(a) Multiply both sides by \(a + 2i\) to clear the fraction:
\(2 + 3ai = \lambda(2 - i)(a + 2i)\).
Expand the right-hand side:
\(2 + 3ai = \lambda(2a + 4i - ai - 2)\).
Separate real and imaginary parts:
Real: \(2 = \lambda(2a - 2)\)
Imaginary: \(3a = \lambda(4 - a)\).
From the real part: \(2 = \lambda(2a - 2) \Rightarrow \lambda = \frac{2}{2a - 2}\).
From the imaginary part: \(3a = \lambda(4 - a) \Rightarrow \lambda = \frac{3a}{4 - a}\).
Equating the two expressions for \(\lambda\):
\(\frac{2}{2a - 2} = \frac{3a}{4 - a}\).
Cross-multiply and simplify:
\(2(4 - a) = 3a(2a - 2)\).
\(8 - 2a = 6a^2 - 6a\).
Rearrange to form a quadratic equation:
\(6a^2 - 4a - 8 = 0\).
Divide by 2:
\(3a^2 + 4a - 4 = 0\).
(b) Solve the quadratic equation \(3a^2 + 4a - 4 = 0\) using the quadratic formula:
\(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3, b = 4, c = -4\).
\(a = \frac{-4 \pm \sqrt{16 + 48}}{6}\).
\(a = \frac{-4 \pm \sqrt{64}}{6}\).
\(a = \frac{-4 \pm 8}{6}\).
\(a = \frac{4}{6} = \frac{2}{3}\) or \(a = \frac{-12}{6} = -2\).
For \(a = -2\), substitute back to find \(\lambda\):
\(3(-2) = \lambda(4 + 2) \Rightarrow -6 = 6\lambda \Rightarrow \lambda = -1\).
For \(a = \frac{2}{3}\), substitute back to find \(\lambda\):
\(3\left(\frac{2}{3}\right) = \lambda\left(4 - \frac{2}{3}\right) \Rightarrow 2 = \lambda\left(\frac{10}{3}\right) \Rightarrow \lambda = \frac{3}{5}\).
