Discover Linear Algebra

Subsection 1.4.1 Row operations versus equation manipulations

Let's examine the operations in Discovery 1.7 in detail, by considering the operations as both equation manipulations and row operations simultaneously.

In each step, notice how the row\(\leftrightarrow\)equation and column\(\leftrightarrow\)variable correspondence is preserved.

Subtract \(3\) times the first row/equation from the second row/equation. For the equation version, we do this by performing the same combination of left- and right-hand sides, collecting terms on the left.

This combination leads to new equation

Notice that when collecting terms, we ended up performing that “subtract \(3\) times the first from the second” on the coefficients of each variable. So we can achieve the same result in the matrix by performing “subtract \(3\) times the entry in the first row from the entry in the second row,” one column at a time.

Next, add \(2\) times the first row/equation to the third row/equation:

leading to new equation

which we will use to replace the old third row/equation.

Finally, multiply the second row/equation by 1/3:

and modify the matrix accordingly:

The final system on the right is much easier to solve: we can see immediately from the third equation that \(z=3\text{,}\) then can use this in the second equation to determine \(y=7\text{,}\) and finally can use both of these in the first equation to determine \(x=0\text{.}\)