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

To solve this problem, we need to determine under what conditions the given system of equations can have an infinite number of solutions. The system is as follows:

\(x + 2y + 3z = 5 \quad \text{(Equation 1)}\)
\(2x + 3y + z = 9 \quad \text{(Equation 2)}\)
\(4x + 3y + \lambda z = \mu \quad \text{(Equation 3)}\)

For the system of linear equations to have infinite solutions, the equations must be consistent and dependent. This means that Equation 3 must be a linear combination of Equation 1 and Equation 2. Let us express this condition:

1. Multiply Equation 1 by some constant \(k_1\), and Equation 2 by some constant \(k_2\)
2. These constants should satisfy:

From \(4 = k_1 + 2k_2\) and \(3 = 2k_1 + 3k_2\), solve these simultaneous equations:

1. From the first equation: \(k_1 = 4 - 2k_2\)
2. Substitute \(k_1\) in the second equation:

Substitute \(k_2 = 5\) back into \(k_1 = 4 - 2k_2\):

\(k_1 = 4 - 2 \times 5 = -6\)

Now, substitute \(k_1\) and \(k_2\) into \(\lambda = k_1 \times 3 + k_2 \times 1\) and \(\mu = k_1 \times 5 + k_2 \times 9\):

\(\lambda = (-6) \times 3 + 1 \times 5 = -18 + 5 = -13\)
\(\mu = (-6) \times 5 + 5 \times 9 = -30 + 45 = 15\)

Finally, we calculate \(\lambda + 2\mu\):

\(\lambda + 2\mu = -13 + 2 \times 15 = -13 + 30 = 17\)

Therefore, the correct answer is 17.

Answer 2

To have an infinite number of solutions, the given system of equations must be dependent, meaning that the third equation must be a linear combination of the first two equations. Let's analyze the system:

\[\begin{align*} 1) & \quad x + 2y + 3z = 5 \\ 2) & \quad 2x + 3y + z = 9 \\ 3) & \quad 4x + 3y + \lambda z = \mu \end{align*}\]

The third equation \(4x + 3y + \lambda z = \mu\) should be a linear combination of the first two equations. Thus, we aim to find constants \(a\) and \(b\) such that:

\[a(x + 2y + 3z) + b(2x + 3y + z) = 4x + 3y + \lambda z\]

Expanding both sides, we have:

\[\begin{align*} ax + 2ay + 3az + 2bx + 3by + bz &= 4x + 3y + \lambda z \\ (a + 2b)x + (2a + 3b)y + (3a + b)z &= 4x + 3y + \lambda z \end{align*}\]

By comparing coefficients, we need the following system of equations:

\[\begin{align*} 1) & \quad a + 2b = 4 \\ 2) & \quad 2a + 3b = 3 \\ 3) & \quad 3a + b = \lambda \end{align*}\]

Solving the first two equations:

1. From equation (1), express \(a\) in terms of \(b\):

\[a = 4 - 2b\]

1. Substitute into equation (2):

\[2(4 - 2b) + 3b = 3 \\ 8 - 4b + 3b = 3 \\ -b = -5 \implies b = 5\]

Substitute \(b = 5\) back to find \(a\):

\[a = 4 - 2 \times 5 = 4 - 10 = -6\]

Now, substitute \(a = -6\) and \(b = 5\) into equation (3) to find \(\lambda\):

\[3(-6) + 5 = \lambda \\ -18 + 5 = \lambda \\ \lambda = -13\]

For the system to be consistent, the right-hand side also needs to satisfy the condition:

1. Linear combination for \(\mu\):

\[-6(5) + 5(9) = \mu \\ -30 + 45 = \mu \\ \mu = 15\]

Thus, \(\lambda = -13\) and \(\mu = 15\). Finally, compute \(\lambda + 2\mu\):

\[\lambda + 2\mu = -13 + 2(15) = -13 + 30 = 17\]

Therefore, the value of \(\lambda + 2\mu\) is 17, which matches the correct answer.