 4! - 16$<br><br>required probability <br><br>$= {{{^6C_4 \times 4!} - 16} \over {{6^4}}} = {{43} \over {162}}$