If p and q are two distinct real values of la | vedclass

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(B) The system of equations has a non-zero solution if and only if the determinant of the coefficient matrix $A$ is zero,i.e.,$|A| = 0$.The matrix $A$

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$9$

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(B) The system of equations has a non-zero solution if and only if the determinant of the coefficient matrix $A$ is zero,i.e.,$|A| = 0$.The matrix $A$ is given by:$A = \begin{bmatrix} \lambda-1 & 3\lambda+1 & 2\lambda \ \lambda-1 & 4\lambda-2 & \lambda+3 \ 2 & 3\lambda+1 & 3(\lambda-1) \end{bmatrix}$Setting $|A| = 0$:$\begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda \ \lambda-1 & 4\lambda-2 & \lambda+3 \ 2 & 3\lambda+1 & 3(\lambda-1) \end{vmatrix} = 0$Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$\begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda \ 0 & \lambda-3 & -\lambda+3 \ -\lambda+3 & 0 & \lambda-3 \end{vmatrix} = 0$Factoring out $(\lambda-3)$ from $R_2$ and $R_3$:$(\lambda-3)^2 \begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda \ 0 & 1 & -1 \ -1 & 0 & 1 \end{vmatrix} = 0$Expanding the determinant:$(\lambda-3)^2 [(\lambda-1)(1-0) - (3\lambda+1)(0-1) + 2\lambda(0 - (-1))] = 0$$(\lambda-3)^2 [\lambda-1 + 3\lambda+1 + 2\lambda] = 0$$(\lambda-3)^2 [6\lambda] = 0$Thus,the distinct real values are $\lambda = 3$ and $\lambda = 0$. So,$p=3$ and $q=0$.Finally,$p^2+q^2-pq = 3^2 + 0^2 - (3)(0) = 9 + 0 - 0 = 9$.

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