Question

If the system of equations
\( 2x + 3y - z = 5 \)  
\( x + \alpha y + 3z = -4 \)  
\( 3x - y + \beta z = 7 \)  
has infinitely many solutions, then \( 13 \alpha \beta \) is equal to:  

• A. 1110
• B. 1120
• C. 1210
• D. 1220

Answer 1

The Correct Option is B

We are given the system of equations:

\[ 2x + 3y - z = 5 \] \[ x + \alpha y + 3z = -4 \] \[ 3x - y + \beta z = 7 \]

We can write this system in terms of a family of planes. Using the family of planes, we have:

\[ 2x + 3y - z = k_1 \left( x + \alpha y + 3z \right) + k_2 \left( 3x - y + \beta z \right) \]

Expanding and simplifying:

\[ 2 = k_1 + 3k_2, \quad 3 = k_1 \alpha - k_2, \quad -1 = 3k_1 + \beta k_2, \quad -5 = 4k_1 - 7k_2 \]

Solving this system, we find:

\[ k_2 = \frac{13}{19}, \quad k_1 = -\frac{1}{19}, \quad \alpha = -70, \quad \beta = -\frac{16}{13} \]

Now, calculate \( 13 \alpha \beta \):

\[ 13 \alpha \beta = 13 \times (-70) \times \left( -\frac{16}{13} \right) = 1120 \]