Let $$S=\left\{M=\left[a_{i j}\right], a_{i j} \in\{0,1,2\}, 1 \leq i, j \leq 2\right\}$$ be a sample space and $A=\{M \in S: M$ is invertible $\}$ be an event. Then $P(A)$ is equal to :

We have, $S=\left\{M=\left[a_{i j}\right], a_{i j} \in\{0,1,2\}, 1 \leq i, j \leq 2\right\}$ <br/><br/>Let $M=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, where $a, b, c, d \in\{0,1,2\}$ <br/><br/>$n(s)=3^4=81$ <br/><br/>If $A$ is invertible, then $|A| \neq 0$ <br/><br/>Now, if $|A|=0$, then $|M|=0$ <br/><br/>$\therefore a d-b c=0$ or $a d=b c$ <br/><br/><b>Case I :</b> When $a d=b c=0$, then <br/><br/>There are five ways when $a d=0$ i.e., <br/><br/>$(a, d)=(0,0),(0,1),(0,2),(1,0),(2,0)$ <br/><br/>Similarly, there are again five ways, when $b c=0$ <br/><br/>$\therefore$ There are total $5 \times 5=25$ ways, when $a d=b c=0$ <br/><br/><b>Case II :</b> When $a d=b c=1$ <br/><br/>There is only one way, when $a d=b c=1$ <br/><br/>$\text { i.e. } \quad a=b=c=d=1$ <br/><br/><b>Case III :</b> When $a d=b c=2$ <br/><br/>There are two ways, when $a d=2$, i.e. <br/><br/>$(a, d)=(1,2) \text { or }(2,1)$ <br/><br/>Similarly, there are two ways <br/><br/>when $b c=2$ i.e., $(b, c)=(1,2)$ or $(2,1)$ <br/><br/><b>Case IV :</b> When $a d-b c=4$ <br/><br/>There is only way, when $a d=b c=4$ <br/><br/>$\text { i.e., } a=b=c=d=2$ <br/><br/>$\therefore$ Total number of ways, when <br/><br/>$$ \begin{aligned} & (\bar{A})=\frac{31}{81}|A|=0 \text { is } 25+1+4+1=31 \\\\ & \text { Hence, } P(A)=1-P(\bar{A})=1-\frac{31}{81}=\frac{50}{81} \end{aligned} $$