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

The Correct Option is D

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

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,

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