Question

Determine which of the following systems have non-trivial solutions: $$ \text { (a) } x+2 y-z=0 } \\ {3 x+y+2 z=0} \\ {x+y=0} \end{array} $$ $$ \text { (b) } \begin{aligned} &2 x-3 y-2 z=0 \\ &3 x+y-3 z=0 \\ &x-7 y-z=0 \end{aligned} $$ $$ \text { (c) } \begin{aligned} x+2 y+3 z &=0 \\ 4 x-3 y-z &=0 \\ 6 x+y+3 z &=0 \end{aligned} $$ $$ \text { (d) } \begin{aligned} &x+3 z=0 \\ &x-y=0 \\ &y+2 z=0 \end{aligned} $$

Short answer

a) \(\begin{cases} x + 2y - z = 0 \\ 3x + y + 2z = 0 \\ x + y = 0 \end{cases}\) b) \(\begin{cases} 2x - 3y - 2z = 0 \\ 3x + y - 3z = 0 \\ x - 7y - z = 0 \end{cases}\) c) \(\begin{cases} x + 2y + 3z = 0 \\ 4x - 3y - z = 0 \\ 6x + y + 3z = 0 \end{cases}\) d) \(\begin{cases} x + 3z = 0 \\ x - y = 0 \\ y + 2z = 0 \end{cases}\) Answer: Systems (a) and (c) have non-trivial solutions.

Step-by-step solution

System a: Coefficient Matrix and Augmented Matrix

Write down the coefficient matrix \(A\) and the augmented matrix \([A|0]\) for system (a): $$ A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 1 & 2 \\ 1 & 1 & 0 \end{bmatrix}, [A|0] = \begin{bmatrix} 1 & 2 & -1 & 0\\ 3 & 1 & 2 & 0\\ 1 & 1 & 0 & 0 \end{bmatrix} $$

System a: Calculate Ranks

Apply Gaussian elimination to the augmented matrix and calculate the ranks: $$ [A|0] \rightarrow \begin{bmatrix} 1 & 2 & -1 & 0\\ 3 & 1 & 2 & 0\\ 1 & 1 & 0 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 2 & -1 & 0\\ 0 & -5 & 5 & 0\\ 0 & -1 & 1 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 2 & -1 & 0\\ 0 & -5 & 5 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} $$ The rank \(r_A\) of the coefficient matrix A is 2, and the rank \(r_{[A|0]}\) of the augmented matrix \([A|0]\) is also 2.

System a: Compare Ranks

In system (a), we have 3 unknowns (x, y and z). Since \(r_A = r_{[A|0]} = 2 < 3\), system (a) has non-trivial solutions. Repeat the process for other systems:

System b:

: Coefficient matrix A and augmented matrix [A|0]: $$A = \begin{bmatrix} 2 & -3 & -2 \\ 3 & 1 & -3 \\ 1 & -7 & -1 \end{bmatrix}, [A|0] = \begin{bmatrix} 2 & -3 & -2 & 0\\ 3 & 1 & -3 & 0\\ 1 & -7 & -1 & 0 \end{bmatrix} $$ Applying Gaussian elimination, we get \(r_A = r_{[A|0]} = 3\). Since we have three unknowns and the rank of both matrices is equal to the number of unknowns, system (b) does not have non-trivial solutions.

System c:

: Coefficient matrix A and augmented matrix [A|0]: $$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & -3 & -1 \\ 6 & 1 & 3 \end{bmatrix}, [A|0] = \begin{bmatrix} 1 & 2 & 3 & 0\\ 4 & -3 & -1 & 0\\ 6 & 1 & 3 & 0 \end{bmatrix} $$ Applying Gaussian elimination, we get \(r_A = r_{[A|0]} = 2\). Since \(r_A = r_{[A|0]} < 3\) (where 3 is the number of unknowns), system (c) has non-trivial solutions.

System d:

: Coefficient matrix A and augmented matrix [A|0]: $$A = \begin{bmatrix} 1 & 0 & 3 \\ 1 & -1 & 0 \\ 0 & 1 & 2 \end{bmatrix}, [A|0] = \begin{bmatrix} 1 & 0 & 3 & 0\\ 1 & -1 & 0 & 0\\ 0 & 1 & 2 & 0 \end{bmatrix} $$ Applying Gaussian elimination, we get \(r_A = r_{[A|0]} = 3\). Since we have three unknowns and the rank of both matrices is equal to the number of unknowns, system (d) does not have non-trivial solutions. In conclusion, systems (a) and (c) have non-trivial solutions, while systems (b) and (d) do not.

Key concepts

Systems of Linear Equations

When we talk about systems of linear equations, we're referring to a collection of equations that we aim to solve simultaneously. Each equation in the system represents a line in two dimensions, a plane in three dimensions, or a hyperplane in higher dimensions. Intersection points of these geometrical figures are the solutions to the system.

A non-trivial solution is a solution where the variables are not all zero; in other words, it's a point, line, or plane that is not the origin. To find a non-trivial solution in a system, we need at least as many equations as variables. However, this does not guarantee a unique solution; it's possible for systems to have infinitely many solutions, especially when the equations describe the same geometrical figure. Thus, identifying the conditions under which a system has non-trivial solutions is crucial for properly solving systems of linear equations.

Gaussian Elimination

The technique known as Gaussian elimination is a systematic approach for solving systems of linear equations. It's a type of algorithm that simplifies a system to a form that is easier to solve, known as row echelon form. The goal is to use row operations to create zeros below the pivot positions, which are typically the leading coefficients in each row.

Three row operations are permitted: swapping two rows, multiplying a row by a non-zero scalar, and adding or subtracting a multiple of one row from another row. By applying these operations, we can simplify our system to where we can read off the solutions, or determine if there's a single solution, infinitely many, or none (aside from the trivial solution of all variables being zero).

Understanding Gaussian Elimination Steps

In the exercise provided, we notice that Gaussian elimination is used to reduce the augmented matrices. By achieving a row with zeros, we observe that there may be dependencies among the equations, which can lead to the existence of non-trivial solutions.

Rank of a Matrix

The concept of the rank of a matrix is closely tied to finding solutions of systems of linear equations. The rank is essentially the number of linearly independent rows (or columns) in a matrix. A row is linearly independent if it cannot be written as a combination of the other rows.

The significance of matrix rank in the context of solving linear systems is that it tells us the maximum number of independent equations in the system. If a system has more variables than the rank, it means that the system has at least one parameter - which is a variable that can be assigned any value - leading to infinitely many solutions. The rank helps us determine if the system is overdetermined (more equations than variables), underdetermined (more variables than equations), or perfectly determined (the same number of variables and equations).

Applying Rank to Determine Non-trivial Solutions

In the exercise, the rank of the matrices was compared with the number of variables to identify non-trivial solutions. Systems (a) and (c) had ranks less than the number of variables, indicating the potential for multiple solutions, while systems (b) and (d) had full rank, pointing to a single unique solution (or no solution) for each system.