Question

Let the system of equations \[x + 2y + 3z = 5, \quad 2x + 3y + z = 9, \quad 4x + 3y + \lambda z = \mu\]have an infinite number of solutions. Then $\lambda + 2\mu$ is equal to:

• A. 28
• B. 17
• C. 22
• D. 15

Answer 1

The Correct Option is B

For the given system of equations to have an infinite number of solutions, the determinant of the coefficient matrix must be zero, and the system must be consistent.
The coefficient matrix is given by:
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda \end{bmatrix} \]

We compute the determinant of \( A \):
\[ \text{Det}(A) = 1 \times (3\lambda - 3) - 2 \times (2\lambda - 4) + 3 \times (6 - 12) \]
Simplifying each term:
\[ \text{Det}(A) = 3\lambda - 3 - 4\lambda + 8 - 18 = -\lambda + 5 - 18 = -\lambda - 13 \]
For the system to have an infinite number of solutions, we must have:
\[ -\lambda - 13 = 0 \implies \lambda = -13 \]

Consistency Condition
For the system to be consistent, the augmented matrix must have a rank less than 3. The augmented matrix is given by:
\[ [A \mid b] = \begin{bmatrix} 1 & 2 & 3 & 5 \\ 2 & 3 & 1 & 9 \\ 4 & 3 & -13 & \mu \end{bmatrix} \]

The first two rows imply:
\[ x + 2y + 3z = 5, \quad 2x + 3y + z = 9 \]
To ensure consistency, the third equation must be a linear combination of the first two. We find \( \mu \) by substituting \( \lambda = -13 \) and expressing the third equation as a linear combination of the first two:
\[ 4x + 3y - 13z = \alpha(x + 2y + 3z) + \beta(2x + 3y + z) \]
Matching coefficients, we get:
\[ 4 = \alpha + 2\beta, \quad 3 = 2\alpha + 3\beta, \quad -13 = 3\alpha + \beta \]
Solving this system of equations yields:
\[ \alpha = 2, \quad \beta = -1, \quad \mu = 17 \]

Calculating \( \lambda + 2\mu \)
\[ \lambda + 2\mu = -13 + 2 \times 17 = -13 + 34 = 17 \]

Conclusion: \( \lambda + 2\mu = 17 \).