Why is the number of pairs with equal to ?

Total possible pairs from is . The pairs where are those where both , which is . Hence, pairs with is .

JEE MainMathematics

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Matrices and Determinants: Mathematics | JEE Main

Let

\text{S} = \{\text{M} = [\text{a}_{\text{ij}}],\ \text{a}_{\text{ij}} \in \{0,1,2\},\ 1 \leq \text{i},\text{j} \leq 2\}

be a sample space and

\text{A} = \{\text{M} \in \text{S}: \text{M}\text{ is invertible}\}

be an event. Then

\text{P}(\text{A})

is equal to

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{50}{81}

\frac{47}{81}

\frac{49}{81}

\frac{16}{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 ·

\frac{50}{81}

None. This is the correct probability. Correct. There are $81$ total matrices and $31$ non-invertible matrices, leaving $50$ invertible matrices.

B ·

\frac{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 $ad = bc$ . For $ad = bc = 0$ , 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 ·

\frac{49}{81}

The student counted $32$ singular matrices instead of $31$ , missing that one configuration was overcounted. List the products $ad = bc \in \{0, 1, 2, 4\}$ . 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 ·

\frac{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 $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ 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., $\det(M) = ad - bc \neq 0$ ).

02approach

Count non-invertible matrices ( $ad - bc = 0 \implies ad = bc$ ) by considering all possible values of the product $ad \in \{0, 1, 2, 4\}$ , then use $P(A) = 1 - P(A')$ .

03execute

Case 1: $ad = bc = 0$ . For $ad = 0$ , 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: $ad = bc = 1$ . Only $(a,d) = (1,1)$ , so $1 \times 1 = 1$ matrix. Case 3: $ad = bc = 2$ . Pairs are $(1,2)$ and $(2,1)$ (2 choices), so $2 \times 2 = 4$ matrices. Case 4: $ad = bc = 4$ . 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: $ad = 0$ (5), $ad = 1$ (1), $ad = 2$ (2), $ad = 4$ (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 $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ 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 $\det(M) = ad - bc = 0$ , and subtract from $1$ .
What are the possible values for the products $ad$ and $bc$ , 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 $\det(M) = ad - bc = 0$ , 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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is the number of pairs with

ad=0

equal to

3^2 - 2^2

Total possible pairs from $\{0,1,2\}$ is $3 \times 3 = 9$ . The pairs where $ad \neq 0$ are those where both $a, d \in \{1,2\}$ , which is $2 \times 2 = 4$ . Hence, pairs with $ad = 0$ 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: $\frac{50}{81}$ - None. This is the correct probability. Correct. There are $81$ total matrices and $31$ non-invertible matrices, leaving $50$ invertible matrices.
B: $\frac{47}{81}$ - 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 $ad = bc$ . For $ad = bc = 0$ , 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: $\frac{49}{81}$ - The student counted $32$ singular matrices instead of $31$ , missing that one configuration was overcounted. List the products $ad = bc \in \{0, 1, 2, 4\}$ . 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: $\frac{16}{27}$ - 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 $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ 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., $\det(M) = ad - bc \neq 0$ ).
approach: Count non-invertible matrices ( $ad - bc = 0 \implies ad = bc$ ) by considering all possible values of the product $ad \in \{0, 1, 2, 4\}$ , then use $P(A) = 1 - P(A')$ .
execute: Case 1: $ad = bc = 0$ . For $ad = 0$ , 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: $\frac{16}{27}$ . Only $(a,d) = (1,1)$ , so $1 \times 1 = 1$ matrix. Case 3: $ad = bc = 2$ . Pairs are $(1,2)$ and $(2,1)$ (2 choices), so $2 \times 2 = 4$ matrices. Case 4: $ad = bc = 4$ . 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: $ad = 0$ (5), $ad = 1$ (1), $ad = 2$ (2), $ad = 4$ (1). Total pairs $(a,d) = 5+1+2+1=9 = 3^2$ , covering all pairs correctly.

Matrices and Determinants: Mathematics | JEE Main

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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