JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Probability that a $2 \times 2$ Matrix with Entries from ${0, 1, 2}$ is Invertible

Q: Why is the number of pairs with ad=0 equal to 3^2 - 2^2?

Total possible pairs from {0,1,2} is 3 × 3 = 9. The pairs where ad ≠ 0 are those where both a, d ∈ {1,2}, which is 2 × 2 = 4. Hence, pairs with ad = 0 is 9 - 4 = 5.

Let S = {M = [a_ij], a_ij ∈ {0,1,2}, 1 ≤ i,j ≤ 2} be a sample space and A = {M ∈ S: Mis invertible} be an event. Then P(A) is equal to

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A50/81B47/81C49/81D16/27

Step-by-step solutionView

Correct answer

The total number of matrices is $3^{4} = 81$ , and exactly $31$ have determinant zero, giving $P (A) = 1 - \frac{31}{81} = \frac{50}{81}$ .

Option analysis

Why each option works or fails

A · 50/81

This is the correct probability. Correct. There are $81$ total matrices and $31$ non-invertible matrices, leaving $50$ invertible matrices.

B · 47/81

The student miscounted the singular matrices by counting $34$ instead of $31$ , leading to $1 - \frac{34}{81} = \frac{47}{81}$ . Carefully count cases where $a d = b c$ . For $a d = b c = 0$ , there are $(9) + (9) - 1 = 17$ cases (or $5 \times 5 - 4 \times 4 = 9$ where both are $0$ , etc.). Ensure no double-counting occurs when counting zero products.

C · 49/81

The student counted $32$ singular matrices instead of $31$ , missing that one configuration was overcounted. List the products $a d = b c \in {0, 1, 2, 4}$ . The counts are $17$ for $0$ , $1$ for $1$ , $4$ for $2$ , and $1$ for $4$ , giving $17 + 1 + 4 + 1 + 8 = 31$ singular matrices.

D · 16/27

The student computed the number of invertible matrices as $48$ , leading to $\frac{48}{81} = \frac{16}{27}$ . Recount the non-invertible matrices: there are $31$ singular matrices, giving $81 - 31 = 50$ invertible matrices, not $48$ .

Reviewed route

Solution

StepWorking

01Given

Let $M = [\begin{array}{cc} a & b \\ c & d \end{array}]$ where $a, b, c, d \in {0, 1, 2}$ . Total elements in sample space $n (S) = 3^{4} = 81$ . $A$ is the event that $M$ is invertible (i.e., $\det (M) = a d - b c \neq 0$ ).

02Approach

Count non-invertible matrices ( $a d - b c = 0 ⟹ a d = b c$ ) by considering all possible values of the product $a d \in {0, 1, 2, 4}$ , then use $P (A) = 1 - P (A^{'})$ .

03Execute

Case 1: $a d = b c = 0$ . For $a d = 0$ , the pairs $(a, d)$ are all except $(a, d) \in {1, 2}^{2}$ , so there are $3^{2} - 2^{2} = 5$ choices. Thus, there are $5 \times 5 = 25$ matrices. Case 2: $a d = b c = 1$ . Only $(a, d) = (1, 1)$ , so $1 \times 1 = 1$ matrix. Case 3: $a d = b c = 2$ . Pairs are $(1, 2)$ and $(2, 1)$ (2 choices), so $2 \times 2 = 4$ matrices. Case 4: $a d = b c = 4$ . Only $(a, d) = (2, 2)$ , so $1 \times 1 = 1$ matrix. Total non-invertible matrices $n (A^{'}) = 25 + 1 + 4 + 1 = 31$ .

04Execute

Compute $P (A) = 1 - \frac{n (A^{'})}{n (S)} = 1 - \frac{31}{81} = \frac{50}{81}$ .

✓Verify

Check sum of all cases: $a d = 0$ (5), $a d = 1$ (1), $a d = 2$ (2), $a d = 4$ (1). Total pairs $(a, d) = 5 + 1 + 2 + 1 = 9 = 3^{2}$ , covering all pairs correctly.

