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7.2 Systems of Linear Equations

Triangular form and back-substitution

A linear equation is a polynomial equation in which each term is of degree \(0\) or \(1\). (The degree of a term of a polynomial is found by adding the exponents on all the variables of the term.) The following is an example of a system of linear equations. More particularly, it is an example of a triangular system of linear equations because of the triangular arrangement of the variables.

Example:

\(\mathbb{S}_1\)

\(\begin{eqnarray*} x + y + 2 z &=& 5\\ y\, -\, z &=& - 2\\ z &=& 3 \end{eqnarray*}\)

A triangular system is easily solved by the method of back-substitution. The value \(z = 3\) is back-substituted into \(E_2\) to yield the equivalent system

\(\mathbb{S}_2\)

\(\begin{eqnarray*} x + y + 2 z &=& 5\\ y \quad\quad &=& 1\\ z &=& 3 \end{eqnarray*}\)

Next, the values \(z = 3\) and \(y = 1\) are back-substituted into \(E_1\) to get the equivalent system

\(\mathbb{S}_3\)

\(\begin{eqnarray*} x \quad\quad &=& -2\\ y \quad &=& 1\\ z &=& 3 \end{eqnarray*}\)

for which the solution \(( -2, 1, 3 )\) is apparent.

Exercise 7.2.1

Solve the triangular system by the method of back substitution:

\(\mathbb{S}_1\)

\(\begin{eqnarray*} x + 2 y - z &=& 3\\ y - z &=& 1\\ z &=& 2 \end{eqnarray*}\)

See Solution

Triangularizing a linear system

In order to use the method of back substitution, the system must be in upper triangular form.

But what if the system is not in upper triangular form?

If a system is not in upper triangular form, it is possible to replace it by an equivalent system which is in upper triangular form by a specific process which we are about to describe.

In order to describe the process of triangularizing a system of linear equations, it is necessary to learn what operations on a system of equations produce equivalent systems.

Operations producing equivalent systems

One of the simplest and most obvious such operations is switching two equations in the system. For example, if one is given the following system

\(\mathbb{S}_1\)

\(\begin{eqnarray*} x + 2 y - z &=& 3\\ z &=& 2\\ y - z &=& 1 \end{eqnarray*}\)

The operation \(E_2 \leftrightarrow E_3\) merely switches equations two and three of the system to produce an upper triangular system:

\(\mathbb{S}_2\)

\(\begin{eqnarray*} x + 2 y - z &=& 3\\ y - z &=& 1\\ z &=& 2\\ \end{eqnarray*}\)

We can symbolize the Switching Principle by saying that operations of the form \(E_n \leftrightarrow E_m\) always produce equivalent systems.

In the last section, we noted the principle that any equation in a system can be replaced by an equivalent equation. One way to do this is to replace an equation by a non-zero multiple of itself. This principle can be symbolized by saying that operations of the form \(E_n \rightarrow c E_n\) where \(c \ne 0\) produce equivalent systems.

The final of the three principles which we will need is more abstract and less obvious than the first two. It states that any equation in the system may be replaced by its sum with some multiple of one of the other equations. This may be symbolized by saying that operations of the form \(E_n \rightarrow c E_m + E_n\) produce equivalent systems.

Summarizing, the following three types of operation produce equivalent systems:

  1. \(E_n \leftrightarrow E_m\)
  2. \(E_n \rightarrow c E_n\)
  3. \(E_n \rightarrow c E_m + E_n\)

Exercise 7.2.2

Perform the following operations on the system:

  1. \(E_2 \rightarrow - 2 E_1 + E_2\)
  2. \(E_3 \rightarrow 3 E_1 + E_3\)

\(\mathbb{S}_1\)

\(\begin{eqnarray*} x + 2 y - z &=& 0\\ 2 x + y + z &=& 0\\ - 3 x + y + 2 z &=& - 6 \end{eqnarray*}\)

Notice that the operations have the effect of removing the \(x\) terms from \(E_2\) and \(E_3\), making the system closer to the form of a triangular system.

See Solution

Exercise 7.2.3

Perform the operation \(E_2 \rightarrow -\frac{1}{3} E_2\) on \(\mathbb{S}_1\) to produce \(\mathbb{S}_2\). Then perform the operation \(E_3 \rightarrow - 7 E_2 + E_3\) on \(\mathbb{S}_2\) to produce a equivalent system \(\mathbb{S}_3\).

The system \(\mathbb{S}_3\) should be upper triangular. Solve it by back substitution.

\(\mathbb{S}_1\)

\(\begin{eqnarray*} x + 2 y - z &=& 0\\ - 3 y + 3 z &=& 0\\ 7 y - z &=& - 6 \end{eqnarray*}\)

See Solution

Exercise 7.2.4

Triangularize the system and solve by back substitution.

\(\mathbb{S}_1\)

\(\begin{eqnarray*} x + 2 y - z &=& 1\\ x \quad\quad + z &=& 3\\ y + z &=& 1 \end{eqnarray*}\)

See Solution