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

The given system of equations is: \begin{align} x + 5y - z &= 1 \quad &(1) \\4x + 3y - 3z &= 7 \quad &(2) \\24x + y + \lambda z &= \mu \quad &(3) \end{align} For the system to have infinitely many solutions, the third equation must be a linear combination of the first two. Let equation (3) be $a \times (1) + b \times (2)$: \[ a(x + 5y - z) + b(4x + 3y - 3z) = a + 7b \] \[ (a + 4b)x + (5a + 3b)y + (-a - 3b)z = a + 7b \] Comparing coefficients with equation (3), we get: \begin{align} a + 4b &= 24 5a + 3b &= 1 \\-a - 3b &= \lambda \\a + 7b &= \mu \end{align} Solving the first two equations, we find $a = -4$ and $b = 7$. Substituting these values, we get $\lambda = -(-4) - 3(7) = -17$ and $\mu = -4 + 7(7) = 45$. Now, we find the integer solutions of the first two equations. Multiply equation (1) by 3 and subtract it from equation (2): \[ (4x + 3y - 3z) - 3(x + 5y - z) = 7 - 3 \] \[ x - 12y = 4 \implies x = 12y + 4 \] Substitute $x$ in equation (1): \[ (12y + 4) + 5y - z = 1 \] \[ 17y + 4 - z = 1 \implies z = 17y + 3 \] The integer solutions are of the form $(12y + 4, y, 17y + 3)$ for any integer $y$. We are given the condition $7 \leq x + y + z \leq 77$. \[ x + y + z = (12y + 4) + y + (17y + 3) = 30y + 7 \] So, $7 \leq 30y + 7 \leq 77$. \[ 0 \leq 30y \leq 70 \] \[ 0 \leq y \leq \frac{70}{30} = \frac{7}{3} \] Since $y$ must be an integer, the possible values for $y$ are $0, 1, 2$. For each value of $y$, we have a unique integer solution $(x, y, z)$:

• If $y = 0$, $(x, y, z) = (4, 0, 3)$, and $x + y + z = 7$.
• If $y = 1$, $(x, y, z) = (16, 1, 20)$, and $x + y + z = 37$.
• If $y = 2$, $(x, y, z) = (28, 2, 37)$, and $x + y + z = 67$.

There are 3 integer solutions that satisfy the given condition. 
Final Answer: The final answer is $3$