Question

Let S be the set of all \( 2 \times 2 \) symmetric matrices whose entries are either zero or one. A matrix X is chosen from S. The probability that the determinant of X is not zero is:

• A. \( \frac{1}{3} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{1}{4} \)
• E. \( \frac{2}{9} \)

Hint:

For \(2 \times 2\) matrices, determinant becomes zero when rows or columns become linearly dependent.

Answer 1

The Correct Option is B

Concept: A symmetric \(2 \times 2\) matrix has the form \[ \begin{pmatrix} a & b \\ b & c \end{pmatrix} \] where each entry is either \(0\) or \(1\). The determinant is \[ |A| = ac-b^2 \]

Step 1: Find total number of symmetric matrices.
Since \(a,b,c\) can independently take two values each, \[ \text{Total matrices} = 2^3 = 8 \]

Step 2: Find matrices having zero determinant.
We need \[ ac-b^2=0 \] Case 1: \(b=0\) Then, \[ ac=0 \] Possible cases: \[ (a,c)=(0,0),(1,0),(0,1) \] So, favorable matrices \(=3\). Case 2: \(b=1\) Then, \[ ac-1=0 \Rightarrow ac=1 \] Possible only when \[ a=1,\quad c=1 \] So, favorable matrices \(=1\). Hence total matrices with determinant zero: \[ 3+1=4 \]

Step 3: Calculate probability.
\[ P=\frac{4}{8}=\frac{1}{2} \] \[ \boxed{\frac{1}{2}} \]