Section 2.4 Examples
Here we use our procedures to use matrices to reduce and solve the systems from Discovery guide 2.1. Here are a few things to note about our method.
- We only use the three elementary row operations. It sometimes is possible to reduce a bit faster using non-elementary operations such as adding a multiple of a row to a multiple of another row, but remember we are not interested in short-cuts, and using non-elementary operations will get us into trouble in later topics.
- We sometimes perform more than one operation at the same time. This is an acceptable short-cut, as long as we never simultaneously modify a row and also use that row to modify another row.
- We write down the row operation(s) we are using in that reduction step to the right of the matrix, to keep track of what we are doing.
- We don’t always have to multiply a row by a fraction to get a leading one — we can sometimes use a difference between entries in a column, and avoid fractions that way.
- There are many different sequences of operations one could use to get from initial augmented matrix to an RREF matrix. The reductions in the examples below are not the only way, nor are they necessarily the best way to proceed. As long as we steadily progress toward RREF, that’s all that matters.
- We never write an equals sign between matrices when row reducing. We will explore what it means for two matrices to be equal in Discovery guide 4.1, and this here is not it. When you perform a row operation, the result is a different matrix than the original, and the two matrices represent different systems of equations. However, the two matrices are related, and it is to express this relationship that we have the terminology row equivalent.
Subsection 2.4.1 Worked examples from the discovery guide
Example 2.4.1. One unique solution.
From Discovery 2.1:
\begin{equation*}
\left\{\begin{array}{rcrcrcr}
2x \amp \amp \amp - \amp 2z \amp = \amp 4, \\
x \amp - \amp y \amp \amp \amp = \amp 3,\\
4x \amp - \amp 2y \amp - \amp 3z \amp = \amp 7.
\end{array}\right.
\end{equation*}
We form the augmented matrix for the system, and reduce.
\begin{align*}
\amp
\left[\begin{array}{rrr|r}
2 \amp 0 \amp -2 \amp 4 \\
1 \amp -1 \amp 0 \amp 3\\
4 \amp -2 \amp -3 \amp 7
\end{array}\right]
\begin{matrix}R_1\leftrightarrow R_2\\\\\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp -1 \amp 0 \amp 3\\
2 \amp 0 \amp -2 \amp 4 \\
4 \amp -2 \amp -3 \amp 7
\end{array}\right]
\begin{matrix}\phantom{X}\\R_2 - 2R_1\\R_3-4R_1\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp -1 \amp 0 \amp 3\\
0 \amp 2 \amp -2 \amp -2 \\
0 \amp 2 \amp -3 \amp -5
\end{array}\right]
\begin{matrix}\phantom{X}\\\\\frac{1}{2}R_2\\\phantom{x}\\\phantom{x}\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp -1 \amp 0 \amp 3\\
0 \amp 1 \amp -1 \amp -1 \\
0 \amp 2 \amp -3 \amp -5
\end{array}\right]
\begin{matrix}R_1+R_2\\\phantom{X}\\R_3-2R_2\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp 0 \amp -1 \amp 2 \\
0 \amp 1 \amp -1 \amp -1 \\
0 \amp 0 \amp -1 \amp -3
\end{array}\right]
\begin{matrix}\phantom{X}\\\phantom{X}\\-R_3\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp 0 \amp -1 \amp 2\\
0 \amp 1 \amp -1 \amp -1 \\
0 \amp 0 \amp 1 \amp 3
\end{array}\right]
\begin{matrix}R_1+R_3\\R_2+R_3\\\phantom{X}\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp 0 \amp 0 \amp 5 \\
0 \amp 1 \amp 0 \amp 2 \\
0 \amp 0 \amp 1 \amp 3
\end{array}\right]
\end{align*}
Every variable column has a leading one, so there are no free variables and no parameters are required. We can solve for each variable as a specific number, so the system has one unique solution: \(x = 5\text{,}\) \(y = 2\text{,}\) and \(z = 3\text{.}\)
Example 2.4.2. Infinite number of solutions.
From Discovery 2.2:
\begin{equation*}
\left\{\begin{array}{rcrcrcr}
3x \amp + \amp 6y \amp + \amp 5z \amp = \amp -9,\\
2x \amp + \amp 4y \amp + \amp 3z \amp = \amp -5,\\
3x \amp + \amp 6y \amp + \amp 6z \amp = \amp -12.
\end{array}\right.
