The number of square matrices of order 2, each of whose entries is either 0 or 1, that are singular is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 10
Let the matrix be \(\begin{bmatrix}a & b\\ c & d\end{bmatrix}\) with each entry 0 or 1. Each entry has 2 choices:
\(\text{Total} = 2^{4} = 16\)
It is easier to count the non-singular ones, where \(ad - bc \ne 0\). Each of ad and bc is 0 or 1, so \(ad - bc \ne 0\) only when one product is 1 and the other is 0.
ad = 1 means a = d = 1; bc = 0 then leaves 3 choices for (b, c): (0, 0), (0, 1), (1, 0).
\(ad = 1,\ bc = 0:\ 3\) matrices
In the same way:
\(bc = 1,\ ad = 0:\ 3\) matrices
\(\text{Non-singular} = 3 + 3 = 6\)
\(\text{Singular} = 16 - 6 = 10\)
Why the other options are incorrect
Option (1) 6 — Incorrect. This is the number of non-singular matrices, not singular ones.
Option (2) 8 — Incorrect. Only 6 of the 16 matrices have a non-zero determinant, so 10 are singular, not 8.
Option (4) 12 — Incorrect. The 6 matrices with \(ad - bc = \pm 1\) leave 10 singular ones, not 12.