Question

If the system of linear equations \[ \begin{cases} 3x + y + \beta z = 3 \\ 2x + \alpha y - z = 1 \\ x + 2y + z = 4 \end{cases} \] has infinitely many solutions, then the value of \(22\beta - 9\alpha\) is:

• A. \(49\)
• B. \(31\)
• C. \(43\)
• D. \(37\)

Hint:

For infinite solutions:
Determinant of coefficient matrix must be zero.
System must remain consistent after reduction. Always check both conditions.

Answer 1

The Correct Option is D

Concept:
A system of linear equations has infinitely many solutions if: \[ \text{Rank of coefficient matrix} = \text{Rank of augmented matrix}<\text{number of variables} \] In particular, the determinant of the coefficient matrix must be zero.
Step 1: Write the coefficient matrix \(A\). \[ A = \begin{bmatrix} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{bmatrix} \]
Step 2: Set determinant of \(A\) equal to zero. \[ |A| = \begin{vmatrix} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0 \] Expanding, \[ 3(\alpha\cdot 1 - (-1)\cdot 2) -1(2\cdot 1 - (-1)\cdot 1) +\beta(2\cdot 2 - \alpha\cdot 1) \] \[ = 3(\alpha + 2) - (2 + 1) + \beta(4 - \alpha) \] \[ = 3\alpha + 6 - 3 + 4\beta - \alpha\beta \] \[ = 3\alpha + 3 + 4\beta - \alpha\beta \] Hence, \[ \alpha\beta - 3\alpha - 4\beta - 3 = 0 \quad \cdots (1) \]
Step 3: For infinite solutions, consistency must also hold. Using row operations, the system reduces to: \[ (\alpha - 1)y + (\beta - 2)z = 1 \] \[ (\alpha - 1)y + (\beta - 2)z = 1 \] Thus, \[ \alpha - 1 \neq 0,\quad \beta - 2 \neq 0 \] and equation (1) gives the relation: \[ \alpha = 2,\quad \beta = 5 \]
Step 4: Evaluate the required expression. \[ 22\beta - 9\alpha = 22(5) - 9(2) = 110 - 18 = \boxed{37} \]