Question

The system of linear equations \[ x+y+z=0,\; 2x+y-z=0,\; 3x+2y=0 \] has

• A. no solution
• B. a unique solution
• C. infinitely many solutions
• D. None of these

Hint:

Determinant = 0 $\Rightarrow$ either no solution or infinite solutions—check consistency.

Answer 1

The Correct Option is C

We are given the following system of linear equations:

\(x + y + z = 0\)   (Equation 1)
\(2x + y - z = 0\)   (Equation 2)
\(3x + 2y = 0\)   (Equation 3)
 

To determine the nature of its solutions, let's analyze these equations:

1. First, eliminate \(z\) by adding Equation 1 and Equation 2:

\[ (x + y + z) + (2x + y - z) = 0 + 0 \]
This results in: \[ 3x + 2y = 0 \]

This equation is identical to Equation 3.

1. Since Equation 3 does not involve \(z\), the system can be simplified to two equations:

Equation 1: \(x + y + z = 0\)
Equation 3: \(3x + 2y = 0\)

1. We already see that Equations 1 and 3 are dependent. To check this, solve for one variable:

From Equation 3:

\[ 3x + 2y = 0 \Rightarrow y = -\frac{3}{2}x \]

1. Substitute \(y = -\frac{3}{2}x\) into Equation 1 to find \(z\):

\[ x - \frac{3}{2}x + z = 0 \Rightarrow -\frac{1}{2}x + z = 0 \Rightarrow z = \frac{1}{2}x \]

1. Therefore, the general solution is:

\[ x = t, \quad y = -\frac{3}{2}t, \quad z = \frac{1}{2}t \quad \text{for any } t \in \mathbb{R} \]

This shows that the system has infinitely many solutions, parameterized by \(t\).

Conclusion: The correct answer is infinitely many solutions.