Analyzing Linear Equations for Infinitely Many Solutions

We are given a system of three linear equations:

1. $x + y + z = 6$
2. $x + 2y + 5z = 10$
3. $2x + 3y + \lambda z = \mu$

For this system to have infinitely many solutions, two conditions must be met:

• The determinant of the coefficient matrix must be zero.
• The system must be consistent (the ranks of the coefficient matrix and the augmented matrix must be equal).

Calculating the Determinant

The coefficient matrix is:

$ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda \end{bmatrix} $

The determinant is calculated as:

$ \det(A) = 1 \begin{vmatrix} 2 & 5 \\ 3 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 5 \\ 2 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} $ $ \det(A) = 1(2\lambda - 15) - 1(\lambda - 10) + 1(3 - 4) $ $ \det(A) = 2\lambda - 15 - \lambda + 10 - 1 $ $ \det(A) = \lambda - 6 $

For infinitely many solutions, $det(A) = 0$.

$ \lambda - 6 = 0 \implies \lambda = 6 $

Checking Consistency with Gaussian Elimination

Substitute $\lambda = 6$ into the augmented matrix and perform row operations:

$ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 5 & | & 10 \\ 2 & 3 & 6 & | & \mu \end{bmatrix} $

Apply $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - 2R_1$:

$ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 4 & | & 4 \\ 0 & 1 & 4 & | & \mu - 12 \end{bmatrix} $

Apply $R_3 \rightarrow R_3 - R_2$:

$ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 4 & | & 4 \\ 0 & 0 & 0 & | & (\mu - 12) - 4 \end{bmatrix} $ $ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 4 & | & 4 \\ 0 & 0 & 0 & | & \mu - 16 \end{bmatrix} $

For infinitely many solutions, the last row must be entirely zeros:

$ \mu - 16 = 0 \implies \mu = 16 $

Final Calculation

We found $\lambda = 6$ and $\mu = 16$.

Therefore, the value of $\lambda + \mu$ is:

$ \lambda + \mu = 6 + 16 = 22 $