Q. 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 :

A) $\frac3{16}$
B) $\frac58$
C) $\frac38$
D) $\frac18$
Explanation

[Solution by Dr. Abubacker] There are $2^4=16$ matrices. $|A|=a_{11}a_{22}-a_{12}a_{21}\ne0$ requires exactly one of the products to be $1$: $a_{11}a_{22}=1$ and $a_{12}a_{21}=0$ ($3$ ways), or the reverse ($3$ ways). So $6$ matrices are invertible and $P(E)=\frac6{16}=\frac38$. Answer (3).

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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