If the system of equations x+y+z=2,2x+4y-z=6, | vedclass

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(C) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the augmente

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$2\lambda+\mu=14$

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(C) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the augmented determinants $\Delta_x, \Delta_y, \Delta_z$ must also be $0$.First,calculate $\Delta = 0$:$\Delta = \begin{vmatrix} 1 & 1 & 1 \ 2 & 4 & -1 \ 3 & 2 & \lambda \end{vmatrix} = 1(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 0$$4\lambda + 2 - 2\lambda - 3 - 8 = 0$$2\lambda - 9 = 0 \Rightarrow \lambda = \frac{9}{2}$.Next,calculate $\Delta_x = 0$:$\Delta_x = \begin{vmatrix} 2 & 1 & 1 \ 6 & 4 & -1 \ \mu & 2 & \lambda \end{vmatrix} = 2(4\lambda + 2) - 1(6\lambda + \mu) + 1(12 - 4\mu) = 0$Substitute $\lambda = \frac{9}{2}$:$2(18 + 2) - (27 + \mu) + 12 - 4\mu = 0$$40 - 27 - \mu + 12 - 4\mu = 0$$25 - 5\mu = 0 \Rightarrow \mu = 5$.Now,check the options with $\lambda = 4.5$ and $\mu = 5$:$2\lambda + \mu = 2(4.5) + 5 = 9 + 5 = 14$.Thus,the correct option is $C$.

If the system of equations $x+y+z=2$,$2x+4y-z=6$,and $3x+2y+\lambda z=\mu$ has infinitely many solutions,then:

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