Let for some real numbers α and β, a = α - iβ. If the system of equations 4ix + (1 + i)y = 0 and 8(c

Q: Why is the determinant zero? Can't we just solve for x and y?

A homogeneous system of linear equations $Ax = 0$ has non-trivial solutions (solutions other than $x=0, y=0$) if and only if the matrix $A$ is singular, which means its determinant must be zero. If the determinant were non-zero, the only solution would be the trivial one, $x=0$ and $y=0$.

Q: Why do we equate the real and imaginary parts to zero separately?

A complex number $z = x + iy$ is equal to zero if and only if both its real part $x$ and its imaginary part $y$ are independently zero. Since our determinant expansion results in a complex expression of the form $A + iB = 0$, we must have $A=0$ and $B=0$ to satisfy the equation.

Q: Can we use Cramer's rule here?

Cramer's rule is typically used for systems with a unique solution, where the determinant is non-zero. In this problem, we are looking for the condition where there is *more than one* solution, which implies the system is dependent. Therefore, the determinant must be zero, and Cramer's rule is not the appropriate tool.

Question

Let for some real numbers $\alpha$ and $\beta$, $a = \alpha - i\beta$. If the system of equations $4ix + (1 + i)y = 0$ and $8(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})x + \bar{a}y = 0$ has more than one solution then $\frac{\alpha}{\beta}$ is equal to :

Options:

(A) $-2 + \sqrt{3}$

(B) $2 - \sqrt{3}$

(C) $2 + \sqrt{3}$

(D) $-2 - \sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a system of linear equations:
$4ix + (1 + i)y = 0$
$8(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})x + \bar{a}y = 0$

Notice that the constants on the right side are both zero. This is a **homogeneous system**.

In the realm of linear algebra, a homogeneous system $Ax = 0$ is like a balance scale. If the determinant of the coefficient matrix is non-zero, the only solution is the trivial one: $x=0$ and $y=0$.

However, the problem states there is **more than one solution**. This implies our system is singular, and the determinant of the coefficient matrix must be exactly zero.

### Preparing the Ingredients

Before we calculate the determinant, let us prepare our mathematical components. We are given $a = \alpha - i\beta$, which implies the complex conjugate is $\bar{a} = \alpha + i\beta$.

Next, consider the coefficient of $x$ in the second equation: $8e^{i\frac{2\pi}{3}}$. Using Euler's formula, $e^{i	heta} = \cos 	heta + i \sin 	heta$, we convert this into Cartesian form:

$$8 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} ight) = 8 \left( -\frac{1}{2} + i\frac{\sqrt{3}}{2} ight)$$

This transformation turns the exponential expression into a standard complex number, simplifying our upcoming expansion.

### The Algebraic Dance

We set the determinant of the coefficient matrix to zero:

$$\begin{vmatrix} 4i & 1 + i \ 8e^{i\frac{2\pi}{3}} & \bar{a} \end{vmatrix} = 0$$

Expanding this $2 	imes 2$ determinant, we obtain:

$$4i\bar{a} - 8(1 + i)e^{i\frac{2\pi}{3}} = 0$$

Substituting our prepared values for $\bar{a}$ and the exponential term, the equation becomes:

$$4i(\alpha + i\beta) - 8(1 + i)\left( -\frac{1}{2} + i\frac{\sqrt{3}}{2} ight) = 0$$

To simplify, we divide the entire equation by $4$:

$$i(\alpha + i\beta) - 2(1 + i)\left( -\frac{1}{2} + i\frac{\sqrt{3}}{2} ight) = 0$$

Distributing the $2$ into the second bracket yields:

$$i(\alpha + i\beta) - (1 + i)(-1 + i\sqrt{3}) = 0$$

### The Final Simplification

Expanding the terms, the first part $i(\alpha + i\beta)$ becomes $i\alpha + i^2\beta$, which simplifies to $-\beta + i\alpha$. For the second part, we expand the product $(1 + i)(-1 + i\sqrt{3})$:

$$(1 + i)(-1 + i\sqrt{3}) = -1 + i\sqrt{3} - i + i^2\sqrt{3} = (-1 - \sqrt{3}) + i(\sqrt{3} - 1)$$

Combining these results, we have:

$$(-\beta + i\alpha) - [-(1 + \sqrt{3}) + i(\sqrt{3} - 1)] = 0$$

Distributing the negative sign, we get:

$$(-\beta + 1 + \sqrt{3}) + i(\alpha - (\sqrt{3} - 1)) = 0$$

For this complex number to be zero, both the real and imaginary parts must vanish independently. Setting the real part to zero gives $\beta = 1 + \sqrt{3}$, and setting the imaginary part to zero gives $\alpha = \sqrt{3} - 1$.

Finally, we calculate the ratio:

$$\frac{\alpha}{\beta} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$$

Rationalizing the denominator by multiplying by $(\sqrt{3} - 1)$:

$$\frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 1 - 2\sqrt{3}}{2} = \frac{4 - 2\sqrt{3}}{2} = \mathbf{2 - \sqrt{3}}$$

Select Answer:

Visualized Solution

Analyze the System Type

Define Complex Conjugate $\overset{a}{ˉ}$

Convert Euler Form to Cartesian

Expand the Determinant

Substitute $\overset{a}{ˉ}$ and $e^{i \frac{2 π}{3}}$

Simplify the Constants

Expand the First Term

Expand the Product

Combine All Terms

Separate Real and Imaginary Parts

Solve for $α$ and $β$

Calculate the Ratio $\frac{α}{β}$

Final Simplification

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

Preparing the Ingredients

The Algebraic Dance

The Final Simplification

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