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Section 17.3 Matrices and Systems of Linear Equations

Consider the following system of linear equations

\begin{align*} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n \amp= b_1\\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n \amp= b_2\\ \quad\amp \vdots \quad\\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n \amp= b_m \end{align*}

Now that we have introduced matrix multiplication we can write this system as

\begin{equation*} A \mathbf{x} = \mathbf{b} \end{equation*}

where \(A\) is the \(m \times n\) coefficient matrix

\begin{equation*} A = \begin{pmatrix} 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}\\ \end{pmatrix}, \end{equation*}

\(\mathbf{x}\) is the \(n \times 1\) matrix (i.e. the column vector of the variables)

\begin{equation*} \mathbf{x}=\begin{pmatrix}x_1 \\ x_2 \\\vdots \\x_n\end{pmatrix} \end{equation*}

and \(\mathbf{b}\) is the \(m \times 1\) (i.e. the column vector of the constants)

\begin{equation*} \mathbf{b} = \begin{pmatrix}b_1 \\ b_2 \\ \vdots \\ b_m\end{pmatrix}. \end{equation*}

The system of linear equations

\begin{align*} 3x_1 - x_2 + 2x_3 + x_4 \amp= 2\\ -x_1 + 6x_2 - 11x_4 \amp= -22\\ x_1 + x_2 + 7x_3 + 4x_4 \amp= 8 \end{align*}

can be written as

\begin{equation*} A = \begin{pmatrix} 3 \amp -1 \amp 2 \amp 1 \\ -1 \amp 6 \amp 0 \amp -11 \\ 1 \amp 1 \amp 7 \amp 4 \end{pmatrix} \begin{pmatrix}x_1\\ x_2 \\ x_3 \\ x_4\end{pmatrix} = \begin{pmatrix} 2 \\ -22 \\ 8\end{pmatrix}. \end{equation*}

We can now use the properties of matrix multiplication to establish some interesting facts about the solutions to systems of linear equations. For example:

  1. Let \(\mathbf{x} = \mathbf{u}\) and \(\mathbf{x} = \mathbf{v}\) be two distinct solutions to (17.3.1) and let \(\mathbf{w} = \mathbf{u} − \mathbf{v}\text{.}\) By using the properties of matrix multiplication we can see that

    \begin{equation*} A\mathbf{w} = A(\mathbf{u} − \mathbf{v}) = A\mathbf{u} − A\mathbf{v} =\mathbf{b} − \mathbf{b} = \mathbf{0}. \end{equation*}

    So, any vector of the form \(\mathbf{u} + t\mathbf{w}\text{,}\) \(t \in \mathbb{R}\text{,}\) will be a solution to (17.3.1) since

    \begin{equation*} A(\mathbf{u} +t\mathbf{w}) = A\mathbf{u} + A(t\mathbf{w}) = A\mathbf{u} +tA\mathbf{w} = \mathbf{b} +t\mathbf{0} = \mathbf{b}\text{,} \end{equation*}

    i.e. (17.3.1) has an infinite number of solutions.

  2. Let \(\mathbf{x} = \mathbf{u}\) be the unique solution to (17.3.1) and let \(\mathbf{x} = \mathbf{v}\) be a non-zero solution to (17.3.2). Then

    \begin{equation*} A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v} = \mathbf{b} + \mathbf{0} = \mathbf{b} \end{equation*}

    and so \(\mathbf{x} = \mathbf{u} + \mathbf{v}\) is another solution to (17.3.1), which is not possible. Thus there cannot be any non-zero solutions to (17.3.2).

  3. Since (17.3.1) has an infinite number of solutions let \(\mathbf{x} = \mathbf{u}\) and \(\mathbf{x} = \mathbf{v}\) be two distinct solutions to (17.3.1). Now let \(\mathbf{w} = \mathbf{u} − \mathbf{v}\) and \(t \in \mathbb{R}\text{.}\) Then

    \begin{equation*} A(t\mathbf{w}) = tA(\mathbf{u} − \mathbf{v}) = t(A\mathbf{u} − A\mathbf{v}) = t(\mathbf{b} − \mathbf{b}) = \mathbf{0}\text{,} \end{equation*}
    i.e. there are an infinite number of solutions to (17.3.2).

Note that these arguments are general and hold for any system of \(m\) linear equations in \(n\) variables. Note also that if the system consists of \(3\) linear equations in \(3\) variables then result (17.3.2) above can be stated as:

Remark 17.3.3.

Consider the system of linear equations whose augmented matrix is

\begin{equation*} \begin{pmatrix} a_{11} \amp a_{12} \amp a_{13} \amp b_1 \\ a_{21} \amp a_{22} \amp a_{23} \amp b_2 \\ a_{31} \amp a_{32} \amp a_{33} \amp b_3 \end{pmatrix}. \end{equation*}

The following statements are equivalent:

  1. The system has a unique solution.

  2. The planes represented by the rows intersect in a point.

  3. The column vectors of the coefficient matrix are linearly independent.

Exercises Example Tasks

1.

Consider the following system of linear equations:

\begin{align*} x_1 -x_2 -x_3 \amp =1\\ x_2 + 3x_3 \amp =1\\ 2x_1 -x_2 +x_3 \amp =3. \end{align*}

  1. Calculate \(A\mathbf{u}\) where

    \begin{equation*} A=\begin{pmatrix}1 \amp -1 \amp -1 \\ 0 \amp 1 \amp 3 \\ 2 \amp -1 \amp 1\end{pmatrix}, \quad \mathbf{u} = \begin{pmatrix}2 \\ 3 \\ -1\end{pmatrix}\text{.} \end{equation*}

  2. Show that \((1,1,0)\) is a solution to the system of equations.

  3. Write down all solutions to the system of equations.