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(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's ru

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$\lambda=2, \mu 
eq 10$

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(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's rule determinants $(D_1, D_2, D_3)$ must be non-zero.First,calculate $D$:$D = \begin{vmatrix} 1 & 1 & 1 \ 3 & 5 & 5 \ 1 & 2 & \lambda \end{vmatrix} = 1(5\lambda - 10) - 1(3\lambda - 5) + 1(6 - 5) = 5\lambda - 10 - 3\lambda + 5 + 1 = 2\lambda - 4$.Setting $D = 0$,we get $2\lambda - 4 = 0$,which implies $\lambda = 2$.Now,substitute $\lambda = 2$ into the system:$x + y + z = 6$ (Eq. $1$)$3x + 5y + 5z = 26$ (Eq. $2$)$x + 2y + 2z = \mu$ (Eq. $3$)Subtract $3 	imes$ (Eq. $1$) from (Eq. $2$): $(3x + 5y + 5z) - 3(x + y + z) = 26 - 3(6) \implies 2y + 2z = 8 \implies y + z = 4$.Subtract (Eq. $1$) from (Eq. $3$): $(x + 2y + 2z) - (x + y + z) = \mu - 6 \implies y + z = \mu - 6$.For the system to have no solution,these two derived equations must be inconsistent:$4 
eq \mu - 6 \implies \mu 
eq 10$.Thus,the condition for no solution is $\lambda = 2$ and $\mu 
eq 10$.

The values of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6$,$3x+5y+5z=26$,and $x+2y+\lambda z=\mu$ has no solution are:

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