Question ID: #415
Let $A=[a_{ij}]$ be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is an invertible matrix. Then the probability $P(E)$ is:
- (1) $\frac{5}{8}$
- (2) $\frac{3}{16}$
- (3) $\frac{3}{8}$
- (4) $\frac{1}{8}$
Solution:
Total number of $2 \times 2$ matrices with entries $\{0, 1\}$ is:
$$ 2^4 = 16 $$
Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. For $A$ to be invertible, its determinant must be non-zero:
$$ |A| = ad – bc \neq 0 $$
Since entries are only 0 or 1, possible values for $ad$ and $bc$ are 0 or 1.
This implies two mutually exclusive cases:
Case 1: $ad = 1$ and $bc = 0$
$$ ad = 1 \Rightarrow a=1, d=1 \quad (1 \text{ way}) $$
$$ bc = 0 \Rightarrow (b, c) \in \{(0,0), (0,1), (1,0)\} \quad (3 \text{ ways}) $$
Total matrices in this case $= 1 \times 3 = 3$.
Case 2: $ad = 0$ and $bc = 1$
$$ bc = 1 \Rightarrow b=1, c=1 \quad (1 \text{ way}) $$
$$ ad = 0 \Rightarrow (a, d) \in \{(0,0), (0,1), (1,0)\} \quad (3 \text{ ways}) $$
Total matrices in this case $= 1 \times 3 = 3$.
Total favorable outcomes $= 3 + 3 = 6$.
Required probability $P(E)$:
$$ P(E) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{6}{16} = \frac{3}{8} $$
Ans. (3)
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