MATH1120: Mathematics for Engineering, Science and Technology 2

The augmented matrix and its reduced row-echelon equivalent matrix are

Thus the solution is $(x,y)=(\frac{3}{2},2)\text{.}$ Geometrically we can interpret this solution as saying the vector $\left(\begin{array}{c}-3\\ 7\end{array}\right)$ can be written in terms of the vectors $\left(\begin{array}{c}2\\ 2\end{array}\right)$ and $\left(\begin{array}{c}-3\\ 2\end{array}\right)$ via the expression

See Figure 16.2.4. Note that the solution is also saying that there is only one way in which $\left(\begin{array}{c}-3\\ 7\end{array}\right)$ can be written in terms of $\left(\begin{array}{c}2\\ 2\end{array}\right)$ and $\left(\begin{array}{c}-3\\ 2\end{array}\right)\text{.}$

The span of the set $\{{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\}$ is the set of vectors $\mathbf{u}$ of the form

where $x$ and $y$ are scalars. From what we learnt about vectors in Math1110 we know that any vector in the plane will be able to be written as linear combination of these two vectors and hence the span of $\{{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\}$ will be all vectors in ${\mathbb{R}}^2\text{.}$ Alternatively, we could approach this problem algebraically. Let $\mathbf{u}=\left(\begin{array}{c}a\\ b\end{array}\right)$ be a vector in the span of $\{{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\}\text{.}$ Then

Using Gauss Jordon elimination gives

and so the solution is $(x,y)=(\frac{a}{5}+\frac{3b}{10},\frac{b}{5}-\frac{a}{5})\text{.}$ Thus, no matter which vector, $\mathbf{u}\text{,}$ in ${\mathbb{R}}^2$ we choose we can find the scalars so that $\mathbf{u}$ is a linear combination of ${\mathbf{v}}_{\mathbf{1}}$ and ${\mathbf{v}}_{\mathbf{2}}$ and hence in the span of $\{{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\}\text{.}$

Firstly note that the vectors in this set are $3$-dimensional vectors and hence the vectors in the span will also be $3$-dimensional. Now, the span of the set $\{{\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\}$ is the set of vectors $\mathbf{u}$ of the form

We recognise this as the vector equation of the plane through the origin and with direction vectors ${\mathbf{v}}_{\mathbf{1}}$ and ${\mathbf{v}}_{\mathbf{2}}\text{,}$ as shown in Figure 16.2.9.

Thus the Cartesian equation of this plane is

1. This set of vectors is linearly dependent since (from our previous examples) we know that

   $$\left(\begin{array}{c}-3\\ 7\end{array}\right)=\frac{3}{2}\left(\begin{array}{c}2\\ 2\end{array}\right)+2\left(\begin{array}{c}-3\\ 2\end{array}\right)$$
   i.e. ${\mathbf{v}}_{\mathbf{3}}$ can be written as a linear combination of ${\mathbf{v}}_{\mathbf{1}}$ and ${\mathbf{v}}_{\mathbf{2}}\text{.}$

2. This set of vectors is linearly independent since the only solution to

   $$x_1{\mathbf{v}}_{\mathbf{1}}+x_2{\mathbf{v}}_{\mathbf{2}}=\mathbf{0}$$
   is $x_1=x_2=0\text{.}$ Note that in this case this is telling us that ${\mathbf{v}}_{\mathbf{2}}$ is not a scalar multiple of ${\mathbf{v}}_{\mathbf{1}}\text{,}$ i.e. ${\mathbf{v}}_{\mathbf{2}}$ is not parallel to ${\mathbf{v}}_{\mathbf{1}}\text{.}$

The augmented matrix and its reduced row-echelon equivalent matrix are

Thus the solution is $(x,y,z)=(\frac{3}{2},-1,2)\text{.}$ The geometric interpretation of this solution is that the vector $\left(\begin{array}{c}3\\ 6\\ 3\end{array}\right)$ (shown in blue in Figure 16.2.13) can be written as a linear combination of the vectors $\{\left(\begin{array}{c}2\\ 2\\ 4\end{array}\right),\left(\begin{array}{c}4\\ 3\\ 1\end{array}\right),\left(\begin{array}{c}2\\ 3\\ -1\end{array}\right)\}$ (shown in red in Figure 16.2.13) and in only one way, (shown in green in Figure 16.2.13).

The augmented matrix and its reduced row-echelon equivalent matrix are

Thus the solution is $(x,y,z)=(0,0,0)\text{.}$ The geometric interpretatifon of this solution is that the column vectors of the coefficient matrix are linearly independent, i.e. that no one of those vectors can be written in terms of the other two. If the column vectors had turned out to be linearly dependent then we could have written, for example,

This would imply that ${\mathbf{a}}_{\mathbf{3}}$ lies in the plane defined by the vectors ${\mathbf{a}}_{\mathbf{1}}$ and ${\mathbf{a}}_{\mathbf{2}}\text{,}$ or to say the same thing, that ${\mathbf{a}}_{\mathbf{1}}\text{,}$ ${\mathbf{a}}_{\mathbf{2}}$ and ${\mathbf{a}}_{\mathbf{3}}$ are coplanar. Since the vectors are linearly independent we can say these vectors are not coplanar. See Figure 16.2.15.

The augmented matrix and its reduced row-echelon equivalent matrix are

Thus this system has an infinite number of solutions given by

To discuss the geometric interpretation of this solution let

Then we can say that $\mathbf{b}$ can be written as a linear combination of the vectors $\{{\mathbf{a}}_{\mathbf{1}}\mathbf{,}{\mathbf{a}}_{\mathbf{2}}\mathbf{,}{\mathbf{a}}_{\mathbf{3}}\}$ in an infinite number of ways. Notice that with $t=0$ we have

Since $\mathbf{b}$ can written as a linear combination of $\{{\mathbf{a}}_{\mathbf{1}}\mathbf{,}{\mathbf{a}}_{\mathbf{2}}\}\text{,}$ the vectors $\mathbf{b}\text{,}$ ${\mathbf{a}}_{\mathbf{1}}$ and ${\mathbf{a}}_{\mathbf{2}}$ are coplanar. Similarly (with $t=-2$) we can see that $\mathbf{b}\text{,}$ ${\mathbf{a}}_{\mathbf{1}}$ and ${\mathbf{a}}_{\mathbf{3}}$ are coplanar. So, in fact, $\mathbf{b}\text{,}$ ${\mathbf{a}}_{\mathbf{1}}\text{,}$ ${\mathbf{a}}_{\mathbf{2}}$ and ${\mathbf{a}}_{\mathbf{3}}$ are all coplanar.

Finally we can see from the above working that the vectors $\{{\mathbf{a}}_{\mathbf{1}}\mathbf{,}{\mathbf{a}}_{\mathbf{2}}\mathbf{,}{\mathbf{a}}_{\mathbf{3}}\}$ are linearly dependent since if $\mathbf{b}\mathbf{=}\mathbf{0}$ then the reduced row-echelon form would be