\end{equation*}
We form the augmented matrix for the system, and reduce.
\begin{align*}
\amp
\left[\begin{array}{rrr|r}
3 \amp 6 \amp 5 \amp -9\\
2 \amp 4 \amp 3 \amp -5\\
3 \amp 6 \amp 6 \amp -12
\end{array}\right]
\begin{matrix}R_1-R_2\\\phantom{X}\\\phantom{X}\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp 2 \amp 2 \amp -4\\
2 \amp 4 \amp 3 \amp -5\\
3 \amp 6 \amp 6 \amp -12
\end{array}\right]
\begin{matrix}\phantom{X}\\R_2-2R_1\\R_3-3R_1\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp 2 \amp 2 \amp -4\\
0 \amp 0 \amp -1 \amp 3\\
0 \amp 0 \amp 0 \amp 0
\end{array}\right]
\begin{matrix}\phantom{X}\\-R_2\\\phantom{X}\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp 2 \amp 2 \amp -4\\
0 \amp 0 \amp 1 \amp -3\\
0 \amp 0 \amp 0 \amp 0
\end{array}\right]
\begin{matrix}R_1-2R_2\\\phantom{X}\\\phantom{X}\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp 2 \amp 0 \amp 2\\
0 \amp 0 \amp 1 \amp -3\\
0 \amp 0 \amp 0 \amp 0
\end{array}\right]
\end{align*}
The second column does not contain a leading one, so variable \(y\) is free and we assign to it a parameter: \(y=t\text{.}\) We can then use the simplified system
\begin{equation*}
\left\{\begin{array}{rcrcrcr}
x \amp + \amp 2y \amp \amp \amp = \amp 2,\\
\amp \amp \amp \amp z \amp = \amp -3,\\
\amp \amp \amp \amp 0 \amp = \amp 0.
\end{array}\right.
\end{equation*}
to solve for \(x = 2 - 2t\) and \(z=-3\text{.}\) In parametric form, the general solution of the system can be expressed as
\begin{align*}
x \amp= 2-2t, \amp y \amp= t, \amp z = -3,
\end{align*}
and every particular solution to the system can be obtained by choosing a value for \(t\text{.}\) For example, the particular solution associated to \(t=3\) is
\begin{align*}
x \amp= -4, \amp y \amp= 3, \amp z = -3,
\end{align*}
and the particular solution associated to \(t=-\sqrt{2}\) is
\begin{align*}
x \amp= 2+2\sqrt{2}, \amp y \amp= -\sqrt{2}, \amp z = -3.
\end{align*}
Example 2.4.3. No solution.
From Discovery 2.3:
\begin{equation*}
\left\{\begin{array}{rcrcrcr}
x \amp + \amp 2y \amp + \amp z \amp = \amp 2,\\
2x \amp + \amp 5y \amp + \amp 2z \amp = \amp -3,\\
2x \amp + \amp 4y \amp + \amp 2z \amp = \amp -1.
\end{array}\right.
\end{equation*}
We form the augmented matrix for the system, and reduce.
\begin{align*}
\amp
\left[\begin{array}{rrr|r}
1 \amp 2 \amp 1 \amp 2\\
2 \amp 5 \amp 2 \amp -3\\
2 \amp 4 \amp 2 \amp -1
\end{array}\right]
\begin{matrix}\phantom{X}\\R_2-2R_1\\R_3-2R_1\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp 2 \amp 1 \amp 2\\
0 \amp 1 \amp 0 \amp -7\\
0 \amp 0 \amp 0 \amp -5
\end{array}\right]
\begin{matrix}R_1-2R_2\\\phantom{X}\\-\frac{1}{5}R_3\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp 0 \amp 1 \amp 16\\
0 \amp 1 \amp 0 \amp -7\\
0 \amp 0 \amp 0 \amp 1
\end{array}\right]
\begin{matrix} R_1 - 16 R_3 \\ R_2 + 7 R_3 \\ \phantom{X} \end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrr|r}
1 \amp 0 \amp 1 \amp 0\\
0 \amp 1 \amp 0 \amp 0\\
0 \amp 0 \amp 0 \amp 1
\end{array}\right]
\end{align*}
Here we have a leading one in the “equals” column. If we turn that third row back into an equation, we have
\begin{equation*}
0x+0y+0z=1,
\end{equation*}
but there are no possible values of \(x,y,z\) that satisfy this equation. Therefore, the system is inconsistent. (Of course, we could have seen that this would happen right from the second matrix, and could have stopped there. But we went all the way to RREF to have another example demonstrating the row reduction strategy. In practice, we should stop reducing as soon as we can see that the system will be inconsistent.)
Example 2.4.4. A homogeneous system.
From Discovery 2.4:
\begin{equation*}
\left\{\begin{array}{rcrcrcrcrcr}
3x_1 \amp + \amp 6x_2 \amp - \amp 8x_3 \amp + \amp 13x_4 \amp = \amp 0,\\
x_1 \amp + \amp 2x_2 \amp - \amp 2x_3 \amp + \amp 3x_4 \amp = \amp 0,\\
2x_1 \amp + \amp 4x_2 \amp - \amp 5x_3 \amp + \amp 8x_4 \amp = \amp 0.
\end{array}\right.
