Question

Determine which of the following systems have non-trivial solutions: (a) \(x-2 y=0\) \(3 x-6 y=0\) (b) \(3 x+y=0\) \(9 x+2 y=0\) (c) \(4 x-3 y=0\) \(-4 x+3 y=0\) (d) \(6 x-2 y=0\) \(2 x-\frac{2}{3} y=0\) (e) \(y=2 x\) \(x=3 y\)

Short answer

a) x - 2y = 5 and 3x + 6y = 15 b) 3x + y = 2 and 9x - 2y = -1 c) 4x - 3y = -8 and -4x + 3y = -12 d) 6x - 2y = 1 and 2x + (2/3)y = 4 e) y = 2x and x = 3y Answer: Systems (a) and (c) have non-trivial solutions.

Step-by-step solution

System (a)

Find the ratio of the coefficients of x and y in the two equations: Ratio1 = (Coefficient of x in Eq1)/(Coefficient of x in Eq2) = 1/3 Ratio2 = (Coefficient of y in Eq1)/(Coefficient of y in Eq2) = -2/-6 = 1/3 Since Ratio1 = Ratio2, the system is consistent and has non-trivial solutions.

System (b)

Find the ratio of the coefficients of x and y in the two equations: Ratio1 = 3/9 Ratio2 = 1/2 Since Ratio1 ≠ Ratio2, the system is inconsistent and does not have non-trivial solutions.

System (c)

Find the ratio of the coefficients of x and y in the two equations: Ratio1 = 4/-4 = -1 Ratio2 = -3/3 = -1 Since Ratio1 = Ratio2, the system is consistent and has non-trivial solutions.

System (d)

Find the ratio of the coefficients of x and y in the two equations: Ratio1 = 6/2 Ratio2 = -2/(-2/3) = 3 Since Ratio1 ≠ Ratio2, the system is inconsistent and does not have non-trivial solutions.

System (e)

In this case, the given equations are already solved for y. If we eliminate y, we can find the relationship between x and y: First equation: \(y = 2x\) Second equation: \(x = 3y\) Substitute the first equation into the second equation: \(x = 3(2x)\) \(x = 6x\) Solve for x: \(5x = 0\), so \(x = 0\) Now substitute the value of x back into the first equation: \(y = 2(0) \Rightarrow y = 0\) The system is consistent and has only the trivial solution (x=0 and y=0). Thus, it does not have non-trivial solutions.

Key concepts

Linear Algebra

Linear algebra is the branch of mathematics that deals with vectors, matrices, and linear transformations. In the context of solving systems of equations, linear algebra provides methods for finding solutions to a set of linear equations. These equations represent lines, planes, or hyperplanes in multidimensional space. When we have multiple equations, we are looking for points (solutions) where these geometric objects intersect. A solution can either be a single point, a line of points, an entire plane of points, or there might be no intersection at all, which means no solution exists.

Finding solutions is often associated with matrix operations and determinants. Key concepts such as the rank of a matrix, which represents the maximum number of linearly independent row or column vectors, help determine the nature of the solutions. If the rank is less than the number of variables, there might be infinitely many solutions (non-trivial), or no solutions at all. Systems with as many equations as there are variables, where all equations are independent, usually have a single unique solution.

Homogeneous Systems

Homogeneous systems of linear equations are a special set where all the constant terms are zero. In other words, each equation in the system typically looks like 'ax + by + cz = 0'. What's unique about homogeneous systems is that they always have at least one solution, which is the trivial solution where all variables equal zero, representing the intersection at the origin of the coordinate system. However, the interesting cases arise when we look for non-trivial solutions, which are solutions where not all the variables are zero.

Determining Non-Trivial Solutions

To ascertain if non-trivial solutions exist, we often look at the coefficients of the variables. If the ratio of the coefficients of corresponding variables in different equations is the same (as seen in systems (a) and (c) in our exercise), this suggests that the equations are multiples of each other and lay on top of each other, indicating there are infinitely many solutions along a line or plane. Indicatively, we can use tools such as Gaussian elimination or evaluating the determinant of the associated matrix to further investigate the existence of non-trivial solutions.

Systems of Linear Equations

Systems of linear equations consist of two or more equations that are solved simultaneously. Solving these systems means finding all possible values of the variables that satisfy all equations in the system at once. The approach to solving these systems varies based on the number of equations relative to the number of unknowns, and the specific characteristics of the equations involved.

For example, when we have a system of two equations and two unknowns, we can graphically represent these equations as lines on a plane. The solution to the system corresponds to the points where the lines intersect. Depending on the slopes and y-intercepts of these lines, we could have a single point of intersection (unique solution), an infinite number of intersections if the lines overlap (infinitely many solutions), or no intersection if the lines are parallel (no solution).

In our exercise, systems (b) and (d) showed inconsistent ratios, which implies the lines representing the equations would be parallel, thus having no intersection. This means no non-trivial solution exists for these systems. However, it's important to note that for systems with more equations and variables, visualization becomes more complex and requires abstract methods like matrix algebra for solutions.