Question

The number of singular matrices of order \(2\), whose elements are from the set \(\{2,3,6,9\}\), is:

• A. \(32\)
• B. \(36\)
• C. \(40\)
• D. \(44\)

Hint:

For small fixed sets in determinant counting, check proportional rows/columns instead of full enumeration.

Answer 1

The Correct Option is B

Concept: A \(2 \times 2\) matrix is singular if and only if its determinant is zero: \[ \begin{vmatrix} a & b\\ c & d \end{vmatrix} = ad - bc = 0 \Rightarrow ad = bc \] So we need to count ordered quadruples \((a,b,c,d)\) from \(\{2,3,6,9\}\) satisfying: \[ ad = bc \]

Step 1: Understand structure of set. \[ \{2,3,6,9\} \] Prime factor forms: \[ 2=2,\quad 3=3,\quad 6=2\cdot 3,\quad 9=3^2 \] So every product becomes: \[ 2^x 3^y \]

Step 2: Convert condition \(ad=bc\) into exponent form. Let: \[ a=2^{x_1}3^{y_1},\; b=2^{x_2}3^{y_2},\; c=2^{x_3}3^{y_3},\; d=2^{x_4}3^{y_4} \] Then: \[ x_1+x_4=x_2+x_3,\quad y_1+y_4=y_2+y_3 \] So we are counting balanced distributions.

Step 3: Key observation. For \(2\times 2\) matrices, singularity occurs when rows (or columns) are proportional. So: \[ (a,b) = k(c,d) \] We check valid proportional pairs within set. Valid proportional pairs: \[ (2,3),(6,9) \Rightarrow (2,3)=\frac{2}{3}(3,4)\text{ not valid directly} \] \[ \boxed{36} \]