Section 1.4 Examples
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.
\begin{align*}
\amp\left[\begin{array}{rrr|r}
-2 \amp 2 \amp -5 \amp -1 \\
3 \amp 0 \amp 3 \amp 9 \\
1 \amp -1 \amp 3 \amp 2
\end{array}\right]
\amp
\amp\left\{\begin{array}{rcrcrcr}
-2x \amp + \amp 2y \amp - \amp 5z \amp = \amp -1 \\
3x \amp \amp \amp + \amp 3z \amp = \amp 9 \\
x \amp - \amp y \amp + \amp 3z \amp = \amp 2
\end{array}\right.\\
\end{align*}
Interchange the first and third rows/equations.
\begin{align*} \amp\left[\begin{array}{rrr|r} 1 \amp -1 \amp 3 \amp 2 \\ 3 \amp 0 \amp 3 \amp 9 \\ -2 \amp 2 \amp -5 \amp -1 \end{array}\right] \amp \amp\left\{\begin{array}{rcrcrcr} x \amp - \amp y \amp + \amp 3z \amp = \amp 2 \\ 3x \amp \amp \amp + \amp 3z \amp = \amp 9 \\ -2x \amp + \amp 2y \amp - \amp 5z \amp = \amp -1 \end{array}\right. \end{align*}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.
\begin{equation*}
\begin{array}{ccccccc}
(\text{LHS}_2) \amp - \amp 3(\text{LHS}_1) \amp = \amp (\text{RHS}_2) \amp - \amp 3(\text{RHS}_1) \\
(3x + 3z) \amp - \amp 3(x-y+3z) \amp = \amp 9 \amp - \amp 3(2)
\end{array}
\end{equation*}
This combination leads to new equation
\begin{equation*}
0x + 3y - 6z = 3 \text{.}
\end{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.
\begin{align*}
\amp\left[\begin{array}{rrr|r}
1 \amp -1 \amp 3 \amp 2 \\
0 \amp 3 \amp -6 \amp 3 \\
-2 \amp 2 \amp -5 \amp -1
\end{array}\right]
\amp
\amp\left\{\begin{array}{rcrcrcr}
x \amp - \amp y \amp + \amp 3z \amp = \amp 2 \\
\amp \amp 3y \amp - \amp 6z \amp = \amp 3 \\
-2x \amp + \amp 2y \amp - \amp 5z \amp = \amp -1
\end{array}\right.
\end{align*}
Next, add \(2\) times the first row/equation to the third row/equation:
\begin{equation*}
\begin{array}{ccccccc}
(\text{LHS}_3) \amp + \amp 2 (\text{LHS}_1) \amp = \amp (\text{RHS}_3) \amp + \amp 2 (\text{RHS}_1) \\
(-2x+2y-5z) \amp + \amp 2 (x-y+3z) \amp = \amp -1 \amp + \amp 2(2),
\end{array}
\end{equation*}
leading to new equation
\begin{equation*}
z = 3 \text{,}
\end{equation*}
which we will use to replace the old third row/equation.
\begin{align*}
\amp\left[\begin{array}{rrr|r}
1 \amp -1 \amp 3 \amp 2 \\
0 \amp 3 \amp -6 \amp 3 \\
0 \amp 0 \amp 1 \amp 3
\end{array}\right]
\amp
\amp\left\{\begin{array}{rcrcrcr}
x \amp - \amp y \amp + \amp 3z \amp = \amp 2 \\
\amp \amp 3y \amp - \amp 6z \amp = \amp 3 \\
\amp \amp \amp \amp z \amp = \amp 3
\end{array}\right.
\end{align*}
Finally, multiply the second row/equation by 1/3.
\begin{equation*}
\begin{array}{ccc}
(1/3)(\text{LHS}_2) \amp = \amp (1/3)(\text{RHS}_2) \\
(1/3)(3y-6z) \amp = \amp (1/3)(3) \\
y - 2z \amp = \amp 1
\end{array}
\end{equation*}
The matrix is modified accordingly.
\begin{align*}
\amp\left[\begin{array}{rrr|r}
1 \amp -1 \amp 3 \amp 2 \\
0 \amp 1 \amp -2 \amp 1 \\
0 \amp 0 \amp 1 \amp 3
\end{array}\right]
\amp
\amp\left\{\begin{array}{rcrcrcr}
x \amp - \amp y \amp + \amp 3z \amp = \amp 2 \\
\amp \amp y \amp - \amp 2z \amp = \amp 1 \\
\amp \amp \amp \amp z \amp = \amp 3
\end{array}\right.
\end{align*}
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 \(x=0\text{.}\)
A look ahead.
In Chapter 2, we will develop a systematic method of simplifying a system in this manner, but working exclusively with augmented matrices. Also, we will take the process a few steps further to make the system as simple as possible. Notice how “back-solving” the system proceeds from bottom-right to top-left. We will use the same process when solving systems using matrices.
