Let \(A = [a_{ij}]\) be a square matrix of order 2 with elements either 0 or 1. Then the difference between the possible number of singular and non-singular matrices is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4
Let \(A = \begin{bmatrix}a & b\\c & d\end{bmatrix}\) with each entry 0 or 1. Each entry has 2 choices, so there are
\(2^4 = 16 \text{ matrices}\)
A is non-singular when \(|A| = ad - bc \ne 0\). Each of ad and bc is 0 or 1, so \(|A| \ne 0\) only when exactly one of ad, bc is 1.
ad = 1 needs a = d = 1 (one way), and bc = 0 happens in 3 of the 4 choices of (b, c). So
\(|A| = 1:\ 1 \times 3 = 3 \text{ matrices}\)
\(|A| = -1:\ 3 \times 1 = 3 \text{ matrices (in the same way)}\)
\(\text{Non-singular} = 6,\quad \text{Singular} = 16 - 6 = 10\)
\(\text{Difference} = 10 - 6 = 4\)