Matrices & linear systems

A matrix (plural matrices) is a rectangular array of numbers arranged in rows and columns. They have a variety of uses, but for us they are just a convenient way of staying organized while solving linear systems. Solving a system of two linear equations by elimination is easy, but it becomes much harder once you have four or five equations. Consider this linear system:

$x+3y=4$ and $-2x+3y=10\text{.}$

The first step is to move the constant term to the right-hand side. In this case, it’s already done. Now, we take the coefficients and we put them into an augmented matrix:

$\left[\begin{array}{ccc}1& 3& 4\\ -2& 3& 10\end{array}\right]\text{.}$

The first two columns represent the x and y coefficients respectively; the last column, separated by a bar, represents the right-hand sides of the equations. To solve the system, we will use Gauss-Jordan elimination, which is a method of transforming the matrix to reduced row-echelon form. We want the columns left of the bar to have a diagonal of ones and zeros everywhere else. For example, if we were solving a linear system of four equations, the end result would look like this:

$\left[\begin{array}{ccccc}1& 0& 0& 0& x\\ 0& 1& 0& 0& y\\ 0& 0& 1& 0& z\\ 0& 0& 0& 1& w\end{array}\right]\text{.}$

The values of x, y, z, and w would be the solution. To perform Gauss-Jordan elimination, we use elementary operations until we get to the reduced row-echelon form. There are just three operations—we can

• interchange any two rows;
• multiply one row by a nonzero scalar;
• add a multiple of one row to another.

Example

Let’s return to our original example. We had

$\left[\begin{array}{ccc}1& 3& 4\\ -2& 3& 10\end{array}\right]\text{.}$

We can subtract the second row from the first to get

$\left[\begin{array}{ccc}3& 0& -6\\ -2& 3& 10\end{array}\right]\text{.}$

We can add two-thirds of the first row to the second to get

$\left[\begin{array}{ccc}3& 0& -6\\ 0& 3& 6\end{array}\right]\text{.}$

Finally, we can divide both rows by three:

$\left[\begin{array}{ccc}1& 0& -2\\ 0& 1& 2\end{array}\right]\text{.}$

The values $x=-2$ and $y=2$ are indeed the solution to this system.