# 13. Simultaneous equations and row operations

## 13. Simultaneous equations and row operations

Thus far in the course, we have focused on the geometric aspects of matrices and the transformations they determine. Now we'll approach the subject from the point of view of simultaneous equations.

### Simultaneous equations

A system of simultaneous linear equations, for example $x-y=-1,\qquad x+y=3,$ conmprises a finite list of linear equations in some number of variables. Linear means that each term is either constant of else linear in the variables (i.e. the terms are things like $3x$ or $-4y$ , not $x^{2}$ or $xye^{x}$ ).

A system of simultaneous linear equations is a matrix equation in disguise. For example, the system above can be written as $\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}-1\\ 3\end{pmatrix}.$ Indeed, if we multiply this expression out, we get $\begin{pmatrix}x-y\\ x+y\end{pmatrix}=\begin{pmatrix}-1\\ 3\end{pmatrix},$ and these two vectors are equal if and only if their components are equal, which is equivalent to the two equations we started with.

We will abbreviate such a matrix equation by writing a so-called augmented matrix: we write the matrix of coefficients, then a vertical bar, then the column of constants: $\begin{pmatrix}1&-1&|&-1\\ 1&1&|&3\end{pmatrix}$ This is really just a shorthand for the equation $\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}-1\\ 3\end{pmatrix}.$

### Solving these equations

To solve this system, we will manipulate the equations one at a time. We will see what happens to the augmented matrix as we perform these manipulations. We start with: $x-y=-1,\qquad x+y=3,\qquad\begin{pmatrix}1&-1&|&-1\\ 1&1&|&3\end{pmatrix}.$ We can eliminate $x$ from the second equation by subtracting equation 1 from equation 2: $x-y=-1,\qquad 2y=4,\qquad\begin{pmatrix}1&-1&|&-1\\ 0&2&|&4\end{pmatrix}.$ This has had the effect on the augmented matrix of subtracting the first row from the second. We will write this operation as $R_{2}\mapsto R_{2}-R_{1}.$ Next, we can divide equation 2 by $2$ : $x-y=-1,\qquad y=2,\qquad\begin{pmatrix}1&-1&|&-1\\ 0&1&|&2\end{pmatrix}.$ This has the effect of dividing row 2 by 2, which we write as $R_{2}\mapsto\frac{1}{2}R_{2}.$ Finally, we add equation 2 to equation 1: $x=1,\qquad y=2,\qquad\begin{pmatrix}1&0&|&1\\ 0&1&|&2\end{pmatrix}.$ This is the row operation $R_{1}\mapsto R_{1}+R_{2}.$ We have now solved our system of equations.

### Row operations

This process of solving simultaneous equations can therefore be understood as performing a sequence of row operations on the augmented matrix.

Definition:
• Type I row operations: Replace row $i$ by row $i$ plus a multiple (say $\lambda$ ) of row $j$ . We write this as $R_{i}\mapsto R_{i}+\lambda R_{j}$ . In terms of equations, this means we're adding/subtracting a multiple of equation $j$ to equation $i$ .

• Type II row operations: Replace row $i$ by a nonzero multiple (say $\lambda\neq 0$ ) of row $i$ . We write this as $R_{i}\mapsto\lambda R_{i}$ . In terms of equations, this means we're multiplying an equation by a nonzero constant.

• Type III row operations: Swap row $i$ and row $j$ . This corresponds to reordering your equations.

Remark:

We don't allow ourselves to multiply an equation by zero: this will change our system of equations by effectively ignoring some of them.

In our example, we "solved" the equation when we reached $x=1,y=2$ . This meant that the augmented matrix had the identity matrix on the left of the vertical bar. So the aim of the row operations is to put the augmented matrix in the form $(I|b)$ where $I$ is the identity matrix and $b$ is a column vector (of "constants"). Of course, this will sometimes fail:

Example:

Consider the system $x+y=1$ (one equation, two variables). The augmented matrix is now $\begin{pmatrix}1&1&|&1\end{pmatrix}.$ No matter what you do, you can't put this into the form $(I|b)$ because the matrix on the left hand side of the vertical bar isn't a square matrix (so can't be turned into the identity matrix by row operations). Nonetheless, the equation is easy to solve: we have $x=1-y$ , so given any $y$ we get a solution $\begin{pmatrix}1-y\\ y\end{pmatrix}$ . So we don't need to get to the identity matrix so say our system is "solved". In the next few videos, we'll see the correct thing to aim for is to put the matrix on the left of the bar into reduced echelon form.