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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 :

JEE · Math · previous-year question

  1. A.$$\frac{47}{81}$$
  2. B.$$\frac{49}{81}$$
  3. C.$$\frac{50}{81}$$correct
  4. D.$$\frac{16}{27}$$

Answer

C. $$\frac{50}{81}$$

Explanation

We have, $S=\left\{M=\left[a_{i j}\right], a_{i j} \in\{0,1,2\}, 1 \leq i, j \leq 2\right\}$ 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 $|A| \neq 0$ 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 $a d=b c=1$ $$ \text { i.e. } \quad a=b=c=d=1 $$ 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 $$ \begin{aligned} & (\bar{A})=\frac{31}{81}|A|=0 \text { is } 25+1+4+1=31 \\\\ & \text { Hence, } P(A)=1-P(\bar{A})=1-\frac{31}{81}=\frac{50}{81} \end{aligned} $$

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