Question

Let the system of equations $ x + 5y - z = 1 $ $ 4x + 3y - 3z = 7 $ $ 24x + y + \lambda z = \mu $ where $ \lambda, \mu \in \mathbb{R} $, have infinitely many solutions. Then the number of the solutions of this system, if $x, y, z$ are integers and satisfy $7 \leq x + y + z \leq 77$, is:

• A. 3
• B. 6
• C. 5
• D. 4

Hint:

To find the number of solutions for systems with infinite solutions, examine the dependency of equations and check if the rank condition holds.

Answer 1

The Correct Option is A

To determine the conditions for infinitely many solutions for the given system of equations, the system must be consistent and dependent. Consider the equations:

Equation 1: \(x + 5y - z = 1\)

Equation 2: \(4x + 3y - 3z = 7\)

Equation 3: \(24x + y + \lambda z = \mu\)

For infinite solutions, Equation 3 must be a linear combination of Equations 1 and 2. We express Equation 3 as:

\(24x + y + \lambda z = k_1(x + 5y - z) + k_2(4x + 3y - 3z)\)

Equating coefficients yields:

For \(x\): \(24 = k_1 + 4k_2\) (Equation i)

For \(y\): \(1 = 5k_1 + 3k_2\) (Equation ii)

For \(z\): \(\lambda = -k_1 - 3k_2\) (Equation iii)

For constants: \(\mu = k_1(1) + k_2(7)\) (Equation iv)

Solving for \(k_1\) and \(k_2\):

1. From Equation i: \(k_1 = 24 - 4k_2\)
2. Substitute into Equation ii: \(1 = 5(24 - 4k_2) + 3k_2\)
3. Simplify: \(1 = 120 - 20k_2 + 3k_2\)
4. Solve for \(k_2\): \(1 = 120 - 17k_2 \Rightarrow 17k_2 = 119\)
5. \(k_2 = \frac{119}{17} = 7\)
6. Substitute \(k_2 = 7\) into \(k_1 = 24 - 4k_2\): \(k_1 = 24 - 4(7) = 24 - 28 = -4\)

Calculate \(\lambda\) and \(\mu\):

From Equation iii: \(\lambda = -k_1 - 3k_2 = -(-4) - 3(7) = 4 - 21 = -17\)

From Equation iv: \(\mu = k_1 + 7k_2 = -4 + 7(7) = -4 + 49 = 45\)

Therefore, the conditions for infinitely many solutions are \(\lambda = -17\) and \(\mu = 45\).

Investigating integer solutions satisfying \(7 \leq x + y + z \leq 77\), we consider the equation \(x + y + z = k\). After verifying boundaries with the derived solutions, the number of viable integer solutions adhering to this constraint is 3.