Let A = begin{bmatrix} b^2+c^2 & a^2 & a^2 b | vedclass

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(D) Given,$a = \sin \frac{\pi}{6} = \frac{1}{2}$,$b = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,and $c = \cot \frac{\pi}{2} = 0$. Substituting these va

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Non-singular matrix

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(D) Given,$a = \sin \frac{\pi}{6} = \frac{1}{2}$,$b = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,and $c = \cot \frac{\pi}{2} = 0$. Substituting these values into the matrix $A$: $A = \begin{bmatrix} (\frac{1}{2})^2 + 0^2 & (\frac{1}{2})^2 & (\frac{1}{2})^2 \ (\frac{1}{\sqrt{2}})^2 & 0^2 + (\frac{1}{2})^2 & (\frac{1}{\sqrt{2}})^2 \ 0^2 & 0^2 & (\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \ \frac{1}{2} & \frac{1}{4} & \frac{1}{2} \ 0 & 0 & \frac{3}{4} \end{bmatrix}$. Now,calculate the determinant $|A|$: $|A| = \frac{1}{4} \left( \frac{1}{4} \cdot \frac{3}{4} - 0 ight) - \frac{1}{4} \left( \frac{1}{2} \cdot \frac{3}{4} - 0 ight) + \frac{1}{4} (0 - 0) = \frac{1}{4} \cdot \frac{3}{16} - \frac{1}{4} \cdot \frac{3}{8} = \frac{3}{64} - \frac{6}{64} = -\frac{3}{64}$. Wait,let us re-evaluate the matrix $A$ construction: $A = \begin{bmatrix} b^2+c^2 & a^2 & a^2 \ b^2 & c^2+a^2 & b^2 \ c^2 & c^2 & a^2+b^2 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/4 & 1/4 \ 1/2 & 1/4 & 1/2 \ 0 & 0 & 3/4 \end{bmatrix}$. Expanding along the third row: $|A| = 0 - 0 + \frac{3}{4} \left( \frac{1}{4} \cdot \frac{1}{4} - \frac{1}{4} \cdot \frac{1}{2} ight) = \frac{3}{4} \left( \frac{1}{16} - \frac{2}{16} ight) = \frac{3}{4} \left( -\frac{1}{16} ight) = -\frac{3}{64} 
eq 0$. Re-checking the question values: $a^2 = 1/4$,$b^2 = 1/2$,$c^2 = 0$. $A = \begin{bmatrix} 1/2 & 1/4 & 1/4 \ 1/2 & 1/4 & 1/2 \ 0 & 0 & 3/4 \end{bmatrix}$. Since $|A| 
eq 0$,it is a non-singular matrix.

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 $a = \sin \frac{\pi}{6}$,$b = \cos \frac{\pi}{4}$,and $c = \cot \frac{\pi}{2}$,then $A$ is:

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