Question

If the system of equations: $$ \begin{aligned} 3x + y + \beta z &= 3 \\2x + \alpha y + z &= 2 \\x + 2y + z &= 4 \end{aligned} $$ has infinitely many solutions, then the value of \( 22\beta - 9\alpha \) is:

• A. 49
• B. 31
• C. 43
• D. 37

Hint:

Infinitely many solutions occur when rank(A) = rank(A|B)<number of variables.

Answer 1

The Correct Option is B

We are given a system of three linear equations and told that it has infinitely many solutions. We need to find the value of the expression \( 22\beta - 9\alpha \).

The system of equations is:

\[ 3x + y + \beta z = 3 \] \[ 2x + \alpha y - z = -3 \] \[ x + 2y + z = 4 \]

Concept Used:

For a system of linear equations \( AX = B \), the condition for having infinitely many solutions is that the determinant of the coefficient matrix \( A \), denoted as \( \Delta \), must be zero, and the determinants of the matrices formed by replacing each column with the constant vector \( B \), denoted as \( \Delta_x, \Delta_y, \Delta_z \), must also be zero.

\[ \Delta = 0, \quad \Delta_x = 0, \quad \Delta_y = 0, \quad \Delta_z = 0 \]

Step-by-Step Solution:

Step 1: Set the determinant of the coefficient matrix (\( \Delta \)) to zero.

The coefficient matrix \( A \) is:

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

The determinant \( \Delta \) is:

\[ \Delta = \begin{vmatrix} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0 \]

Expanding along the first row:

\[ 3(\alpha \cdot 1 - (-1) \cdot 2) - 1(2 \cdot 1 - (-1) \cdot 1) + \beta(2 \cdot 2 - \alpha \cdot 1) = 0 \] \[ 3(\alpha + 2) - 1(2 + 1) + \beta(4 - \alpha) = 0 \] \[ 3\alpha + 6 - 3 + 4\beta - \alpha\beta = 0 \] \[ 3\alpha + 4\beta - \alpha\beta + 3 = 0 \quad \cdots(1) \]

Step 2: Set the determinant \( \Delta_y \) to zero.

To find \( \Delta_y \), we replace the second column of \( A \) with the constant vector \( B = \begin{pmatrix} 3 \\ -3 \\ 4 \end{pmatrix} \):

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

Expanding along the first row:

\[ 3(-3 \cdot 1 - (-1) \cdot 4) - 3(2 \cdot 1 - (-1) \cdot 1) + \beta(2 \cdot 4 - (-3) \cdot 1) = 0 \] \[ 3(-3 + 4) - 3(2 + 1) + \beta(8 + 3) = 0 \] \[ 3(1) - 3(3) + 11\beta = 0 \] \[ 3 - 9 + 11\beta = 0 \] \[ -6 + 11\beta = 0 \implies \beta = \frac{6}{11} \]

Step 3: Substitute the value of \( \beta \) into equation (1) to find \( \alpha \).

Using \( \beta = \frac{6}{11} \) in the equation \( 3\alpha + 4\beta - \alpha\beta + 3 = 0 \):

\[ 3\alpha + 4\left(\frac{6}{11}\right) - \alpha\left(\frac{6}{11}\right) + 3 = 0 \] \[ 3\alpha + \frac{24}{11} - \frac{6\alpha}{11} + 3 = 0 \]

To eliminate the fraction, multiply the entire equation by 11:

\[ 33\alpha + 24 - 6\alpha + 33 = 0 \] \[ 27\alpha + 57 = 0 \] \[ 27\alpha = -57 \] \[ \alpha = -\frac{57}{27} = -\frac{19}{9} \]

Final Computation & Result:

We have found the values \( \alpha = -\frac{19}{9} \) and \( \beta = \frac{6}{11} \). Now we compute the value of \( 22\beta - 9\alpha \):

\[ 22\beta - 9\alpha = 22\left(\frac{6}{11}\right) - 9\left(-\frac{19}{9}\right) \] \[ = 2 \cdot 6 - (-19) \] \[ = 12 + 19 \] \[ = 31 \]

Thus, the value of \( 22\beta - 9\alpha \) is 31.

Answer 2

For infinitely many solutions, the rank of the coefficient matrix and the augmented matrix must be equal and less than the number of variables. Using determinant conditions: 
\[ \Delta = \begin{vmatrix} 3 & 1 & \beta 2 & \alpha & 1 \\ 1 & 2 & 1 \end{vmatrix} = 0 \quad \text{and} \quad \Delta_3 = \begin{vmatrix} 3 & 1 & 3\\ 2 & \alpha & 2 \\1 & 2 & 4 \end{vmatrix} = 0 \] Solving gives \( \alpha = \frac{19}{9}, \beta = \frac{6}{11} \), so: \[ 22\beta - 9\alpha = 31 \]