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

Answer: (B)

1120

Answer 2

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 $13 \alpha \beta$:

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