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(C) For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the augmented matr

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$22$

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(C) For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the augmented matrix must be consistent.First, calculate the determinant $\Delta$ of the coefficient matrix:$\Delta = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 5 \ 2 & 3 & \lambda \end{vmatrix} = 1(2\lambda - 15) - 1(\lambda - 10) + 1(3 - 4) = 0$$2\lambda - 15 - \lambda + 10 - 1 = 0$$\lambda - 6 = 0 \Rightarrow \lambda = 6$.Now, consider the augmented matrix $[A|B] = \begin{pmatrix} 1 & 1 & 1 & | & 6 \ 1 & 2 & 5 & | & 10 \ 2 & 3 & 6 & | & \mu \end{pmatrix}$.Perform row operations: $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - 2R_1$:$\begin{pmatrix} 1 & 1 & 1 & | & 6 \ 0 & 1 & 4 & | & 4 \ 0 & 1 & 4 & | & \mu - 12 \end{pmatrix}$.For infinitely many solutions, the last row must be a zero row, so $\mu - 12 = 4 \Rightarrow \mu = 16$.Thus, $\lambda = 6$ and $\mu = 16$.The value of $\lambda + \mu = 6 + 16 = 22$.

If the system of linear equations: $x + y + z = 6, x + 2y + 5z = 10, 2x + 3y + \lambda z = \mu$ has infinitely many solutions, then the value of $\lambda + \mu$ equals:

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