If $\left| {\begin{array}{*{20}{c}} {\rm{x}}&{\rm{y}}&0\ 0&{\rm{x}}&{\rm{y}}\ {\rm{y}}&0&{\rm{x}} \end{array}} \right| = 0$ , then which one of the following is correct?

Question

Prepp Team · Accepted Answer

Solving the Given Determinant Equation

The problem asks us to analyze the relationship between x and y given that the determinant of a specific 3x3 matrix is equal to zero. The given matrix is:

$\left| {\begin{array}{*{20}{c}} {\rm{x}}&{\rm{y}}&0\\ 0&{\rm{x}}&{\rm{y}}\\ {\rm{y}}&0&{\rm{x}} \end{array}} \right| = 0$

To solve this, we first need to calculate the determinant of the matrix. We can expand the determinant along the first row using the cofactor expansion method. The determinant of a 3x3 matrix $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$ is given by $a(ei - fh) - b(di - fg) + c(dh - eg)$.

Applying this formula to our matrix:

Determinant $ = {\rm{x}} \begin{vmatrix} {\rm{x}} & {\rm{y}} \\ 0 & {\rm{x}} \end{vmatrix} - {\rm{y}} \begin{vmatrix} 0 & {\rm{y}} \\ {\rm{y}} & {\rm{x}} \end{vmatrix} + 0 \begin{vmatrix} 0 & {\rm{x}} \\ {\rm{y}} & 0 \end{vmatrix}$

Now, we calculate the 2x2 determinants:

$\begin{vmatrix} {\rm{x}} & {\rm{y}} \\ 0 & {\rm{x}} \end{vmatrix} = {\rm{x}} \cdot {\rm{x}} - {\rm{y}} \cdot 0 = {\rm{x}}^2 - 0 = {\rm{x}}^2$
$\begin{vmatrix} 0 & {\rm{y}} \\ {\rm{y}} & {\rm{x}} \end{vmatrix} = 0 \cdot {\rm{x}} - {\rm{y}} \cdot {\rm{y}} = 0 - {\rm{y}}^2 = -{\rm{y}}^2$
$\begin{vmatrix} 0 & {\rm{x}} \\ {\rm{y}} & 0 \end{vmatrix} = 0 \cdot 0 - {\rm{x}} \cdot {\rm{y}} = 0 - {\rm{xy}} = -{\rm{xy}}$

Substitute these values back into the determinant expansion:

Determinant $ = {\rm{x}}({\rm{x}}^2) - {\rm{y}}(-{\rm{y}}^2) + 0(-{\rm{xy}})$

Determinant $ = {\rm{x}}^3 + {\rm{y}}^3 + 0$

Determinant $ = {\rm{x}}^3 + {\rm{y}}^3$

The problem states that the determinant is equal to 0, so we have:

${\rm{x}}^3 + {\rm{y}}^3 = 0$

We can rearrange this equation:

${\rm{x}}^3 = -{\rm{y}}^3$

Now, let's consider the case where ${\rm{y}} \neq 0$. We can divide both sides of the equation by ${\rm{y}}^3$:

$\frac{{\rm{x}}^3}{{\rm{y}}^3} = \frac{-{\rm{y}}^3}{{\rm{y}}^3}$

$\left( \frac{{\rm{x}}}{{\rm{y}}} \right)^3 = -1$

This equation tells us that the ratio $ \frac{{\rm{x}}}{{\rm{y}}} $ is a number whose cube is -1. A number whose cube is -1 is defined as a cube root of -1.

Therefore, $ \frac{{\rm{x}}}{{\rm{y}}} $ must be one of the cube roots of -1.

Let's check the options:

Option 1: $ \frac{{\rm{x}}}{{\rm{y}}} $ is one of the cube roots of unity (meaning $ (\frac{x}{y})^3 = 1 $). This contradicts our result $ (\frac{x}{y})^3 = -1 $.
Option 2: x is one of the cube roots of unity (meaning $ x^3 = 1 $). This is not directly implied by $ x^3 + y^3 = 0 $.
Option 3: y is one of the cube roots of unity (meaning $ y^3 = 1 $). This is not directly implied by $ x^3 + y^3 = 0 $.
Option 4: $ \frac{{\rm{x}}}{{\rm{y}}} $ is one of the cube roots of -1 (meaning $ (\frac{x}{y})^3 = -1 $). This matches our derived equation.

