Differential Equations and Linear Algebra, 4th Edition

3.2 Matrices and Gaussian Elimination

In Example 6 of Section 3.1 we applied the method of elimination to solve the linear system

$$\begin{array}{rcr}\mathbf{1}x+\mathbf{2}y+\mathbf{1}z& =& \mathbf{4}\\ \mathbf{3}x+\mathbf{8}y+\mathbf{7}z& =& \mathbf{20}\\ \mathbf{2}x+\mathbf{7}y+\mathbf{9}z& =& \mathbf{23}.\end{array}$$ (1)

There we employed elementary operations to transform this system into the equivalent system

$$\begin{array}{rcl}\mathbf{1}x+\mathbf{2}y+\mathbf{1}z& =& \mathbf{4}\\ \mathbf{0}x+\mathbf{1}y+\mathbf{2}z& =& \mathbf{4}\\ \mathbf{0}x+\mathbf{0}y+\mathbf{1}z& =& \mathbf{3},\end{array}$$ (2)

which we found easy to solve by back substitution. Here we have printed in color the coefficients and constants (including the 0s and 1s that would normally be omitted) because everything else—the symbols x, y, and z for the variables and the $+$ and $=$ signs—is excess baggage that means only extra writing, for we can keep track of these symbols mentally. In effect, in Example 6 we used an appropriate sequence of operations to transform the array

$$[\begin{array}{rrrr}1& 2& 1& 4\end{array}$$