Complex Numbers
System of equations with complex numbers
Grade None
Question:
<p>If \(x\) and \(y\) are complex numbers, then the system of equations \((1+i)x + (1-i)y = 1\), \(2ix + 2y = 1+i\) has</p>
<p>unique solution</p>
<p>no solution</p>
<p>infinite number of solutions</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Treat this as a linear system with complex coefficients and use Gaussian elimination or Cramer's rule to determine if the system has a unique solution, no solution, or infinitely many solutions by checking if the coefficient matrix is singular.
<p><strong>Step 1:</strong> Write the system in matrix form:</p><p>Coefficient matrix: A = <span style='font-family: monospace;'>[(1+i) (1-i)]</span></p><p><span style='font-family: monospace;'>[2i 2 ]</span></p><p><strong>Step 2:</strong> Calculate determinant:</p><p>det(A) = (1+i)(2) - (1-i)(2i)</p><p>= 2 + 2i - (2i - 2i²)</p><p>= 2 + 2i - (2i + 2)</p><p>= 2 + 2i - 2i - 2 = 0</p><p><strong>Step 3:</strong> Since det(A) = 0, the system is either inconsistent or has infinitely many solutions. Check consistency by row reducing:</p><p>Divide equation 2 by 2i: -1·x + (1/i)y = (1+i)/(2i)</p><p>Simplify: -x - iy = (1+i)(-i)/2 = (-i+1)/2</p><p>Multiply equation 1 by appropriate factor and compare. The equations are dependent (one is a scalar multiple of the other).</p><p><strong>Step 4:</strong> Since coefficient matrix is singular and equations are consistent (dependent), the system has infinitely many solutions forming a line in complex space.</p><p>∴ Answer: C (Infinitely many solutions)</p>
Correct Answer: C