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(C) The given system of equations is:$x - 2y + 5z = 3$$2x - y + z = 1$$11x - 7y + pz = q$For the system to have infinitely many solutions,the determin

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$p - q = 2$

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(C) The given system of equations is:$x - 2y + 5z = 3$$2x - y + z = 1$$11x - 7y + pz = q$For the system to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$.$\Delta = \begin{vmatrix} 1 & -2 & 5 \ 2 & -1 & 1 \ 11 & -7 & p \end{vmatrix} = 1(-p + 7) + 2(2p - 11) + 5(-14 + 11) = 0$$-p + 7 + 4p - 22 - 15 = 0$$3p - 30 = 0 \Rightarrow p = 10$Now,for infinitely many solutions,$\Delta_x = \Delta_y = \Delta_z = 0$.$\Delta_x = \begin{vmatrix} 3 & -2 & 5 \ 1 & -1 & 1 \ q & -7 & 10 \end{vmatrix} = 3(-10 + 7) + 2(10 - q) + 5(-7 + q) = 0$$3(-3) + 20 - 2q - 35 + 5q = 0$$-9 + 20 - 35 + 3q = 0$$3q - 24 = 0 \Rightarrow q = 8$Thus,$p - q = 10 - 8 = 2$.

If the system of equations $\begin{bmatrix} 1 & -2 & 5 \\ 2 & -1 & 1 \\ 11 & -7 & p \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ q \end{bmatrix}$ has infinitely many solutions,then:

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