Question

If \( x = \alpha, y = \beta, z = \gamma \) is the solution of the system of equations \( 2x+3y+z=-1 \), \( 3x+y+z=4 \), \( x-3y-2z=1 \), then the value of \( \beta \) is

• A. -2
• B. -1
• C. 2
• D. 1

Hint:

When solving a 3x3 system for only one variable, plan your eliminations. Pick a variable to eliminate first (like \( \gamma \) here) to reduce the system to 2x2, which is much easier to solve.

Answer 1

The Correct Option is A

We are given the following system of linear equations:
(1) \( 2\alpha + 3\beta + \gamma = -1 \)
(2) \( 3\alpha + \beta + \gamma = 4 \)
(3) \( \alpha - 3\beta - 2\gamma = 1 \)
Our goal is to find the value of \( \beta \). We can use the elimination method.
First, let's eliminate \( \gamma \) using equations (1) and (2). Subtracting (1) from (2):
\( (3\alpha + \beta + \gamma) - (2\alpha + 3\beta + \gamma) = 4 - (-1) \)
\( \alpha - 2\beta = 5 \) (Equation 4)
Next, let's eliminate \( \gamma \) using equations (1) and (3). Multiply equation (1) by 2 and add it to equation (3):
\( 2(2\alpha + 3\beta + \gamma) + (\alpha - 3\beta - 2\gamma) = 2(-1) + 1 \)
\( (4\alpha + 6\beta + 2\gamma) + (\alpha - 3\beta - 2\gamma) = -2 + 1 \)
\( 5\alpha + 3\beta = -1 \) (Equation 5)
Now we have a system of two equations with two variables:
(4) \( \alpha - 2\beta = 5 \)
(5) \( 5\alpha + 3\beta = -1 \)
From equation (4), we can express \( \alpha \) in terms of \( \beta \): \( \alpha = 5 + 2\beta \).
Substitute this expression for \( \alpha \) into equation (5):
\( 5(5 + 2\beta) + 3\beta = -1 \)
\( 25 + 10\beta + 3\beta = -1 \)
\( 13\beta = -1 - 25 \)
\( 13\beta = -26 \)
\( \beta = \frac{-26}{13} = -2 \).