QJEEM23Ma0b9d: JEE Main 2023 Mathematics - Class 12 JEEM23Ma0b9d Solution: Let S = { M = [ a i j ] , a i j ∈ { 0 , 1 , 2 } , 1 ≤ i , j ≤ 2 …

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

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

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

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

$\therefore a d-b c=0$ or $a d=b c$

Case I : When $a d=b c=0$ , then

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

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

Similarly, there are again five ways, when $b c=0$

$\therefore$ There are total $5 \times 5=25$ ways, when $a d=b c=0$

Case II : When $a d=b c=1$

There is only one way, when $a d=b c=1$

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

Case III : When $a d=b c=2$

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

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

Similarly, there are two ways

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

Case IV : When $a d-b c=4$

There is only way, when $a d=b c=4$

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

$\therefore$ Total number of ways, when

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