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 \( 13 \alpha \beta \):

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

Answer 2

Given the system of equations has infinitely many solutions:

\[ \begin{aligned} (1) \quad & 2x + 3y - z = 5 \\ (2) \quad & x + \alpha y + 3z = -4 \\ (3) \quad & 3x - y + \beta z = 7 \end{aligned} \]

We need to find \( 13 \alpha \beta \).

Concept Used:

For a system of three linear equations in three variables to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the system must be consistent (all three planes intersect in a line). This means the rank of the coefficient matrix equals the rank of the augmented matrix, and both are less than 3 (here rank = 2). We can use the condition that all 3×3 sub-determinants of the augmented matrix vanish.

Step-by-Step Solution:

Step 1: Write the coefficient matrix \(A\) and augmented matrix \([A|B]\).

\[ A = \begin{bmatrix} 2 & 3 & -1 \\ 1 & \alpha & 3 \\ 3 & -1 & \beta \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ -4 \\ 7 \end{bmatrix} \]

Step 2: For infinitely many solutions, \(\det(A) = 0\).

\[ \det(A) = 2 \begin{vmatrix} \alpha & 3 \\ -1 & \beta \end{vmatrix} - 3 \begin{vmatrix} 1 & 3 \\ 3 & \beta \end{vmatrix} + (-1) \begin{vmatrix} 1 & \alpha \\ 3 & -1 \end{vmatrix} \] \[ = 2(\alpha \beta + 3) - 3(\beta - 9) - 1(-1 - 3\alpha) \] \[ = 2\alpha\beta + 6 - 3\beta + 27 + 1 + 3\alpha \] \[ = 2\alpha\beta + 3\alpha - 3\beta + 34 \]

Set \(\det(A) = 0\):

\[ 2\alpha\beta + 3\alpha - 3\beta + 34 = 0 \quad \text{(Eq. 1)} \]

Step 3: Use the condition that the augmented matrix has rank 2, so any 3×3 submatrix including the constants column has determinant 0.

Consider the determinant formed by columns 1, 2, and constants (replace column 3 of A with B):

\[ D_1 = \begin{vmatrix} 2 & 3 & 5 \\ 1 & \alpha & -4 \\ 3 & -1 & 7 \end{vmatrix} = 0 \]

Expanding:

\[ D_1 = 2 \begin{vmatrix} \alpha & -4 \\ -1 & 7 \end{vmatrix} - 3 \begin{vmatrix} 1 & -4 \\ 3 & 7 \end{vmatrix} + 5 \begin{vmatrix} 1 & \alpha \\ 3 & -1 \end{vmatrix} \] \[ = 2(7\alpha - 4) - 3(7 + 12) + 5(-1 - 3\alpha) \] \[ = 14\alpha - 8 - 3(19) - 5 - 15\alpha \] \[ = (14\alpha - 15\alpha) + (-8 - 57 - 5) \] \[ = -\alpha - 70 \]

So \(-\alpha - 70 = 0 \Rightarrow \alpha = -70\).

Step 4: Similarly, consider columns 1, 3, and constants (replace column 2 of A with B):

\[ D_2 = \begin{vmatrix} 2 & -1 & 5 \\ 1 & 3 & -4 \\ 3 & \beta & 7 \end{vmatrix} = 0 \]

Expanding:

\[ D_2 = 2 \begin{vmatrix} 3 & -4 \\ \beta & 7 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -4 \\ 3 & 7 \end{vmatrix} + 5 \begin{vmatrix} 1 & 3 \\ 3 & \beta \end{vmatrix} \] \[ = 2(21 + 4\beta) + 1(7 + 12) + 5(\beta - 9) \] \[ = 42 + 8\beta + 19 + 5\beta - 45 \] \[ = (42 + 19 - 45) + (8\beta + 5\beta) \] \[ = 16 + 13\beta \]

So \(16 + 13\beta = 0 \Rightarrow \beta = -\frac{16}{13}\).

Step 5: Verify Eq. 1 with these values.

\[ 2\alpha\beta + 3\alpha - 3\beta + 34 = 2(-70)\left(-\frac{16}{13}\right) + 3(-70) - 3\left(-\frac{16}{13}\right) + 34 \] \[ = \frac{2240}{13} - 210 + \frac{48}{13} + 34 \] \[ = \frac{2240 + 48}{13} - 176 \] \[ = \frac{2288}{13} - 176 = 176 - 176 = 0 \]

It is satisfied.

Step 6: Compute \(13\alpha\beta\).

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

Therefore, \(13\alpha\beta = \mathbf{1120}\).