Let S=M=[a_ij],a_ij 0,1,2,1 i,j 2 be a sample space and A=M S:M is invertible be an event.

Question

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

Accepted Answer

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

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

Solutions · 1