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The solution of the equation \[\left( {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }&{\cos \theta }\\{ - \sin \theta }&{\cos \theta }&{\sin \theta }\\{ - \cos \theta }&{ - \sin \theta }&{\cos \theta }\end{array}} \right) = 0\], is
A. \[\theta = n\pi \]
B. \[\theta = 2n\pi \pm \dfrac{\pi }{2}\]
C. \[\theta = n\pi \pm {\left( { - 1} \right)^n}\dfrac{\pi }{4}\]
D. \[\theta = 2n\pi \pm \dfrac{\pi }{4}\]

Answer
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Hint: The first element of the first row is multiplied by a secondary \[2 \times 2\] matrix made up of the other \[3 \times 3\]matrix elements that do not belong to the row or column to which your first chosen element belongs in order to get the determinant of the matrix. It is important to note that the top row's elements, aa, bb, and cc, act as scalar multipliers for the matching\[2 \times 2\]matrix. In order to generate a matrix from the leftover components left behind after drawing vertical and horizontal line segments through the scalar aa, the scalar aa must be multiplied.

Formula Used: A matrix's determinant can be located in a number of methods. To make it simple to find the determinant for a \[3 \times 3\]matrix, we must first divide the provided matrix into \[2 \times 2\] determinants.

Complete step by step solution: After solving the determinant, \[2\cos \theta = 0\]
Then \[\theta = 2n\pi \pm \dfrac{\pi }{2}\]

Option ‘B’ is correct

Note: If \[A\] is the matrix, then \[\left| A \right|\]stands for the matrix's determinant. A square matrix is required to discover any matrix, such as the determinant of a \[2 \times 2\] matrix, a \[3 \times 3\]matrix, or a \[n \times n\]matrix. In other words, there should be an equal number of rows and columns in the matrix. The inverse of a matrix, a system of linear equations, and other problems can all be solved with the aid of finding a matrix's determinants. Top-row entries a, b, and c serve as the scalar multipliers for a corresponding \[2 \times 2\]matrix. The remaining elements that were formed when the vertical and horizontal line segments were drawn through an are multiplied by the scalar element in a \[2 \times 2\] matrix.