Question

Let \[ A= \begin{bmatrix} b^2+c^2 & a^2 & a^2 \\ b^2 & c^2+a^2 & b^2 \\ c^2 & c^2 & a^2+b^2 \end{bmatrix} \] If \[ a=\sin\frac{\pi}{6},\quad b=\cos\frac{\pi}{4},\quad c=\cot\frac{\pi}{2}, \] then \(A\) is

• A. Symmetric matrix
• B. Skew-Symmetric matrix
• C. Singular matrix
• D. Non-singular matrix

Hint:

A square matrix is singular if its determinant is equal to zero. If two rows or two columns become proportional or identical after substitution, the determinant becomes zero.

Answer 1

The Correct Option is C

Step 1: Find the values of \(a,b,c\).
Given, \[ a=\sin\frac{\pi}{6} \] So, \[ a=\frac{1}{2} \] Also, \[ b=\cos\frac{\pi}{4} \] So, \[ b=\frac{1}{\sqrt{2}} \] And, \[ c=\cot\frac{\pi}{2} \] Since \[ \cot\frac{\pi}{2}=0, \] we get \[ c=0 \]

Step 2: Find \(a^2,b^2,c^2\).
\[ a^2=\left(\frac{1}{2}\right)^2=\frac{1}{4} \] \[ b^2=\left(\frac{1}{\sqrt{2}}\right)^2=\frac{1}{2} \] \[ c^2=0^2=0 \]

Step 3: Substitute the values in matrix \(A\).
\[ A= \begin{bmatrix} b^2+c^2 & a^2 & a^2 \\ b^2 & c^2+a^2 & b^2 \\ c^2 & c^2 & a^2+b^2 \end{bmatrix} \] Substituting, \[ A= \begin{bmatrix} \frac{1}{2}+0 & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{2} & 0+\frac{1}{4} & \frac{1}{2} \\ 0 & 0 & \frac{1}{4}+\frac{1}{2} \end{bmatrix} \] Thus, \[ A= \begin{bmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{4} & \frac{1}{2} \\ 0 & 0 & \frac{3}{4} \end{bmatrix} \]

Step 4: Check the determinant of \(A\).
Since the third row has two zeros, \[ |A|=\frac{3}{4} \begin{vmatrix} \frac{1}{2} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{4} \end{vmatrix} \] Now, \[ \begin{vmatrix} \frac{1}{2} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{4} \end{vmatrix} = \frac{1}{2}\cdot \frac{1}{4} - \frac{1}{2}\cdot \frac{1}{4} \] \[ =\frac{1}{8}-\frac{1}{8} \] \[ =0 \] Therefore, \[ |A|=\frac{3}{4}\times 0=0 \]

Step 5: Final conclusion.
Since \[ |A|=0, \] the matrix \(A\) is a singular matrix.
Therefore, \[ \boxed{\text{Singular matrix}} \]