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For what value of k is the matrix \(\begin{bmatrix} 2\cos 2\theta & 2\cos 2\theta & 6 \\ 1 -2 \sin^2\theta & 2 \cos^2\theta -1 & 3 \\ k & 2k & 1 \end{bmatrix}\) singular?
Concept:1). The determinant of a Matrix with two Identical rows or columns is equal to 0.2). A matrix is singular if its determinant is 0. A singular square matrix that does not have a matrix inverse. Formula used:Cos2? = cos2? - sin2? = 1 - 2sin2? = 2cos2? - 1Calculation:\(\begin{bmatrix} 2\cos 2? & 2\cos 2? & 6 \\ 1 -2 \sin^2? & 2 \cos^2? -1 & 3 \\ k & 2k & 1 \end{bmatrix}\)? A = \(\begin{vmatrix} 2\cos 2? & 2\cos 2? & 6 \\ 1 -2 \sin^2? & 2 \cos^2? -1 & 3 \\ k & 2k & 1 \end{vmatrix}\)? A = \(\begin{vmatrix} 2\cos 2? & 2\cos 2? & 6 \\ cos\ 2? & cos\ 2? & 3 \\ k & 2k & 1 \end{vmatrix}\)? A = 2\(\begin{vmatrix} \cos 2? & \cos 2? & 3 \\ cos\ 2? & cos\ 2? & 3 \\ k & 2k & 1 \end{vmatrix}\)So, R1 and R2 become the same. ? A = 0.? The value of the determinant is zero for any real value of k.
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