Question

Solve the given set of equations, or else show that there is no solution. $$ \begin{aligned} x_{1}+2 x_{2}-x_{3} &=0 \\ 2 x_{1}+x_{2}+x_{3} &=0 \\\ x_{1}-x_{2}+2 x_{3} &=0 \end{aligned} $$

Short answer

Based on the step-by-step solution provided: Question: Determine if the following system of linear equations has a unique solution or not: x1 + 2x2 - x3 = 0 2x1 + x2 + x3 = 0 x1 - x2 + 2x3 = 0 Answer: The given system of linear equations does not have a unique solution.

Step-by-step solution

Formulate the Equations as Matrices

Firstly, let's write the given system of equations in matrix form \textrm{A} . \textrm{x} = \textrm{b} where A is the matrix of coefficients, x is the vector of variables \[x_{1}, x_{2}, x_{3}\], and b is the vector of the constants on the right hand side of the equations, which are 0 in this case. So, A: \[ \begin{pmatrix} 1 & 2 & -1 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \\ \end{pmatrix} \] x: \[ \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{pmatrix} \] b: \[ \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} \]

Find the Determinant of Matrix A

Next, the determinant of A (denoted as |A|) needs to be found. This is a determinant of a 3 x 3 matrix and can be evaluated as follows: \[ |A| = 1 * (1*2 - 1*(-1)) - 2 * (2*2 - 1*1) - (-1) * (2*(-1) - 1*1) \] Calculate this, and the result is 0.

Check for a Solution

Since the determinant of A is equal to 0, it indicates the system of equations has no unique solution. Therefore, the given set of equations does not have a solution.

Key concepts

Matrix Representation of Linear Equations

When dealing with linear equations, a convenient and compact way to represent them is through the use of matrices. Matrix representation turns a system of linear equations into a matrix equation of the form \textbf{A.x = b}. Here's how the process works:

In the given exercise, we have a system of three equations with three variables, namely, \( x_1, x_2, \) and \( x_3 \). To form a matrix representation, we need to isolate the coefficients of these variables and arrange them into a square matrix, known as the coefficient matrix \( \textbf{A} \). Similarly, we represent the variables by a column vector \( \textbf{x} \), and the constant terms by another column vector \( \textbf{b} \).

In this particular case, the system is homogeneous since all the constant terms are zero, leading to a homogeneous matrix equation \( \textbf{A.x = 0} \), which simplifies the analysis of the system. Using matrix representation can make some operations more straightforward, for example, determining if the system has a unique solution, no solution, or infinitely many solutions via manipulation or calculation of the determinant.

Determinant of a Matrix

The determinant of a matrix is a special scalar value that provides important information about the matrix and the system it represents. For a 3x3 matrix \( \textbf{A} \), the determinant can be calculated using a method called expansion by cofactors.

The determinant \( |\textbf{A}| \), when non-zero, indicates that the matrix \( \textbf{A} \) is invertible and the system has a unique solution. However, if the determinant equates to zero, as it does in the provided exercise, this implies that the matrix \( \textbf{A} \) is singular and the system may have either no solutions or an infinite number of solutions. It is an essential step to determine the nature of the solutions of a system of linear equations. Calculating the determinant can initially seem tricky, but with practice, it becomes a systematic process of multiplying and adding certain entries of the matrix.

Homogeneous Equations

Homogeneous equations are systems where all of the constant terms are zero. They take on the form \( \textbf{A.x = 0} \), where \(\textbf{0}\) denotes the zero vector. A fundamental property of homogeneous systems is that they always have at least one solution, namely, the trivial solution where all variables are zero (\( x_{1} = x_{2} = x_{3} = 0 \)).

However, the presence of additional solutions depends on the determinant of the coefficient matrix \(\textbf{A}\). If the determinant of \(\textbf{A}\) is zero (as in our example), it signals the possibility of non-trivial solutions existing, which means there could be infinitely many solutions that satisfy the system. Solving homogeneous equations usually involves finding the null space or kernel of the matrix, which consists of all the vectors \(\textbf{x}\) that make the matrix equation \(\textbf{A.x}\) equal to the zero vector.