Question

If the system of linear equations
$x - 2y + z = -4$
$2x + \alpha y + 3z = 5$
$3x - y + \beta z = 3$
has infinitely many solutions, then $12\alpha + 13\beta$ is equal to

• A. 60
• B. 64
• C. 54
• D. 58

Answer 1

Answer: (D)

58

Answer 2

$D = \begin{vmatrix} 1 & -2 & 1 \\\ \alpha & 3 & 3 \\\ 3 & -1 & \beta \end{vmatrix}$ $= 1(\alpha \beta + 3) + 2(2\beta - 9) + 1(-2 - 3\alpha)$ $= \alpha \beta + 3 + 4\beta - 18 - 2 - 3\alpha$

For infinite solutions, $D = 0$, $D_1 = 0$, $D_2 = 0$, and $D_3 = 0$

$D = 0 \quad \implies \quad \alpha \beta - 3\alpha + 4\beta = 17 \quad \text{....(1)}$ $D_1 = \begin{vmatrix} -4 & -2 & 1 \\\ 5 & 3 & 3 \\\ 3 & -1 & \beta \end{vmatrix} = 0, \quad D_2 = \begin{vmatrix} 2 & 1 & 1 \\\ 5 & \alpha & 3 \\\ 3 & 3 & \beta \end{vmatrix} = 0$

$D_2 = 1(5\beta - 9) + 4(2\beta - 9) + 1(6 - 15) = 0$ $13\beta - 9 - 36 - 9 = 0 \quad \implies \quad 13\beta = 54, \quad \beta = \frac{54}{13}$

Substitute in (1):

$\frac{54}{13}\alpha - 3\alpha + 4\left(\frac{54}{13}\right) = 17$ $\frac{54}{13}\alpha - 39\alpha + 216 = 221$ $15\alpha = 5 \quad \implies \quad \alpha = \frac{1}{3}$

Now,

$12\alpha + 13\beta = 12 \cdot \frac{1}{3} + 13 \cdot \frac{54}{13}$ $= 4 + 54 = 58$