Skip to main content Contents Prev Up Next \(\newcommand{\avec}{{\mathbf a}}
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\)
Section 4.1 The equation \(A\xvec = \bvec\)
So far, we have begun with a matrix \(A\) and a vector \(\xvec\) and formed their product \(A\xvec = \bvec\text{.}\) We would now like to turn this around: beginning with a matrix \(A\) and a vector \(\bvec\text{,}\) we will ask if we can find a vector \(\xvec\) such that \(A\xvec = \bvec\text{.}\) This will naturally give us a set of linear equations called a linear system.
Definition 4.1.1 .
A linear equation in the unknowns \(x_1,x_2,\ldots,x_n\) may be written in the form
\begin{equation*}
a_1x_1 + a_2x_2 + \ldots + a_nx_n = b\text{,}
\end{equation*}
where \(a_1,a_2,\ldots,a_n\) are real numbers known as coefficients . We also say that \(x_1,x_2,\ldots,x_n\) are the variables in the equation.
By a system of linear equations or a linear system , we mean a set of linear equations written in a common set of unknowns.
For instance,
\begin{equation*}
\begin{alignedat}{4}
2x_1 \amp {} + \amp {} 1.2x_2 \amp {}-{} \amp 4x_3 \amp {}={}
\amp 3.7 \\
-0.1x_1 \amp {} \amp {} \amp {} + {} \amp x_3 \amp {}={}
\amp 2 \\
x_1 \amp {}+{} \amp x_2 \amp {}-{} \amp x_3 \amp {}={} \amp
1.4 \\
\end{alignedat}
\end{equation*}
is an example of a linear system.
Definition 4.1.2 .
A solution to a linear system is simply a set of numbers \(x_1 = s_1, x_2 = s_2, \ldots, x_n=s_n\) that satisfy all the equations in the system.
For instance, consider the linear system
\begin{equation*}
\begin{alignedat}{3}
-x \amp {}+{} \amp y \amp {} = {} \amp 1 \\
-2x \amp {}+{} \amp y \amp {} = {} \amp -1. \\
\end{alignedat}
\end{equation*}
To check that \((x,y) = (2,3)\) is a solution, we verify that the following equations are true.
\begin{equation*}
\begin{alignedat}{3}
-2 \amp {}+{} \amp 3 \amp {} = {} \amp 1 \\
-2(2) \amp {}+{} \amp 3 \amp {} = {} \amp -1. \\
\end{alignedat}
\end{equation*}
Definition 4.1.3 .
We call the set of all solutions the solution space of the linear system.
Activity 4.1.1 . The equation \(A\xvec = \bvec\) .
Suppose that
\begin{equation*}
A = \left[
\begin{array}{rrrr}
1 \amp 2 \\
-1 \amp 1 \\
\end{array}
\right],
\bvec = \left[
\begin{array}{r}
6 \\ 0
\end{array}
\right]\text{.}
\end{equation*}
Is there a vector \(\xvec=\left[\begin{array}{r}
x\\ y
\end{array}
\right]\) such that \(A\xvec = \bvec\text{?}\) Provide the linear system that answers this question. Verify that \((x,y) = (2,2)\) is a solution to the system. Verify that \(\xvec=\left[\begin{array}{r}
2\\ 2
\end{array}
\right]\) satisfies the equation \(A\xvec = \bvec\text{.}\)
Consider the linear system
\begin{equation*}
\begin{alignedat}{4}
2x \amp {}+{} \amp y \amp {}-{} \amp 3z \amp {}={} \amp 4 \\
-x \amp {}+{} \amp 2y \amp {}+{} \amp z \amp {}={} \amp 3 \\
3x \amp {}-{} \amp y \amp \amp \amp {}={} \amp -4. \\
\end{alignedat}
\end{equation*}
Identify the matrix \(A\) and vector \(\bvec\) to express this system in the form \(A\xvec = \bvec\text{.}\)
If \(A\) and \(\bvec\) are as below, write the linear system corresponding to the equation \(A\xvec=\bvec\) where \(\xvec=\left[\begin{array}{r}
x \\ y \\ z
\end{array}
\right]\)
\begin{equation*}
A = \left[\begin{array}{rrr}
3 \amp -1 \amp 0 \\
-2 \amp 0 \amp 6
\end{array}
\right],~~~
\bvec = \left[\begin{array}{r}
-6 \\ 2
\end{array}
\right].
\end{equation*}
If \(A\) and \(\bvec\) are as below, write the linear system corresponding to the equation \(A\xvec=\bvec\) where \(\xvec=\left[\begin{array}{r}
x_1 \\ x_2 \\ x_3 \\ x_4
\end{array}
\right]\)
\begin{equation*}
\left[
\begin{array}{rrrr}
1 \amp 2 \amp 0 \amp -1 \\
2 \amp 4 \amp -3 \amp -2 \\
-1 \amp -2 \amp 6 \amp 1 \\
\end{array}
\right]
\xvec
=
\left[\begin{array}{r}
-1 \\ 1 \\ 5
\end{array}
\right]\text{.}
\end{equation*}
Suppose \(A\) is an \(m\times n\) matrix. What can you guarantee about the solution space of the equation \(A\xvec
= \zerovec\text{?}\)
