## Section1.3Another Look at Systems

### Subsection1.3.1A Graphical Look at Systems

###### Investigation1.3.1. A Two Dimensional Example.

Suppose we have the system:

\begin{align*} 3x+2y\amp =7\\ -2x-7y\amp = 9 \end{align*}

we saw in the previous section that we can rewrite this as the vector equation

\begin{equation*} \left[ \begin{array}{r} 3\\ -2 \end{array} \right] x+ \left[ \begin{array}{r} 2\\ -7 \end{array} \right] y= \left[ \begin{array}{r} 7\\ 9 \end{array} \right]. \end{equation*}

Then we can interpret the problem graphically, given vectors

\begin{equation*} \vec{v}= \left[ \begin{array}{r} 3\\ -2 \end{array} \right] \end{equation*}

and

\begin{equation*} \vec{u}= \left[ \begin{array}{r} 2\\ -7 \end{array} \right] \end{equation*}

find a linear combination that equals

\begin{equation*} \vec{b}= \left[ \begin{array}{r} 7\\ 9 \end{array} \right]. \end{equation*}

Solving this like before we get

\begin{align*} \left[ \begin{array}{rr|r} 3 \amp 2 \amp 7\\ -2 \amp -7 \amp 9 \end{array} \right] \amp\amp\stackrel{R_1+R_2}{\Huge \leadsto}\amp\amp\amp \left[ \begin{array}{rr|r} 1 \amp -5 \amp 16\\ -2 \amp -7 \amp 9 \end{array} \right]\\ \amp\amp\stackrel{R_2+2\, R_1}{\Huge \leadsto}\amp\amp\amp \left[ \begin{array}{rr|r} 1 \amp -5 \amp 16\\ 0 \amp -17 \amp 41 \end{array} \right]\\ \amp\amp\stackrel{-\frac{1}{17}R_2}{\Huge \leadsto}\amp\amp\amp \left[ \begin{array}{rr|r} 1 \amp -5 \amp 16\\ 0 \amp 1 \amp -\frac{41}{17} \end{array} \right]\\ \amp\amp\stackrel{R_1+5\, R_2}{\Huge \leadsto}\amp\amp\amp \left[ \begin{array}{rr|r} 1 \amp 0 \amp \frac{67}{17}\\ 0 \amp 1 \amp -\frac{41}{17} \end{array} \right] \end{align*}

so that $x=67/17$ and $y=-41/17\text{.}$

\begin{align*} x\, \vec{v}+y\, \vec{u} \amp = \frac{67}{17} \left[ \begin{array}{r} 3\\ -2 \end{array} \right] - \frac{41}{17} \left[ \begin{array}{r} 2\\ -7 \end{array} \right] \\ \amp = \left[ \begin{array}{r} \frac{201}{17}\\ -\frac{134}{17} \end{array} \right] - \left[ \begin{array}{r} \frac{82}{17}\\ -\frac{287}{17} \end{array} \right] \\ \amp = \left[ \begin{array}{r} \frac{119}{17}\\ \frac{153}{17} \end{array} \right]\\ \amp = \left[ \begin{array}{r} 7\\ 9 \end{array} \right] \end{align*}
###### Investigation1.3.2. A Three Dimensional Example.

Now let's look at

\begin{gather*} \left[ \begin{array}{r} 1 \\ 0 \\ 3 \end{array} \right] x+ \left[ \begin{array}{r} 2 \\ 0 \\ 0 \end{array} \right] y+ \left[ \begin{array}{r} 1 \\ -1 \\ 0 \end{array} \right] z= \left[ \begin{array}{r} 1 \\ 1 \\ 1 \end{array} \right] \end{gather*}

Solving as before

\begin{align*} \left[ \begin{array}{rrr|r} 1 \amp 2 \amp 1 \amp 1\\ 0 \amp 0 \amp -1 \amp 1\\ 3 \amp 0 \amp 0 \amp 1 \end{array} \right]\amp\amp\stackrel{swap\ R_2, R_3}{\Huge \leadsto}\amp\amp\amp \left[ \begin{array}{rrr|r} 1 \amp 2 \amp 1 \amp 1\\ 3 \amp 0 \amp 0 \amp 1\\ 0 \amp 0 \amp -1 \amp 1 \end{array} \right]\\ \amp\amp\stackrel{R_2-3R_1, -R_3}{\Huge \leadsto}\amp\amp\amp \left[ \begin{array}{rrr|r} 1 \amp 2 \amp 1 \amp 1\\ 0 \amp -6 \amp -3 \amp -2\\ 0 \amp 0 \amp 1 \amp -1 \end{array} \right]\\ \amp\amp\stackrel{-\frac{1}{6}R_2}{\Huge \leadsto}\amp\amp\amp \left[ \begin{array}{rrr|r} 1 \amp 2 \amp 1 \amp 1\\ 0 \amp 1 \amp \frac{1}{2} \amp \frac{1}{3}\\ 0 \amp 0 \amp 1 \amp -1 \end{array} \right]\\ \amp\amp\stackrel{R_2-\frac{1}{2}R_3}{\Huge \leadsto}\amp\amp\amp \left[ \begin{array}{rrr|r} 1 \amp 2 \amp 1 \amp 1\\ 0 \amp 1 \amp 0 \amp \frac{5}{6}\\ 0 \amp 0 \amp 1 \amp -1 \end{array} \right]\\ \amp\amp\stackrel{R_1-2R_2-R_3}{\Huge \leadsto}\amp\amp\amp \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 0 \amp \frac{1}{3}\\ 0 \amp 1 \amp 0 \amp \frac{5}{6}\\ 0 \amp 0 \amp 1 \amp -1 \end{array} \right] \end{align*}

