Matrices and Probability
Random matrices and combinatorial probability
GRB_1000_MCQ
Grade Class 12
Question:
Let $S$ be the set of all $3 \times 3$ matrices having 3 entries equal to 1 and 6 entries equal to 0. A matrix $M$ is picked uniformly at random from the set $S$. Then the correct statement(s) is(are):
total number of matrices in the set $S$ is 84
probability that $M$ is non-singular $= \dfrac{1}{14}$
probability that $M$ is identity matrix $= \dfrac{1}{14}$
probability that $M$ has trace equal to $0 = \dfrac{5}{21}$
Step-by-Step Solution
Key Concept: The key idea here is to apply combinatorial principles (combinations for selecting positions, permutations for ordered arrangements) to count different types of $3 \times 3$ matrices based on their specific entry constraints (three 1s, six 0s) and given algebraic properties like singularity, trace, and identity.
Step 1: Count total matrices. We need to place exactly 3 ones in a $3\times 3$ matrix (9 entries). Total = $\binom{9}{3} = 84$. So option (a) is correct.
Step 2: Count non-singular matrices. A $3\times 3$ matrix with exactly one 1 per row and one 1 per column is a permutation matrix, which is non-singular. The number of such matrices = $3! = 6$. So probability of non-singular $= \dfrac{6}{84} = \dfrac{1}{14}$. Option (b) is correct.
Step 3: Check option (c). The identity matrix has 1s on the diagonal — it is one specific permutation matrix. Probability $= \dfrac{1}{84} \neq \dfrac{1}{14}$. Option (c) is incorrect.
Step 4: Count matrices with trace = 0. Trace = 0 means all three 1s are off the diagonal. The diagonal entries are $(1,1), (2,2), (3,3)$. Off-diagonal entries = $9 - 3 = 6$. Number of ways to choose 3 ones from 6 off-diagonal entries $= \binom{6}{3} = 20$. Probability $= \dfrac{20}{84} = \dfrac{5}{21}$. Option (d) is correct.
Correct Answer: 1, 2, 4