The value of ` lambda and mu ` for which the system of linear equation
`x+y+z=2`
`x+2y+3z=5`
`x+3y+lambda z= mu`
has infinitely many solutions are , respectively :

Options

A. 5 and 8
B. 5 and 7
C. 4 and 9
D. 6 and 8

Answer

To find the values of \( \lambda \) and \( \mu \) for which the given system of linear equations has infinitely many solutions, we can follow these steps:

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \( x + y + z = 2 \)
2. \( x + 2y + 3z = 5 \)
3. \( x + 3y + \lambda z = \mu \)

We can represent this system in augmented matrix form as follows:

\[ \begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 1 & 2 & 3 & | & 5 \\ 1 & 3 & \lambda & | & \mu \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix
For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero. The coefficient matrix is:

\[ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{bmatrix} \]

Calculating the determinant:

\[ \text{Det} = 1 \cdot \begin{vmatrix} 2 & 3 \\ 3 & \lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 3 \\ 1 & \lambda \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 3 \\ 3 & \lambda \end{vmatrix} = 2\lambda - 9 \)
2. \( \begin{vmatrix} 1 & 3 \\ 1 & \lambda \end{vmatrix} = \lambda - 3 \)
3. \( \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = 1 \)

Putting it all together:

\[ \text{Det} = 1(2\lambda - 9) - 1(\lambda - 3) + 1(1) = 2\lambda - 9 - \lambda + 3 + 1 = \lambda + 2 - 9 = \lambda - 5 \]

### Step 3: Set the determinant to zero
For the system to have infinitely many solutions, we set the determinant equal to zero:

\[ \lambda - 5 = 0 \implies \lambda = 5 \]

### Step 4: Substitute \( \lambda \) into the third equation
Now we substitute \( \lambda = 5 \) into the third equation and find \( \mu \) such that the system remains consistent.

The augmented matrix now looks like this:

\[ \begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 1 & 2 & 3 & | & 5 \\ 1 & 3 & 5 & | & \mu \end{bmatrix} \]

### Step 5: Find the condition for \( \mu \)
To ensure consistency, we can perform row operations or find the condition under which the third row is a linear combination of the first two rows.

Using the first two rows, we can express the third row as follows:

From Row 1 and Row 2, we can express the third row as:

\[ \text{Row 3} = \text{Row 1} + 2 \cdot (\text{Row 2} - \text{Row 1}) \]

This gives us:

\[ \mu = 2 + 2(5 - 2) = 2 + 2 \cdot 3 = 2 + 6 = 8 \]

### Final Answer
Thus, the values of \( \lambda \) and \( \mu \) for which the system has infinitely many solutions are:

\[ \lambda = 5, \quad \mu = 8 \]