Question

Show that the following homogeneous system of equations has a unique solution: $$ \begin{aligned} \mathrm{x}_{1}+2 \mathrm{x}_{2}+\mathrm{x}_{3} &=0 \\ \mathrm{x}_{2}-3 \mathrm{x}_{3} &=0 \\ -\mathrm{x}_{1}+\mathrm{x}_{2}-\mathrm{x}_{3} &=0 \end{aligned} $$

Short answer

To show that the given homogeneous system of equations has a unique solution, we first write the coefficient matrix A as: $$ A=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -3 \\ -1 & 1 & -1 \end{bmatrix} $$ Then, we calculate the determinant of A: $$ det(A) = 1 \cdot \begin{vmatrix} 1 & -3 \\ 1 & -1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 0 & -3 \\ -1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 1 \\ -1 & 1 \end{vmatrix} = 4 + 6 + 1 = 11 $$ Since det(A) = 11 ≠ 0, the given homogeneous system of equations has a unique solution.

Step-by-step solution

Write down the coefficient matrix A and the constant matrix B

The given system of equations can be written in the form of AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. Write down the coefficient matrix A and the constant matrix B from the given system: $$ A=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -3 \\ -1 & 1 & -1 \end{bmatrix}, \quad X=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \quad B=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

Calculate the determinant of matrix A

In order to determine if the system has a unique solution, we need to calculate the determinant of the coefficient matrix A. If det(A) ≠ 0, the system has a unique solution. The determinant of A can be calculated as follows: $$ det(A) = \begin{vmatrix} 1 & 2 & 1 \\ 0 & 1 & -3 \\ -1 & 1 & -1 \end{vmatrix} = 1 \cdot \begin{vmatrix} 1 & -3 \\ 1 & -1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 0 & -3 \\ -1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 1 \\ -1 & 1 \end{vmatrix} $$

Simplify and check if the determinant is non-zero

Now, let's simplify the determinant we obtained in step 2 and check if it is non-zero: $$ det(A) = 1 \cdot ((1 \cdot (-1)) - (-3 \cdot 1)) - 2 \cdot ((0 \cdot (-1)) - (-3 \cdot (-1))) + 1 \cdot ((0 \cdot 1) - (1 \cdot (-1))) \\ det(A) = 1 \cdot (1 + 3) - 2 \cdot (0 - 3) + 1 \cdot (0 + 1) \\ det(A) = 4 + 6 + 1 \\ det(A) = 11 $$ Since det(A) = 11 ≠ 0, the given homogeneous system of equations has a unique solution.

Key concepts

Coefficient Matrix

The coefficient matrix is a crucial element in understanding systems of linear equations. In our example, the matrix is formed by aligning the coefficients of the variables from each equation in the system. It is expressed in the form of a square or rectangular array, where the entry in the i-th row and the j-th column represents the coefficient of the j-th variable in the i-th equation.

One key property of the coefficient matrix is that it can tell us much about the system of equations it represents, especially when it comes to solving the system. If the determinant of this matrix (which we will discuss shortly) is non-zero, the system has a unique solution, implying that the equations are linearly independent and intersect at a single point.

In our exercise, the coefficient matrix is:
\[A=\begin{bmatrix}1 & 2 & 1 \0 & 1 & -3 \-1 & 1 & -1\end{bmatrix}\]

Determinant Calculation

Determinant calculation is a method used to derive a single number, the determinant, from a square matrix. The value of the determinant can inform us about the matrix's properties and the system of equations it reflects. Such properties include whether the matrix is invertible or if the system has a unique solution.

The determinant of a 2x2 matrix is straightforward to calculate, but when it comes to a 3x3 matrix or larger, the process often involves breaking down the matrix into smaller 2x2 matrices, which is precisely what was done in the step-by-step solution. This process is known as the expansion by minors.
\[\text{For matrix A, the determinant is calculated as follows:} \]\[det(A) = 1 (1 \times (-1) - (-3 \times 1)) - 2 (0 \times (-1) - (-3 \times (-1))) + 1 (0 \times 1 - 1 \times (-1))\]
After computation, we find that \(\det(A) = 11\), which is indeed non-zero. This non-zero determinant indicates that the coefficient matrix is invertible and, consequently, that the corresponding system of linear equations has a unique solution.

Unique Solution

The existence of a unique solution to a homogeneous system of equations indicates that, with all variables set to zero, the only solution to the system is the trivial solution where all variables are equal to zero. This condition holds true when the determinant of the coefficient matrix is non-zero, as a non-zero determinant signifies that the rows of the matrix are linearly independent—no row is a scalar multiple or a linear combination of the other rows.

When solving the equations, this characteristic prevents any overlaps or infinite solutions, which often occur in cases where the determinant is zero, leading to dependent rows and consequently, dependent equations. A unique solution to a homogeneous system is vital in various fields, including engineering and physics, where it can represent a stable and predictable state of a system.

In our exercise, the unique solution is confirmed by the non-zero determinant of the coefficient matrix, ensuring that each variable can be singularly determined. Hence, the solution is consistent and unique, underscoring the reliability of the system's behavior under the given constraints.