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Section 1.3 Systems of Linear Equations

Consider a rectangular plate that has been heated, with heat sensors placed in a grid fashion as below:
A figure of heat sensors placed along a rectangular plate.
Figure 1.3.1. Heat sensors placed on a rectangular plate.
We wish to estimate the temperature of the plate at the points indicated by \(T_1\) and \(T_2\) using information from the surrounding sensors. It's reasonable to approximate the temperature at these points by taking an average of the surrounding temperatures. This gives us the system of equations
\begin{align*} T_1 \amp = \frac{10 + 35 + 30 + T_2}{4} \\ T_2 \amp = \frac{25 + T_1 + 40 + 30}{4} \end{align*}
Both equations are relatively simple since we're not doing anything overly complicated to our unknowns \(T_1\) and \(T_2\text{.}\) In fact, each equation determines a line in \(\RR^2\text{.}\) This suggests the following terminology.

Definition 1.3.2. Linear Equations and Linear Systems.

A linear equation in the variables \(x_1, x_2, \ldots, x_n\) is an equation of the form
\begin{equation*} \sum_{i=1}^n a_i x_i = b\text{,} \end{equation*}
where \(a_1, \ldots, a_n, b\in\RR\) are constants. A system of linear equations, or a linear system, is a system of equations in which each equation is a linear equation. A solution of a linear system is a point \((s_1, s_2, \ldots, s_n)\) whose coordinates satisfy each equation in the system if they are substituted into the corresponding variables \(x_1, x_2,\ldots, x_n\text{.}\) The set of all solutions of a linear system is the solution set. A system is inconsistent if its solution set is empty and consistent if its solution set contains at least one point. A system is homogeneous if it contains no nonzero constant terms, otherwise it is non-homogeneous. A homogeneous system always has \(\mathbf{0}\) as a solution (called the trivial solution), and is therefore always consistent.
The solution set of a linear system in two variables can be viewed as the intersection of a system of lines in the \(xy\)-plane. In higher dimensions, we replace lines with planes and “hyperplanes”.

Example 1.3.3. Determining if a System Is Consistent.

Is the system
\begin{align*} 3x_1 + 6x_2 \amp = -3 \\ 5x_1 + 7x_2 \amp = 10 \end{align*}
consistent?
Solution.
We can try graphing each equation to see if they correspond to intersecting lines. If we do so, we obtain the following plot:
Figure 1.3.4. Graphing a linear system.
Since these lines intersect, the system has a solution. A solution can also be found algebraically by using the method of elimination.
When using elimination to solve a linear system, the variables themselves aren't particularly important. Instead it is the coefficients that determine any solutions. We therefore introduce the use of matrices to express and solve linear systems. In particular, any linear system can be rewritten as the matrix equation \(A\xx = \bb\text{,}\) where \(A\) is the coefficient matrix. The solution process then amount to simplifying the so-called augmented matrix \(\mqty[A \amp \bb]\) using a series of row operations.

Definition 1.3.5. Elementary Row Operations.

Let \(A\) be a matrix. The elementary row operations are defined as follows:
  1. Row replacement: replace one row of \(A\) by the sum of itself and a scalar multiple of another row of \(A\text{.}\)
  2. Interchange: swap two rows.
  3. Scaling: multiply a single row by a nonzero scalar.
If we can obtain the matrix \(B\) from \(A\) by performing a series of elementary row operations, we say that \(B\) is row equivalent to \(A\) and write \(A\sim B\text{.}\)
If the augmented matrices of two linear systems are row equivalent, then the systems have the same solution set. This observation will be fundamental primary tool for solving linear systems.

Example 1.3.6. Solving a System by Row Reduction.

Solve the system
\begin{align*} 2x_1 - 6x_3 \amp = -8 \\ x_2 + 2x_3 \amp = 3 \\ 3x_1 + 6x_2 - 2x_3 \amp = -4 \end{align*}
Solution.
First we set up the augmented matrix:
\begin{equation*} A = \mqty[2 \amp 0 \amp -6 \amp -8 \\ 0 \amp 1 \amp 2 \amp 3 \\ 3 \amp 6 \amp -2 \amp -4]\text{.} \end{equation*}
Now we reduce this matrix using a sequence of elementary row operations. This is easily handled using Octave's rref command as shown below.
The resulting matrix is row equivalent to the original augmented matrix but now corresponds to the system
\begin{align*} x_1 \amp = 2 \\ x_2 \amp = -1 \\ x_3 \amp = 2 \text{.} \end{align*}
The solution of the original system is therefore \((2, -1, 2)\text{.}\)