If the system of linear equations
x+ y +z = 5
x+2y +2z = 6
`x + 3y + lambdaz = mu, (lambda, mu in R)` has infinitely many solutions, then the value of `lambda + mu` is

Options

A. 7
B. 9
C. 10
D. 12

Explanation

To solve the system of equations given by:

1. \( x + y + z = 5 \) (Equation 1)
2. \( x + 2y + 2z = 6 \) (Equation 2)
3. \( x + 3y + \lambda z = \mu \) (Equation 3)

We need to find the values of \( \lambda \) and \( \mu \) such that the system has infinitely many solutions. For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the ratios of the determinants formed by the constants must also be equal.

### Step 1: Write the coefficient matrix and the constant matrix

The coefficient matrix \( A \) and the constant matrix \( B \) can be represented as follows:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ 6 \\ \mu \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix \( A \)

We calculate the determinant \( D \) of matrix \( A \):

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we have:

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

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 2 & 2 \\ 3 & \lambda \end{vmatrix} = 2\lambda - 6 \)

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

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

Thus, we have:

\[ D = 1(2\lambda - 6) - 1(\lambda - 2) + 1(1) \]
\[ D = 2\lambda - 6 - \lambda + 2 + 1 = \lambda - 3 \]

### Step 3: Set the determinant to zero for infinitely many solutions

For the system to have infinitely many solutions, we set \( D = 0 \):

\[ \lambda - 3 = 0 \implies \lambda = 3 \]

### Step 4: Calculate the determinants \( D_1 \) and \( D_2 \)

Next, we calculate \( D_1 \) and \( D_2 \) to ensure the ratios are equal.

1. **For \( D_1 \)**:

\[ D_1 = \begin{vmatrix} 5 & 1 & 1 \\ 6 & 2 & 2 \\ \mu & 3 & \lambda \end{vmatrix} \]

Calculating \( D_1 \):

\[ D_1 = 5 \begin{vmatrix} 2 & 2 \\ 3 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 6 & 2 \\ \mu & \lambda \end{vmatrix} + 1 \begin{vmatrix} 6 & 2 \\ \mu & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 2 & 2 \\ 3 & \lambda \end{vmatrix} = 2\lambda - 6 \)

2. \( \begin{vmatrix} 6 & 2 \\ \mu & \lambda \end{vmatrix} = 6\lambda - 2\mu \)

3. \( \begin{vmatrix} 6 & 2 \\ \mu & 3 \end{vmatrix} = 18 - 2\mu \)

Thus, we have:

\[ D_1 = 5(2\lambda - 6) - (6\lambda - 2\mu) + (18 - 2\mu) \]
Substituting \( \lambda = 3 \):

\[ D_1 = 5(2(3) - 6) - (6(3) - 2\mu) + (18 - 2\mu) \]
\[ D_1 = 5(6 - 6) - (18 - 2\mu) + (18 - 2\mu) \]
\[ D_1 = 0 - 18 + 2\mu + 18 - 2\mu = 0 \]

2. **For \( D_2 \)**:

\[ D_2 = \begin{vmatrix} 1 & 5 & 1 \\ 1 & 6 & 2 \\ 1 & \mu & \lambda \end{vmatrix} \]

Calculating \( D_2 \):

\[ D_2 = 1 \begin{vmatrix} 6 & 2 \\ \mu & \lambda \end{vmatrix} - 5 \begin{vmatrix} 1 & 2 \\ 1 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 6 \\ 1 & \mu \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 6 & 2 \\ \mu & \lambda \end{vmatrix} = 6\lambda - 2\mu \)

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

3. \( \begin{vmatrix} 1 & 6 \\ 1 & \mu \end{vmatrix} = \mu - 6 \)

Thus, we have:

\[ D_2 = 1(6\lambda - 2\mu) - 5(\lambda - 2) + (\mu - 6) \]
Substituting \( \lambda = 3 \):

\[ D_2 = 6(3) - 2\mu - 5(3 - 2) + (\mu - 6) \]
\[ D_2 = 18 - 2\mu - 5 + \mu - 6 \]
\[ D_2 = 7 - \mu - 6 = 1 - \mu \]

### Step 5: Set \( D_1 = D_2 = 0 \)

Setting \( D_2 = 0 \):

\[ 1 - \mu = 0 \implies \mu = 1 \]

### Step 6: Find \( \lambda + \mu \)

Now that we have \( \lambda = 3 \) and \( \mu = 1 \):

\[ \lambda + \mu = 3 + 1 = 4 \]

### Final Answer

The value of \( \lambda + \mu \) is \( 4 \).

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