The homogeneous system of equations \(\left|\begin{array}{ccc}2 & \alpha+\beta+\gamma+\delta & \alpha \beta+\gamma \delta \\ \alpha+\beta+\gamma+\delta & \alpha(\alpha+\beta)(\gamma+\delta) & \alpha \beta(\gamma+\delta)+\gamma \delta(\alpha+\beta) \\ \alpha \beta+\gamma \delta & \alpha \beta(\gamma+\delta)+\gamma \delta(\alpha+\beta) & 2 \alpha \beta \gamma \delta\end{array}\right|\left|\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\\ \mathrm{z}\end{array}\right|=0\) has non-trivial solutions only if..... (a) \(\alpha+\beta+\gamma+\delta=0\) (b) for any \(\alpha, \beta, \gamma, \delta\) (c) \(\alpha \beta+\gamma \delta=0\) (d) \(\alpha \beta(\gamma+\delta)+\gamma \delta(\alpha+\beta)\)

The given matrix is: \(\begin{bmatrix} 2 & \alpha+\beta+\gamma+\delta & \alpha\beta+\gamma\delta \\ \alpha+\beta+\gamma+\delta & \alpha(\alpha+\beta)(\gamma+\delta) & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) \\ \alpha\beta+\gamma\delta & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) & 2\alpha\beta\gamma\delta \end{bmatrix}\)

Using the cofactor expansion method, let's compute the determinant of the given matrix: \(D = \begin{vmatrix} 2 & \alpha+\beta+\gamma+\delta & \alpha\beta+\gamma\delta \\ \alpha+\beta+\gamma+\delta & \alpha(\alpha+\beta)(\gamma+\delta) & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) \\ \alpha\beta+\gamma\delta & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) & 2\alpha\beta\gamma\delta \end{vmatrix}\) Expanding the determinant along the first row, we get: \(D = 2 * \begin{vmatrix} \alpha(\alpha+\beta)(\gamma+\delta) & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) \\ \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) & 2\alpha\beta\gamma\delta \end{vmatrix} - (\alpha+\beta+\gamma+\delta) * \begin{vmatrix} \alpha+\beta+\gamma+\delta & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) \\ \alpha\beta+\gamma\delta & 2\alpha\beta\gamma\delta \end{vmatrix} + (\alpha\beta+\gamma\delta) * \begin{vmatrix} \alpha+\beta+\gamma+\delta & \alpha(\alpha+\beta)(\gamma+\delta) \\ \alpha\beta+\gamma\delta & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) \end{vmatrix}\) Now let's find the sub-determinants: - First Sub-Determinant: \(S1 = \begin{vmatrix} \alpha(\alpha+\beta)(\gamma+\delta) & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) \\ \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) & 2\alpha\beta\gamma\delta \end{vmatrix} = 2\alpha\beta\gamma\delta(\alpha(\alpha+\beta)(\gamma+\delta) - (\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta)))\) - Second Sub-Determinant: \(S2 = \begin{vmatrix} \alpha+\beta+\gamma+\delta & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) \\ \alpha\beta+\gamma\delta & 2\alpha\beta\gamma\delta \end{vmatrix} = 2\alpha\beta\gamma\delta(\alpha+\beta+\gamma+\delta) - (\alpha\beta+\gamma\delta)(\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta))\) - Third Sub-Determinant: \(S3 = \begin{vmatrix} \alpha+\beta+\gamma+\delta & \alpha(\alpha+\beta)(\gamma+\delta) \\ \alpha\beta+\gamma\delta & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) \end{vmatrix} = (\alpha+\beta+\gamma+\delta)(\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta)) - (\alpha\beta+\gamma\delta)\alpha(\alpha+\beta)(\gamma+\delta)\) Let's substitute these sub-determinants back in the full determinant expression: \(D = 2S1 - (\alpha+\beta+\gamma+\delta)S2 + (\alpha\beta+\gamma\delta)S3\) Now, we need to determine which options cause the determinant to be 0.

Using the determinant expression, let's check which of the options would lead to a non-trivial solution (i.e., determinant D being 0) Option (a): If \(\alpha+\beta+\gamma+\delta = 0\), then S2 becomes 0. Thus, D = 0, which leads to a non-trivial solution. Option (b): Since we have already found a specific condition for D = 0 in option (a), the statement in option (b) is not true. Option (c): If \(\alpha\beta+\gamma\delta = 0\), this only sets S1, S2, and S3 to 0 independently and does not guarantee D = 0. Option (d): This option is incorporated in the sub-determinant expressions; it doesn't guarantee a direct D = 0 by itself.

The given homogeneous system of equations has a non-trivial solution only if option (a) is true: \(\alpha+\beta+\gamma+\delta = 0\)