Let the system of equations x+2y+3z=5,2x+3y+z | vedclass

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(B) Given the system of equations:$x+2y+3z=5$$2x+3y+z=9$$4x+3y+\lambda z=\mu$For the system to have an infinite number of solutions,the determinant of

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$17$

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(B) Given the system of equations:$x+2y+3z=5$$2x+3y+z=9$$4x+3y+\lambda z=\mu$For the system to have an infinite number of solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and $\Delta_1 = \Delta_2 = \Delta_3 = 0$.First,calculate $\Delta$:$\Delta = \begin{vmatrix} 1 & 2 & 3 \ 2 & 3 & 1 \ 4 & 3 & \lambda \end{vmatrix} = 1(3\lambda - 3) - 2(2\lambda - 4) + 3(6 - 12) = 0$$3\lambda - 3 - 4\lambda + 8 - 18 = 0$$-\lambda - 13 = 0 \Rightarrow \lambda = -13$Now,calculate $\Delta_1$ using $\lambda = -13$:$\Delta_1 = \begin{vmatrix} 5 & 2 & 3 \ 9 & 3 & 1 \ \mu & 3 & -13 \end{vmatrix} = 5(-39 - 3) - 2(-117 - \mu) + 3(27 - 3\mu) = 0$$5(-42) + 234 + 2\mu + 81 - 9\mu = 0$$-210 + 315 - 7\mu = 0$$105 - 7\mu = 0 \Rightarrow \mu = 15$Finally,calculate $\lambda + 2\mu = -13 + 2(15) = -13 + 30 = 17$.

Let the system of equations $x+2y+3z=5$,$2x+3y+z=9$,and $4x+3y+\lambda z=\mu$ have an infinite number of solutions. Then $\lambda+2\mu$ is equal to:

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