thus $x=1/3\text{,}$ $y=5/6\text{,}$ and $z=-1\text{.}$

We should end this investigation with a note on looking before calculating. The vector we want has a 1 in all three coordinates but of the three vectors on the right that we wish to combine only the first is non-zero in the third coordinate and only the third is non-zero in the second. Therefore, we can conclude immediately that $x=1/3$ and $z=-1\text{,}$ from there it is easy to conclude that $y=5/6\text{.}$ You would have likely noticed this if I had written this as a system of equations like this

\begin{align*} x+2y+z\amp =1\\ -z\amp =1\\ 3x\amp =1 \end{align*}

and so emphasizes the importance of viewing the same problem from multiple perspectives.

### Subsection1.3.2Matrices and Vectors

We have one more way to look at a system of equations, as a matrix equation. Before we do that we need to briefly look at how we multiply a matrix by a vector, we will explore this more deeply in Section 3.2. Consider the matrix

\begin{equation*} A= \left[ \begin{array}{rr} 2 \amp 3\\ 1 \amp 0 \end{array} \right] \end{equation*}

and vector

\begin{equation*} \vec{v}= \left[ \begin{array}{r} 7 \\ -4 \end{array} \right] \end{equation*}

the ploduct $A\vec{x}$ is given by

\begin{align*} A\vec{x}\amp = \amp\amp \left[ \begin{array}{rr} 2 \amp 3\\ 1 \amp 0 \end{array} \right] \left[ \begin{array}{r} 7 \\ -4 \end{array} \right]\\ \amp = \amp\amp \left[ \begin{array}{r} 2 \\ 1 \end{array} \right]\, 7 + \left[ \begin{array}{r} 3 \\ 0 \end{array} \right] (-4)\\ \amp = \amp\amp \left[ \begin{array}{r} 14 \\ 7 \end{array} \right] + \left[ \begin{array}{r} -12 \\ 0 \end{array} \right]\\ \amp = \amp\amp \left[ \begin{array}{r} 2 \\ 7 \end{array} \right] \end{align*}

so that we make a linear combination of the columns of $A$ using the entries in $\vec{v}$ for coefficients. In general if

\begin{equation*} A= \left[ A_1\ A_2\ \cdots\ A_n \right] \end{equation*}

where $A_i$'s' are the columns of $A$ and

\begin{equation*} \vec{v}= \left[ \begin{array}{r} x_1\\ x_2\\ \vdots\\ x_n \end{array} \right] \end{equation*}

then their product is the linear combination

\begin{equation*} A\vec{v}=A_1\, x_1\, + A_2\, x_2\, +\cdots+ A_n\, x_n \end{equation*}

Note that for this to make sense we need the number of entries in the vector to match the number of columns in the matrix, also for reasons that will be more apparent in Section 3.2 we will always write this type of product as $A\vec{v}$ with the vector written as a column and not $\vec{v}A\text{.}$

To practice your products click Evaluate (Sage).

### Subsection1.3.3Matrix Equations

We now have yet another way of viewing a system of equations. In Section 1.2 we saw that the system

\begin{align*} 7x+3y-9z\amp=2\\ -x+2y+5z\amp=13\\ 2x- y\amp=0 \end{align*}

could be rewritten as the vector equation

\begin{equation*} \left[ \begin{array}{r} 7 \\ -1 \\ 2 \end{array} \right] x + \left[ \begin{array}{r} 3 \\ 2 \\ -1 \end{array} \right] y + \left[ \begin{array}{r} -9 \\ 5 \\ 0 \end{array} \right] z = \left[ \begin{array}{r} 2 \\ 13 \\ 0 \end{array} \right]. \end{equation*}

Now that we know something about multiplying matrices and vectors we can recognize that the left hand side of this may be written as a matrix product:

\begin{equation*} \left[ \begin{array}{rrr} 7 \amp 3\amp -9\\ -1 \amp 2\amp 5\\ 2 \amp -1 \amp 0 \end{array} \right] \left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{r} 2 \\ 13 \\ 0 \end{array} \right]. \end{equation*}

This final format is called a matrix equation.

To practice rewriting systems click Evaluate (Sage).

We now have three different ways to write and think about systems of equations each with their on advantages and disadvantages. Which works best depends on the situation and recognizing which to use where is something that comes with time and experience. On a final note, as we will see more in Section 3.2, using matrix equations will allow us to recognize that solving the equations we are looking at now can be approached the same way as solving simple linear equations such as $ax+b=c$ which makes them good for big picture ideas.

Matrix Equation