Thus, based on the calculation, the correct statement is that $ \frac{{\rm{x}}}{{\rm{y}}} $ is one of the cube roots of -1.

Note: If ${\rm{y}} = 0$, the original equation $ {\rm{x}}^3 + {\rm{y}}^3 = 0 $ becomes $ {\rm{x}}^3 + 0^3 = 0 $, which means $ {\rm{x}}^3 = 0 $, so $ {\rm{x}} = 0 $. In this case, $ \frac{{\rm{x}}}{{\rm{y}}} $ is undefined, which isn't covered by the options. The options imply ${\rm{y}} \neq 0$ for the ratio $ \frac{{\rm{x}}}{{\rm{y}}} $ to exist.

Let's summarize the steps:

Calculate the determinant of the given matrix.
Set the determinant equal to zero as per the problem statement.
Simplify the resulting algebraic equation.
Rearrange the equation to find a relationship between x and y, specifically involving the ratio $ \frac{{\rm{x}}}{{\rm{y}}} $.
Interpret the equation involving $ \frac{{\rm{x}}}{{\rm{y}}} $ in terms of roots of a number.
Compare the result with the given options.

The determinant calculation led to the equation $ {\rm{x}}^3 + {\rm{y}}^3 = 0 $, which simplifies to $ (\frac{x}{y})^3 = -1 $ when $ y \neq 0 $. This directly indicates that $ \frac{x}{y} $ is a cube root of -1.

Revision Table: Determinant and Roots Concepts

Concept	Description	Relevance to Problem
Determinant of a Matrix	A scalar value computed from the elements of a square matrix. It provides information about the matrix, e.g., if the determinant is 0, the matrix is singular (not invertible).	The problem starts with the determinant being zero, which leads to an equation involving x and y.
Cofactor Expansion	A method to calculate the determinant of a matrix by summing the products of elements of a row or column with their corresponding cofactors.	Used to calculate the determinant of the 3x3 matrix in the problem.
Cube Root of a Number	A number z such that $z^3 = a$ for a given number a. There are three cube roots for any non-zero complex number (one real, two complex conjugate pairs for real a ≠ 0).	The final equation $ (\frac{x}{y})^3 = -1 $ shows that $ \frac{x}{y} $ is a cube root of -1.
Cube Roots of Unity	The numbers z such that $z^3 = 1$. These are 1, $ \omega = e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} $, and $ \omega^2 = e^{i4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} $.	Mentioned in option 1 and option 2, but not the correct result of this problem.
Cube Roots of -1	The numbers z such that $z^3 = -1$. These are -1, $ -\omega = -e^{i2\pi/3} = \frac{1}{2} - i\frac{\sqrt{3}}{2} $, and $ -\omega^2 = -e^{i4\pi/3} = \frac{1}{2} + i\frac{\sqrt{3}}{2} $. Note that $ -1 = (-1)^3 $. Also, $ (-\omega)^3 = (-1)^3 \omega^3 = -1 \cdot 1 = -1 $, and $ (-\omega^2)^3 = (-1)^3 (\omega^2)^3 = -1 (\omega^3)^2 = -1 \cdot 1^2 = -1 $.	This is the core concept directly matching the problem's result.

Additional Information: Properties of Cube Roots of -1

The equation $ z^3 = -1 $ can be solved algebraically or using complex numbers.

Algebraically:

$ z^3 + 1 = 0 $

This is a sum of cubes, which factors as $ (z+1)(z^2 - z + 1) = 0 $.

The solutions are:

$ z+1 = 0 \Rightarrow z = -1 $ (the real root)
$ z^2 - z + 1 = 0 $

Using the quadratic formula $ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ for $ az^2 + bz + c = 0 $:

$ z = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(1)}}{2(1)} $

$ z = \frac{1 \pm \sqrt{1 - 4}}{2} $

$ z = \frac{1 \pm \sqrt{-3}}{2} $

$ z = \frac{1 \pm i\sqrt{3}}{2} $

So the complex roots are $ \frac{1 + i\sqrt{3}}{2} $ and $ \frac{1 - i\sqrt{3}}{2} $.

These are $ -\omega^2 $ and $ -\omega $ respectively, where $ \omega $ is a principal cube root of unity $ e^{i2\pi/3} $.

The three cube roots of -1 are $ -1, \frac{1 + i\sqrt{3}}{2}, \frac{1 - i\sqrt{3}}{2} $. The problem states that $ \frac{x}{y} $ is one of these three values.