If the system of equations `x+y+z=6, x+2y+lambdaz=10` and `x+2y+3z=mu` has infinite solutions, then the value of `lambda+2mu` is equal to

Options

A. 20
B. 22
C. 23
D. 25

Answer

To solve the problem, we need to determine the values of \( \lambda \) and \( \mu \) such that the system of equations has infinite solutions. The equations given are:

1. \( x + y + z = 6 \)
2. \( x + 2y + \lambda z = 10 \)
3. \( x + 2y + 3z = \mu \)

For the system to have infinite solutions, the determinant of the coefficients must be zero, which can be expressed as:

\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda \\ 1 & 2 & 3 \end{vmatrix} = 0 \]

### Step 1: Calculate the determinant \( \Delta \)

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

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

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 2 & \lambda \\ 2 & 3 \end{vmatrix} = (2)(3) - (2)(\lambda) = 6 - 2\lambda \)
2. \( \begin{vmatrix} 1 & \lambda \\ 1 & 3 \end{vmatrix} = (1)(3) - (1)(\lambda) = 3 - \lambda \)
3. \( \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = (1)(2) - (1)(2) = 0 \)

Substituting these back into the determinant:

\[ \Delta = 1(6 - 2\lambda) - 1(3 - \lambda) + 1(0) \]

This simplifies to:

\[ \Delta = 6 - 2\lambda - 3 + \lambda = 3 - \lambda \]

Setting the determinant equal to zero for infinite solutions:

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

### Step 2: Calculate \( \Delta_1 \) for \( \mu \)

Next, we need to find \( \mu \) using the determinant of the modified matrix where the first column is replaced with the constants from the right-hand side:

\[ \Delta_1 = \begin{vmatrix} 6 & 1 & 1 \\ 10 & 2 & \lambda \\ \mu & 2 & 3 \end{vmatrix} \]

Calculating this determinant:

\[ \Delta_1 = 6 \cdot \begin{vmatrix} 2 & \lambda \\ 2 & 3 \end{vmatrix} - 1 \cdot \begin{vmatrix} 10 & \lambda \\ \mu & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 10 & 2 \\ \mu & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 2 & \lambda \\ 2 & 3 \end{vmatrix} = 6 - 2\lambda \)
2. \( \begin{vmatrix} 10 & \lambda \\ \mu & 3 \end{vmatrix} = 30 - 3\lambda - 10\mu \)
3. \( \begin{vmatrix} 10 & 2 \\ \mu & 2 \end{vmatrix} = 20 - 2\mu \)

Substituting these back into \( \Delta_1 \):

\[ \Delta_1 = 6(6 - 2\lambda) - (30 - 3\lambda - 10\mu) + (20 - 2\mu) \]

Substituting \( \lambda = 3 \):

\[ \Delta_1 = 6(6 - 2 \cdot 3) - (30 - 3 \cdot 3 - 10\mu) + (20 - 2\mu) \]

This simplifies to:

\[ \Delta_1 = 6(0) - (30 - 9 - 10\mu) + (20 - 2\mu) \]
\[ = - (21 - 10\mu) + (20 - 2\mu) = -21 + 10\mu + 20 - 2\mu = 8\mu - 1 \]

Setting \( \Delta_1 = 0 \):

\[ 8\mu - 1 = 0 \implies 8\mu = 1 \implies \mu = \frac{1}{8} \]

### Step 3: Calculate \( \lambda + 2\mu \)

Now we can find \( \lambda + 2\mu \):

\[ \lambda + 2\mu = 3 + 2 \cdot \frac{1}{8} = 3 + \frac{2}{8} = 3 + \frac{1}{4} = \frac{12}{4} + \frac{1}{4} = \frac{13}{4} \]

Thus, the final answer is:

\[ \lambda + 2\mu = \frac{13}{4} \]