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

Given a system of three linear equations with infinitely many solutions, we are tasked with determining 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:

A system of linear equations \( AX = B \) has infinitely many solutions if and only if the determinant of the coefficient matrix \( A \), denoted by \( \Delta \), is zero, and the determinants of the matrices obtained by replacing each column of \( A \) with the constant vector \( B \) (denoted as \( \Delta_x, \Delta_y, \Delta_z \)) are also zero.

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

Step-by-Step Solution:

Step 1: Equate 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 the determinant yields:

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

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

The determinant \( \Delta_y \) is obtained by replacing 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 this determinant:

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

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

Substituting \( \beta = \frac{6}{11} \) into \( 3\alpha + 4\beta - \alpha\beta + 3 = 0 \):

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

Multiplying the equation by 11 to clear fractions:

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

Final Computation & Result:

The determined values are \( \alpha = -\frac{19}{9} \) and \( \beta = \frac{6}{11} \). We now compute \( 22\beta - 9\alpha \):

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

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