Question

The system of simultaneous linear equations : 

 

\[ \begin{array}{rcl} x - 2y + 3z &=& 4 \\ 2x + 3y + z &=& 6 \\ 3x + y - 2z &=& 7 \end{array} \]

• A. Infinitely many solutions
• B. No solution
• C. Unique solution having z = 2
• D. Unique solution having  z =1⁄2

Hint:

Using Gaussian elimination can help quickly identify the nature of the solutions (unique, none, or infinite) for a system of linear equations.

Answer 1

The Correct Option is D

We will use elimination to solve the system. Multiply equation (1) by 2 and subtract it from equation (2): \[ (2x + 3y + z) - 2(x - 2y + 3z) = 6 - 2(4) \] \[ 2x + 3y + z - 2x + 4y - 6z = 6 - 8 \] \[ 7y - 5z = -2 \quad (4) \] Multiply equation (1) by 3 and subtract it from equation (3): \[ (3x + y - 2z) - 3(x - 2y + 3z) = 7 - 3(4) \] \[ 3x + y - 2z - 3x + 6y - 9z = 7 - 12 \] \[ 7y - 11z = -5 \quad (5) \] Subtract equation (5) from equation (4): \[ (7y - 5z) - (7y - 11z) = -2 - (-5) \] \[ 7y - 5z - 7y + 11z = -2 + 5 \] \[ 6z = 3 \] \[ z = \frac{3}{6} = \frac{1}{2} \] Substitute $z = \frac{1}{2}$ into equation (4): \[ 7y - 5\left(\frac{1}{2}\right) = -2 \] \[ 7y - \frac{5}{2} = -2 \] \[ 7y = -2 + \frac{5}{2} = \frac{-4 + 5}{2} = \frac{1}{2} \] \[ y = \frac{1}{14} \] Substitute $y = \frac{1}{14}$ and $z = \frac{1}{2}$ into equation (1): \[ x - 2\left(\frac{1}{14}\right) + 3\left(\frac{1}{2}\right) = 4 \] \[ x - \frac{1}{7} + \frac{3}{2} = 4 \] \[ x = 4 + \frac{1}{7} - \frac{3}{2} = \frac{56 + 2 - 21}{14} = \frac{37}{14} \] Thus, we have a unique solution: \[ x = \frac{37}{14}, \quad y = \frac{1}{14}, \quad z = \frac{1}{2} \] The unique solution has $z = \frac{1}{2}$. Therefore, the correct answer is (4). Final Answer: The final answer is $\boxed{(4)}$