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(D) To find the number of positive integral solutions,we first evaluate the determinant: $\Delta = (1 - \lambda) [\lambda(1 + \lambda) - (-2)(-2)] - 2

Vedclass · Accepted Answer

$1$

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(D) To find the number of positive integral solutions,we first evaluate the determinant: $\Delta = (1 - \lambda) [\lambda(1 + \lambda) - (-2)(-2)] - 2 [(-3)(1 + \lambda) - (-2)(2)] + 1 [(-3)(-2) - \lambda(2)] = 0$ $\Delta = (1 - \lambda) [\lambda + \lambda^2 - 4] - 2 [-3 - 3\lambda + 4] + 1 [6 - 2\lambda] = 0$ $\Delta = (1 - \lambda)(\lambda^2 + \lambda - 4) - 2(1 - 3\lambda) + (6 - 2\lambda) = 0$ $\Delta = (\lambda^2 + \lambda - 4 - \lambda^3 - \lambda^2 + 4\lambda) - 2 + 6\lambda + 6 - 2\lambda = 0$ $\Delta = -\lambda^3 + 5\lambda - 4 - 2 + 6\lambda + 6 - 2\lambda = 0$ $\Delta = -\lambda^3 + 9\lambda = 0$ $\lambda^3 - 9\lambda = 0$ $\lambda(\lambda^2 - 9) = 0$ $\lambda(\lambda - 3)(\lambda + 3) = 0$ The roots are $\lambda = 0, 3, -3$. The positive integral solution is only $\lambda = 3$. Thus,the number of positive integral solutions is $1$.

The number of positive integral solutions of the equation $\left| \begin{array}{ccc} 1 - \lambda & 2 & 1 \\ -3 & \lambda & -2 \\ 2 & -2 & 1 + \lambda \end{array} \right| = 0$ is

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