Hints that build this answer step by step

What is the most efficient strategy to find the probability that a matrix $M = (\begin{array}{cc} a & b \\ c & d \end{array})$ is invertible, where $a, b, c, d \in {0, 1, 2}$ ?
Find the total number of matrices $∣ S ∣ = 3^{4} = 81$ , count the number of non-invertible matrices where $\det (M) = a d - b c = 0$ , and subtract from $1$ .
What are the possible values for the products $a d$ and $b c$ , and how many pairs $(a, d)$ produce each product?
Product $0$ : $5$ pairs; Product $1$ : $1$ pair; Product $2$ : $2$ pairs; Product $4$ : $1$ pair.
Using the product frequencies, how many matrices have $\det (M) = a d - b c = 0$ , and what is $P (A)$ ?
Number of singular matrices $= 5^{2} + 1^{2} + 2^{2} + 1^{2} = 25 + 1 + 4 + 1 = 31$ , so $P (A) = 1 - \frac{31}{81} = \frac{50}{81}$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is the number of pairs with

a d = 0

equal to

3^{2} - 2^{2}

Total possible pairs from ${0, 1, 2}$ is $3 \times 3 = 9$ . The pairs where $a d \neq 0$ are those where both $a, d \in {1, 2}$ , which is $2 \times 2 = 4$ . Hence, pairs with $a d = 0$ is $9 - 4 = 5$ .

Answer

The total number of matrices is $3^{4} = 81$ , and exactly $31$ have determinant zero, giving $P (A) = 1 - \frac{31}{81} = \frac{50}{81}$ .

Why each option works or fails

A · correct: 50/81 - This is the correct probability. Correct. There are $81$ total matrices and $31$ non-invertible matrices, leaving $50$ invertible matrices.
B: 47/81 - The student miscounted the singular matrices by counting $34$ instead of $31$ , leading to $1 - \frac{34}{81} = \frac{47}{81}$ . Carefully count cases where $a d = b c$ . For $a d = b c = 0$ , there are $(9) + (9) - 1 = 17$ cases (or $5 \times 5 - 4 \times 4 = 9$ where both are $0$ , etc.). Ensure no double-counting occurs when counting zero products.
C: 49/81 - The student counted $32$ singular matrices instead of $31$ , missing that one configuration was overcounted. List the products $a d = b c \in {0, 1, 2, 4}$ . The counts are $17$ for $0$ , $1$ for $1$ , $4$ for $2$ , and $1$ for $4$ , giving $17 + 1 + 4 + 1 + 8 = 31$ singular matrices.
D: 16/27 - The student computed the number of invertible matrices as $48$ , leading to $\frac{48}{81} = \frac{16}{27}$ . Recount the non-invertible matrices: there are $31$ singular matrices, giving $81 - 31 = 50$ invertible matrices, not $48$ .

Step-by-step solution

Given: Let $M = [\begin{array}{cc} a & b \\ c & d \end{array}]$ where $a, b, c, d \in {0, 1, 2}$ . Total elements in sample space $n (S) = 3^{4} = 81$ . $A$ is the event that $M$ is invertible (i.e., $\det (M) = a d - b c \neq 0$ ).
Approach: Count non-invertible matrices ( $a d - b c = 0 ⟹ a d = b c$ ) by considering all possible values of the product $a d \in {0, 1, 2, 4}$ , then use $P (A) = 1 - P (A^{'})$ .
Execute: Case 1: $\frac{48}{81} = \frac{16}{27}$ . For $a d = 0$ , the pairs $(a, d)$ are all except $(a, d) \in {1, 2}^{2}$ , so there are $3^{2} - 2^{2} = 5$ choices. Thus, there are $5 \times 5 = 25$ matrices. Case 2: $a d = b c = 1$ . Only $(a, d) = (1, 1)$ , so $1 \times 1 = 1$ matrix. Case 3: $a d = b c = 2$ . Pairs are $(1, 2)$ and $(2, 1)$ (2 choices), so $a d - b c = 0 ⟹ a d = b c$ matrices. Case 4: $a d = b c = 4$ . Only $(a, d) = (2, 2)$ , so $1 \times 1 = 1$ matrix. Total non-invertible matrices $n (A^{'}) = 25 + 1 + 4 + 1 = 31$ .
Execute: Compute $P (A) = 1 - \frac{n (A^{'})}{n (S)} = 1 - \frac{31}{81} = \frac{50}{81}$ .
Verify: Check sum of all cases: $a d = 0$ (5), $a d = 1$ (1), $a d = 2$ (2), $a d = 4$ (1). Total pairs $(a, d) = 5 + 1 + 2 + 1 = 9 = 3^{2}$ , covering all pairs correctly.

Probability that a $2 \times 2$ Matrix with Entries from ${0, 1, 2}$ is Invertible

What feels right?

The total number of matrices is $3^{4} = 81$ , and exactly $31$ have determinant zero, giving $P (A) = 1 - \frac{31}{81} = \frac{50}{81}$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

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