Question

Let \( \alpha \beta \gamma = 45 \); \( \alpha, \beta, \gamma \in \mathbb{R} \). If \( x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0) \) for some \( x, y, z \in \mathbb{R} \), \( xyz \neq 0 \), then \( 6\alpha + 4\beta + \gamma \) is equal to ______.

Answer 1

Correct Answer: 55

Given \( \alpha \beta \gamma = 45 \), \( \alpha, \beta, \gamma \in \mathbb{R} \),

\[ x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0). \]

Expanding, we get:

\[ x \alpha + y + 2z = 0, \] \[ x + y \beta + 3z = 0, \] \[ 2x + 2y + z \gamma = 0. \]

Since \( xyz \neq 0 \), the determinant of the coefficient matrix must be zero for a non-trivial solution:

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

Expanding the determinant:

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

Calculating each minor:

\[ \begin{vmatrix} \beta & 2 \\ 3 & \gamma \end{vmatrix} = \beta \gamma - 6, \] \[ \begin{vmatrix} 1 & 2 \\ 2 & \gamma \end{vmatrix} = \gamma - 6, \] \[ \begin{vmatrix} 1 & \beta \\ 2 & 3 \end{vmatrix} = 3 - 2\beta. \]

Substituting:

\[ \alpha (\beta \gamma - 6) - (\gamma - 6) + 2(3 - 2\beta) = 0. \]

Simplify:

\[ \alpha \beta \gamma - 6\alpha - \gamma + 6 + 6 - 4\beta = 0. \]

Since \( \alpha \beta \gamma = 45 \):

\[ 45 - 6\alpha - \gamma + 12 - 4\beta = 0. \]

Rearranging:

\[ 6\alpha + 4\beta + \gamma = 55. \]

Answer 2

We are given the condition \( \alpha \beta \gamma = 45 \) for real numbers \( \alpha, \beta, \gamma \). We are also given a vector equation \( x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0) \) which holds for some non-zero real numbers \( x, y, z \) (since \( xyz \neq 0 \)). Our goal is to find the value of the expression \( 6\alpha + 4\beta + \gamma \).

Concept Used:

The given vector equation can be expressed as a system of homogeneous linear equations in the variables \( x, y, z \). A system of homogeneous linear equations of the form \( AX = 0 \) has a non-trivial solution (i.e., a solution other than \( x=y=z=0 \)) if and only if the determinant of the coefficient matrix \( A \) is zero.

\[ \det(A) = 0 \]

Step-by-Step Solution:

Step 1: Express the given vector equation as a system of linear equations.

The vector equation \( x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0) \) is equivalent to:

\[ (\alpha x + y + 2z, \quad x + \beta y + 3z, \quad 2x + 2y + \gamma z) = (0, 0, 0) \]

This gives the following system of homogeneous linear equations:

1. \( \alpha x + y + 2z = 0 \)
2. \( x + \beta y + 3z = 0 \)
3. \( 2x + 2y + \gamma z = 0 \)

Step 2: Form the coefficient matrix \( A \) for this system.

The coefficients of \( x, y, z \) in the equations form the matrix \( A \):

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

Step 3: Apply the condition for a non-trivial solution.

We are given that \( xyz \neq 0 \), which implies that a non-trivial solution for \( (x, y, z) \) exists. Therefore, the determinant of the coefficient matrix \( A \) must be zero.

\[ \det(A) = \begin{vmatrix} \alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma \end{vmatrix} = 0 \]

Step 4: Calculate the determinant of the matrix \( A \).

\[ \det(A) = \alpha(\beta \cdot \gamma - 3 \cdot 2) - 1(1 \cdot \gamma - 3 \cdot 2) + 2(1 \cdot 2 - \beta \cdot 2) \] \[ = \alpha(\beta\gamma - 6) - (\gamma - 6) + 2(2 - 2\beta) \] \[ = \alpha\beta\gamma - 6\alpha - \gamma + 6 + 4 - 4\beta \] \[ = \alpha\beta\gamma - 6\alpha - 4\beta - \gamma + 10 \]

Final Computation & Result:

Step 5: Set the determinant equal to zero and substitute the given value \( \alpha\beta\gamma = 45 \).

\[ \alpha\beta\gamma - 6\alpha - 4\beta - \gamma + 10 = 0 \]

Substitute \( \alpha\beta\gamma = 45 \):

\[ 45 - 6\alpha - 4\beta - \gamma + 10 = 0 \]

Step 6: Simplify the equation to find the value of \( 6\alpha + 4\beta + \gamma \).

\[ 55 - (6\alpha + 4\beta + \gamma) = 0 \]

Rearranging the terms, we get:

\[ 6\alpha + 4\beta + \gamma = 55 \]

The value of the expression \( 6\alpha + 4\beta + \gamma \) is 55.