Problem 54 - Differential Equations and Linear Algebra

Question

Determine the solution set to the system \(A \mathbf{x}=0\) for the given matrix \(A\). $$A=\left[\begin{array}{rrr} 1 & 0 & 3 \\ 3 & -1 & 7 \\ 2 & 1 & 8 \\ 1 & 1 & 5 \\ -1 & 1 & -1 \end{array}\right]$$

Step-by-Step Solution

Answer

The solution set for the system \(A\textbf{x}=0\) is the trivial solution \(\textbf{x} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\), where \(x_1 = 0\), \(x_2 = 0\), and \(x_3 = 0\).

Step 1: Set up the augmented matrix of the linear system

To do this, we need to rewrite the given matrix \(A\) into an augmented matrix. To set up the augmented matrix, we write the coefficients of the variables and their constants. In this case, since \(A\textbf{x} = 0\), we will add a column of zeroes to the given matrix \(A\): $$ \left[\begin{array}{rrr|r} 1 & 0 & 3 & 0 \\ 3 & -1 & 7 & 0 \\ 2 & 1 & 8 & 0 \\ 1 & 1 & 5 & 0 \\ -1 & 1 & -1 & 0 \end{array}\right]. $$

Step 2: Perform elementary row operations to get the row-echelon form

Now we will perform elementary row operations to get the row-echelon form of the augmented matrix: First operation: \(\texttt{R2} = \texttt{R2} - 3 \texttt{R1}\), \(\texttt{R3} = \texttt{R3} - 2 \texttt{R1}\), \(\texttt{R4} = \texttt{R4} - \texttt{R1}\) and \(\texttt{R5} = \texttt{R5} + \texttt{R1}\): $$ \left[\begin{array}{rrr|r} 1 & 0 & 3 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 1 & 2 & 0 \end{array}\right]. $$ Next operation: \(\texttt{R3} = \texttt{R3} + \texttt{R2}\), \(\texttt{R4} = \texttt{R4} + \texttt{R2}\) and \(\texttt{R5} = \texttt{R5} + \texttt{R2}\): $$ \left[\begin{array}{rrr|r} 1 & 0 & 3 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 3 & 0 \end{array}\right]. $$ Finally, operation: \(\texttt{R3} = \frac{1}{3} \texttt{R3}\): $$ \left[\begin{array}{rrr|r} 1 & 0 & 3 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 3 & 0 \end{array}\right]. $$

Step 3: Interpret the row-echelon form to find the solution set

Now, we'll write out the system of linear equations from the row-echelon form: $$ \begin{cases} x_1 + 3x_3 &= 0 \\ -x_2 + x_3 &= 0 \\ x_3 &= 0 \end{cases}. $$ Since \(x_3 = 0\), the second equation in the system becomes \(-x_2 + 0 = 0,\) which simplifies to \(x_2 = 0\). Now, substituting \(x_3 = 0\) and \(x_2 = 0\) into the first equation, we obtain \(x_1 + 3(0) = 0\), which simplifies to \(x_1 = 0\). So, the solution set is \(x_1 = 0, \, x_2 = 0, \, x_3 = 0\). The only solution for the given system is the trivial solution \(\textbf{x} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\).

Key Concepts

Elementary Row Operations

In linear algebra, elementary row operations are crucial when solving systems of linear equations. They help simplify the matrices by changing them into a form that's easier to understand, such as the row-echelon form. There are three types of elementary row operations: swapping two rows, multiplying a row by a nonzero scalar, and adding or subtracting a multiple of one row to another row.

These operations are straightforward but very powerful tools. For instance, they allow us to systematically reduce a matrix to reveal key components of the solution of a system of equations.

• Swapping rows can help reposition pivotal elements to simplify calculations.
• Multiplying rows ensures that we don't alter the inherent solutions of the equations.
• Adding or subtracting rows facilitates the elimination of variables, making it easier to solve for remaining unknowns.

In the provided exercise solution, these operations transformed a complex system into a simpler row-echelon form, making it feasible to determine the zero or trivial solutions.

System of Linear Equations

A system of linear equations consists of two or more linear equations with the same set of variables. The goal is to find values for these variables that satisfy all the equations simultaneously. Such systems can be represented compactly in matrix form, which is precisely where linear algebra shines, utilizing matrices and vectors.

In general, different types of solutions can exist for a system of linear equations:

• A single unique solution.
• No solution.
• Infinitely many solutions.

In the case of the problem provided, the system is homogeneous, expressed as \(A\mathbf{x} = 0\), indicating the search for solutions where the result vector is zero. Homogeneous systems always have at least one solution and are easier due to their property that any combination of solutions is also a solution.

Trivial Solution

The trivial solution in a system of linear equations refers to the simplest case where all variables are equal to zero. In context, when working with a homogeneous system like \(A\mathbf{x} = 0\), the trivial solution \(\mathbf{x} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}\) always exists.

This solution is called 'trivial' because regardless of the matrix, this zero vector is input as a straightforward solution for the equation \(A\mathbf{x} = 0\). It shows that by not altering the system, it satisfies the equation.

However, it's important to check if other, nontrivial solutions exist, which could imply additional research or computation into the matrix's rank. Nontrivial solutions imply that there are free variables resulting in more complex solutions than the trivial one. In the original exercise, the absence of nontrivial solutions was confirmed, meaning the trivial solution was solely sufficient for the given scenario.