\end{equation*}
For a homogeneous system, we only reduce the coefficient matrix, since elementary row operations will never change an “equals” columns that contains all zeros.
\begin{align*}
\amp
\left[\begin{array}{rrrr}
3 \amp 6 \amp -8 \amp 13 \\
1 \amp 2 \amp -2 \amp 3 \\
2 \amp 4 \amp -5 \amp 8
\end{array}\right]
\begin{matrix}R_1\leftrightarrow R_2\\\\\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrrr}
1 \amp 2 \amp -2 \amp 3 \\
3 \amp 6 \amp -8 \amp 13 \\
2 \amp 4 \amp -5 \amp 8
\end{array}\right]
\begin{matrix}\phantom{X}\\R_2-3R_1\\R_3-2R_1\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrrr}
1 \amp 2 \amp -2 \amp 3 \\
0 \amp 0 \amp -2 \amp 4 \\
0 \amp 0 \amp -1 \amp 2
\end{array}\right]
\begin{matrix}\\-\frac{1}{2}R_2\\\phantom{X}\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrrr}
1 \amp 2 \amp -2 \amp 3 \\
0 \amp 0 \amp 1 \amp -2 \\
0 \amp 0 \amp -1 \amp 2
\end{array}\right]
\begin{matrix}R_1+2R_3\\\phantom{X}\\R_3+R_2\end{matrix}\\
\longrightarrow \amp
\left[\begin{array}{rrrr}
1 \amp 2 \amp 0 \amp -1 \\
0 \amp 0 \amp 1 \amp -2 \\
0 \amp 0 \amp 0 \amp 0
\end{array}\right]
\end{align*}
To solve, remember that this is just the coefficient matrix for the simplified system, so all columns correspond to a variable, and the “equals” column is still all zeros but does not appear. We have two free variables, corresponding to the lack of leading one in the second and fourth columns. So set parameters \(x_2=s\) and \(x_4=t\text{.}\) The first two rows turn into equations
\begin{equation*}
\begin{array}{rcrcrcrcrcr}
x_1 \amp + \amp 2x_2 \amp \amp \amp - \amp x_4 \amp = \amp 0,\\
\amp \amp \amp \amp x_3 \amp - \amp 2x_4 \amp = \amp 0,
\end{array}
\end{equation*}
from which we obtain the general solution in parametric form
\begin{align*}
x_1 \amp= -2x_2 + x_4 \amp x_2 \amp= s, \amp x_3 \amp= 2x_4 \amp x_4 \amp= t.\\
\amp= -2s + t, \amp \amp \amp \amp= 2t, \amp \amp
\end{align*}
Example 2.4.5. Correspondence between the solutions to homogeneous and nonhomogeneous systems with the same coefficient matrix.
Consider the homogeneous system
\begin{equation*}
\left\{\begin{array}{rcrcrcr}
3x \amp + \amp 6y \amp + \amp 5z \amp = \amp 0,\\
2x \amp + \amp 4y \amp + \amp 3z \amp = \amp 0,\\
3x \amp + \amp 6y \amp + \amp 6z \amp = \amp 0.
\end{array}\right.
\end{equation*}
As in the previous example, to solve we work with just the coefficient matrix
\begin{equation*}
\left[\begin{array}{rrr}
3 \amp 6 \amp 5 \\
2 \amp 4 \amp 3 \\
3 \amp 6 \amp 6
\end{array}\right].
\end{equation*}
But notice that this is the same coefficient matrix as for the system in Discovery 2.2, and the same row reduction sequence we used to solve that system in Example 2.4.2 would reduce this coefficient matrix to
\begin{equation*}
\left[\begin{array}{rrr}
1 \amp 2 \amp 0 \\
0 \amp 0 \amp 1 \\
0 \amp 0 \amp 0
\end{array}\right].
\end{equation*}
And from here we also take the same steps as in Example 2.4.2 to solve this system. Assign parameter \(t\) to free variable \(y\text{,}\) and use the simplified homogeneous system
\begin{equation*}
\left\{\begin{array}{rcrcrcr}
x \amp + \amp 2y \amp \amp \amp = \amp 0,\\
\amp \amp \amp \amp z \amp = \amp 0,\\
\amp \amp \amp \amp 0 \amp = \amp 0.
\end{array}\right.
\end{equation*}
to solve for \(x = -2t\) and \(z=0\text{.}\) Let’s compare the parametric forms of the solutions to the original nonhomogeneous system from Discovery 2.2 and the corresponding homogeneous system solved here.
We have added some zeros and \(t\)s to emphasize the similarity between the solutions. To interpret this similarity, remember that every value of \(t\) provides a particular solution to the systems. When \(t=0\text{,}\) the corresponding solutions are \(\{x=2,y=0,z=-3\}\) for the nonhomogeneous system and the trivial solution for the homogeneous system. For every other value of \(t\text{,}\) it seems that the corresponding solution for the nonhomogeneous system is equal to that “initial” particular solution \(\{x=2,y=0,z=-3\}\) plus the corresponding homogeneous solution values. In Chapter 4 we will see that this pattern emerges in every nonhomogeneous system (see Lemma 4.5.4).
| Nonhomogeneous | Homogeneous | ||
| \(\begin{array}{rcrcr} x \amp = \amp 2 \amp + \amp (-2)t \\ y \amp = \amp 0 \amp + \amp t \\ z \amp = \amp -3 \amp + \amp 0t \end{array}\) | \(\begin{array}{rcr} x \amp = \amp (-2)t \\ y \amp = \amp t \\ z \amp = \amp 0t \end{array}\) |
