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

Options

A. 1110
B. 1120
C. 1210
D. 1220

Answer

To find the value of \( 13 \alpha \beta \) for the given system of equations, we need to analyze the conditions under which the system has infinitely many solutions. This occurs when the determinant of the coefficient matrix is zero.

The system of equations is:

1. \( 2x + 3y - z = 5 \)
2. \( x + \alpha y + 3z = -4 \)
3. \( 3x - y + \beta z = 7 \)

### Step 1: Write the coefficient matrix and the augmented matrix

The coefficient matrix \( A \) is given by:

\[ A = \begin{bmatrix} 2 & 3 & -1 \\ 1 & \alpha & 3 \\ 3 & -1 & \beta \end{bmatrix} \]

The augmented matrix is:

\[ \begin{bmatrix} 2 & 3 & -1 & | & 5 \\ 1 & \alpha & 3 & | & -4 \\ 3 & -1 & \beta & | & 7 \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix

To find the condition for infinitely many solutions, we need to set the determinant \( D \) of the coefficient matrix \( A \) to zero.

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

Calculating the determinant using the first row:

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

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} \alpha & 3 \\ -1 & \beta \end{vmatrix} = \alpha \beta + 3 \)

2. \( \begin{vmatrix} 1 & 3 \\ 3 & \beta \end{vmatrix} = \beta - 9 \)

3. \( \begin{vmatrix} 1 & \alpha \\ 3 & -1 \end{vmatrix} = -1 - 3\alpha = -3\alpha - 1 \)

Substituting these back into the determinant:

\[ D = 2(\alpha \beta + 3) - 3(\beta - 9) - (-3\alpha - 1) \]

Expanding this gives:

\[ D = 2\alpha \beta + 6 - 3\beta + 27 + 3\alpha + 1 \]
\[ D = 2\alpha \beta + 3\alpha - 3\beta + 34 \]

### Step 3: Set the determinant to zero

For the system to have infinitely many solutions, we set \( D = 0 \):

\[ 2\alpha \beta + 3\alpha - 3\beta + 34 = 0 \]

### Step 4: Calculate \( D_y \) and \( D_z \)

We also need to calculate the determinants \( D_y \) and \( D_z \) to find the values of \( \alpha \) and \( \beta \).

1. For \( D_y \):

\[ D_y = \begin{vmatrix} 2 & 3 & -1 \\ 1 & \alpha & 3 \\ 3 & -1 & \beta \end{vmatrix} \]

Replace the last column with the constants:

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

Calculating \( D_y \) similarly will yield another equation involving \( \alpha \) and \( \beta \).

2. For \( D_z \):

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

Replace the last column with the constants:

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

### Step 5: Solve the equations

By solving the equations obtained from \( D = 0 \), \( D_y = 0 \), and \( D_z = 0 \), we can find the values of \( \alpha \) and \( \beta \).

### Step 6: Calculate \( 13 \alpha \beta \)

Once we have the values of \( \alpha \) and \( \beta \), we can calculate \( 13 \alpha \beta \).