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 determine the conditions for an infinite number of solutions for the given system of equations, we analyze its consistency and dependency. The system is:

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

For infinite solutions, Equation 3 must be a linear combination of Equation 1 and Equation 2. This implies:

1. Let Equation 3 be formed by multiplying Equation 1 by \(k_1\) and Equation 2 by \(k_2\):
2. The coefficients must satisfy:

We set up simultaneous equations for the coefficients of x and y:

\(4 = k_1 + 2k_2\)

\(3 = 2k_1 + 3k_2\)

Solving these equations:

1. From the first equation, isolate \(k_1\): \(k_1 = 4 - 2k_2\)
2. Substitute this expression for \(k_1\) into the second equation:

Substituting \(k_2 = 5\) back into the expression for \(k_1\):

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

Now, we use these values of \(k_1\) and \(k_2\) to find \(\lambda\) and \(\mu\):

\(\lambda = k_1 \times 3 + k_2 \times 1 = (-6) \times 3 + 1 \times 5 = -18 + 5 = -13\)

\(\mu = k_1 \times 5 + k_2 \times 9 = (-6) \times 5 + 5 \times 9 = -30 + 45 = 15\)

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

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

The result is 17.