JEE Mathematics 2025 Question with Solution

Given: $A = [a_{ij}]$ is a $2 \times 2$ matrix and each entry is either $0$ or $1$.

Find: The probability that $A$ is invertible.

A $2 \times 2$ matrix is invertible if its determinant is non-zero. The total number of such matrices is

$$2^{2 \times 2} = 2^4 = 16$$

For

$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

the determinant is

$$\det(A) = ad - bc$$

From the solution working, the total number of invertible matrices is $10$. Therefore,

$$P(E) = \frac{\text{Number of invertible matrices}}{\text{Total number of matrices}} = \frac{10}{16} = \frac{5}{8}$$

Therefore, the correct option is C and the probability is $\frac{5}{8}$.

Given: $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ where $a,b,c,d \in \{0,1\}$.

Find: The probability that $A$ is invertible.

Step 1: Count all possible matrices.

$$2^4 = 16$$

Step 2: Use the determinant condition. A matrix is invertible if

$$\det(A) \ne 0$$

that is,

$$ad - bc \ne 0$$

the solution states the favorable count of invertible matrices is $10$.

Step 3: Compute the probability.

$$P(E) = \frac{10}{16} = \frac{5}{8}$$

Therefore, the probability is $\frac{5}{8}$ and the correct option is C.