JEE Main 2023 — Probability Question with Solution

We have, $S=\left\{M=\left[a_{ij}\right],a_{ij}\in \{0,1,2\},1\le i,j\le 2\right\}$

Let $M=\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]$, where $a,b,c,d\in \{0,1,2\}$

$$n(s)=3^4=81$$

If $A$ is invertible, then $|A|\ne 0$

Now, if $|A|=0$, then $|M|=0$

$\therefore ad-bc=0$ or $ad=bc$

Case I : When $ad=bc=0$, then

There are five ways when $ad=0$ i.e.,

$(a,d)=(0,0),(0,1),(0,2),(1,0),(2,0)$

Similarly, there are again five ways, when $bc=0$

$\therefore$ There are total $5\times 5=25$ ways, when $ad=bc=0$

Case II : When $ad=bc=1$

There is only one way, when $ad=bc=1$

$$\text{ i.e. }a=b=c=d=1$$

Case III : When $ad=bc=2$

There are two ways, when $ad=2$, i.e.

$$(a,d)=(1,2)\text{ or }(2,1)$$

Similarly, there are two ways

when $bc=2$ i.e., $(b,c)=(1,2)$ or $(2,1)$

Case IV : When $ad-bc=4$

There is only way, when $ad=bc=4$

$$\text{ i.e., }a=b=c=d=2$$

$\therefore$ Total number of ways, when

$$\begin{array}{rl}& (\bar{A})=\frac{31}{81}|A|=0\text{ is }25+1+4+1=31\\ & \\ & \text{ Hence, }P(A)=1-P(\bar{A})=1-\frac{31}{81}=\frac{50}{81}\end{array}$$