Question

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

Hint:

Alternatively, observe that the third equation could be a linear combination of the first two. If $L_3 = k_1 L_1 + k_2 L_2$, compare coefficients to find parameters.

Answer 1

Correct Answer: 5

Step 1: Understanding the Concept:
For a system of linear equations in three variables to have infinitely many solutions, the determinant of the coefficient matrix (\(D\)) must be zero, and the determinants \(D_x\), \(D_y\), and \(D_z\) must also be zero.
Step 2: Detailed Explanation:
The determinant of coefficients \(D\) is:

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

\[ D = -2\beta + 6 - \beta - 3 - 6 = -3\beta - 3 \]

For infinitely many solutions, \(D = 0 \implies -3\beta - 3 = 0 \implies \beta = -1\).
Now, check \(D_z\):

\[ D_z = \begin{vmatrix} 2 & 1 & 3 \\ 1 & -1 & \alpha \\ 3 & 3 & 3 \end{vmatrix} = 2(-3 - 3\alpha) - 1(3 - 3\alpha) + 3(3 + 3) \]

\[ D_z = -6 - 6\alpha - 3 + 3\alpha + 18 = 9 - 3\alpha \]

For infinitely many solutions, \(D_z = 0 \implies 9 - 3\alpha = 0 \implies \alpha = 3\).
The value of \(\alpha + \beta - \alpha\beta = 3 + (-1) - (3)(-1) = 3 - 1 + 3 = 5\).
Step 3: Final Answer:
The value of \(\alpha + \beta - \alpha\beta\) is 5.