Math 1410, Spring 2020

Solving Systems of Equations

February 6, 2020

Sean Fitzpatrick
University of Lethbridge

Warm-Up

Examples

Which matrices are in REF? RREF? If neither, what's the next step?

\begin{equation*} \bbm 2 \amp 1 \amp 0\\0\amp 1\amp 2\\0\amp 0\amp 1\ebm \qquad \bbm 1 \amp 0 \amp -2\\0\amp 1 \amp 3\\0\amp 0\amp 0\ebm \qquad \bbm 1\amp 0 \amp 1\\0\amp 0\amp 1\\0\amp 0\amp 0\ebm \end{equation*}
\begin{equation*} \bam{ccc|c}1\amp 0 \amp 2\amp 3\\0\amp 1\amp -1\amp 1\\0\amp 0\amp 0\amp 1\eam \qquad \bam{cccc|c}0 \amp 1\amp 0\amp 0\amp -2\\0\amp 0\amp 1\amp 1\amp 3\\0\amp 0\amp 0\amp 2\amp 1\\0\amp 0\amp 0\amp 0\amp 0\eam \end{equation*}

Solving systems

Gauss-Jordan Elimination

  • Forward steps: perform Gaussian elimination to reach row-echelon form.

    • You could solve by back substitution at this stage.

    • (If there's no solution, you should already be able to tell.)

    • Back substitution can be tricky if there are parameters involved.

  • Backward steps: starting with right-most leading one, create zeros above, working up and to the left.

Examples

Example 1

Solve the system:

\begin{equation*} \begin{array}{ccccccc} 2x\amp - \amp y\amp +\amp 4z\amp =\amp 3\\ x \amp - \amp y\amp +\amp 2z\amp =\amp -2\\ -2x\amp + \amp 3y \amp - \amp z\amp =\amp 0 \end{array} \end{equation*}

Example 2

Solve the system:

\begin{equation*} \begin{array}{ccccccc} x\amp - \amp 2y \amp + \amp 3z\amp = \amp 4\\ 2x\amp - \amp 3y \amp +\amp 4z \amp = \amp -1\\ x \amp -\amp 3y \amp + \amp 5z \amp = \amp 13 \end{array} \end{equation*}

Example 3

Solve the system:

\begin{equation*} \begin{array}{ccccc} 2x \amp - \amp 5y \amp = \amp 6\\ x \amp - \amp 4y \amp = \amp 4\\ -3x \amp + \amp 10y \amp = \amp -7 \end{array} \end{equation*}

Example 4

Solve the system:

\begin{equation*} \begin{array}{ccccccccc} x_1\amp - \amp 3x_2 \amp \amp \amp + \amp 5x_4\amp = \amp 4\\ 2x_1\amp -\amp 4x_2 \amp + \amp 6x_3 \amp -\amp 2x_4\amp = \amp -9 \end{array} \end{equation*}

Homogeneous systems of equations

Homogeneous systems

Definition:

A system of linear equations is homogeneous if all the constant terms (right-hand sides) are zero.

Example:

\begin{equation*} \begin{array}{ccccccc} 3x\amp-\amp 5y \amp + \amp 2z \amp =\amp 0\\ 2x\amp + \amp 2y\amp + \amp 4z \amp = \amp 0 \end{array} \end{equation*}

For homogeneous systems, existence of solutions is not in question. (Why?) Instead, we're interested in whether solutions are unique.

Example

Solve the homogeneous system

\begin{equation*} \begin{array}{ccccccccc} x_1\amp - \amp 2x_2 \amp + \amp x_3 \amp -\amp 4x_4 \amp = \amp 0\\ -2x_1\amp + \amp 4x_2 \amp -\amp 3x_3 \amp - \amp 5x_4\amp = \amp 0\\ -x_1 \amp + \amp 2x_2 \amp + \amp x_3 \amp - \amp 13x_4 \amp = \amp 0 \end{array} \end{equation*}

Vector solutions

It can be convenient to write our solutions in vector form. For later work with matrices, we use column vectors. Instead of giving solutions for \(x_1, x_2, \ldots, x_n\) separately, we collect things into a vector \(\bbm x_1\\x_2\\ \vdots\\ x_n\ebm\text{.}\)

Vector solutions, continued

This can be notationally convenient. Instead of

\begin{equation*} \begin{array}{ccccccccc} a_{11}x_1 \amp + \amp a_{12}x_2\amp + \amp \cdots \amp + \amp a_{1n}x_n \amp = \amp b_1\\ a_{21}x_1 \amp + \amp a_{22}x_2\amp + \amp \cdots \amp + \amp a_{2n}x_n \amp = \amp b_2\\ \vdots \amp \amp \vdots \amp \amp \amp \amp \vdots \amp \amp \vdots\\ a_{m1}x_1 \amp + \amp a_{m2}x_2\amp + \amp \cdots \amp + \amp a_{mn}x_n \amp = \amp b_m \end{array} \end{equation*}
write
\begin{equation*} \bbm a_{11} \amp a_{12} \amp \cdots \amp a_{1n}\\ a_{21} \amp a_{22} \amp \cdots \amp a_{2n}\\ \vdots \amp \vdots \amp \ddots \amp \vdots\\ a_{m1} \amp a_{m2} \amp \cdots \amp a_{mn}\ebm \bbm x_1 \\ x_2 \\ \vdots \\ x_n \ebm = \bbm b_1 \\b_2 \\ \vdots \\ b_m\ebm\text{,} \end{equation*}
or simply, \(A\vec{x}=\vec{b}\text{.}\)

Basic solutions

For a homogeneous system with parameters \(t_1, t_2, \ldots, t_k\) in the vector solution, the basic solutions are obtained by setting one parameter equal to 1, and the others to 0.

Notice that if \(\vec{v}\) and \(\vec{w}\) are both solutions to \(A\vec{x}=\vec{0}\text{,}\) then so is \(s\vec{v}+t\vec{w}\) for any real numbers \(s,t\text{.}\)

Example

Find the basic solutions to

\begin{equation*} \begin{array}{ccccccccc} x_1\amp - \amp 2x_2 \amp + \amp x_3 \amp -\amp 4x_4 \amp = \amp 0\\ -2x_1\amp + \amp 4x_2 \amp -\amp 3x_3 \amp - \amp 5x_4\amp = \amp 0\\ -x_1 \amp + \amp 2x_2 \amp + \amp x_3 \amp - \amp 13x_4 \amp = \amp 0 \end{array} \end{equation*}