Question

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

Hint:

To find the number of singular matrices, remember that the determinant condition \( ad = bc \) must hold for each combination of matrix elements. This simplifies to finding matching pairs of products from the given set.

Answer 1

Correct Answer: 36

Let the general form of a \(2 \times 2\) matrix be: \[ \begin{bmatrix} a & b c & d \end{bmatrix} \] The matrix is singular if its determinant is zero: \[ \det = ad - bc = 0 \Rightarrow ad = bc \] Each entry \( a, b, c, d \) is chosen from the set \( \{2, 3, 6, 9\} \), which has 4 elements. 
The total number of \(2 \times 2\) matrices that can be formed is: \[ 4^4 = 256 \] We now count how many of these satisfy \( ad = bc \). 
We do this by checking all possible 4-tuples \( (a, b, c, d) \in \{2, 3, 6, 9\}^4 \), and count those for which \( ad = bc \). 
Using brute-force checking (e.g., via code or enumeration), we find that: \[ \text{Number of singular matrices} = 36 \]

Answer 2

Given determinant: \[ \begin{vmatrix} a & d \\ b & c \end{vmatrix} = ab - bc = 0 \Rightarrow ad = bc \] Case I: Exactly 1 number is used All matrices will be singular. \[ \Rightarrow {}^4C_1 = 4 \] Case II: Exactly 2 numbers are used \[ {}^4C_2 \times 2 \times 2 = 6 \times 4 = 24 \] However, only those with \(ad = bc\) will be singular. So, 6 matrices possible. Case III: Exactly 3 numbers are used None will be singular. \[ \Rightarrow 0 \text{ matrices.} \] --- ### Case IV: Exactly 4 numbers are used For \(ab = cd\): \[ 2 \times 9 = 3 \times 6 \] \[ \Rightarrow {}^4C_1 \times 2! = 8 \text{ matrices.} \] --- Therefore, \[ 4 + 24 + 0 + 8 = 36 \] \[ \boxed{\text{Total number of singular matrices = 36}} \]