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Discover Linear Algebra: A first course in linear algebra

Jeremy Sylvestre

Exercises 1.5 Exercises

1. Recognizing linear equations.

Which of the following is a linear equation?
  1. \(\displaystyle 2 a + 3 b - 6 c = 5 \)
  2. \(\displaystyle 2 a^2 + 3 b - 6 c = 5 \)
  3. \(\displaystyle 2 a = 5 - 3 b + 6 c \)
  4. \(\displaystyle w + 2 x + 3 t y + 4 z = 0 \)
Answer.
Equations a and c are linear, though we would typically rearrange the equation in c so that all unknowns appear on the left-hand side. Equation b is not linear because of the degree-\(2\) term \(2 a^2\text{.}\) Whether equation d is considered linear or not depends on our perspective: we would probably consider \(w\text{,}\) \(x\text{,}\) \(y\text{,}\) and \(z\) to be unknowns, but what is the status of \(t\text{?}\) If the role of \(t\) is also to be an unknown, then the equation is not linear because of the degree-\(2\) term \(3 t y\text{.}\) But if the role of \(t\) is as an unspecified constant value, then the equation is linear in the other four variables.

2. Testing equation solutions.

For each collection of variable values, determine whether it constitutes a solution of the linear equation
\begin{equation*} -4 x + y + 9 z = 3\text{.} \end{equation*}

(a)

\(x = 2, y = 2, z = 1\text{.}\)
Solution.
Substitute the variable values into the left-hand side of the equation and compute
\begin{align*} \textrm{LHS} \amp = -4 x + y + 9 z \\ \amp = -4 \cdot 2 + 2 + 9 \cdot 1 \\ \amp = -8 + 2 + 9 \\ \amp = 3 \\ \amp = \textrm{RHS} \text{.} \end{align*}
As this collection of variable values satisfy the equation, it does constitute a solution of the equation.

(b)

\(x = 1, y = -2, z = 2\text{.}\)
Solution.
Substitute the variable values into the left-hand side of the equation and compute
\begin{align*} \textrm{LHS} \amp = -4 x + y + 9 z \\ \amp = -4 \cdot 1 + (-2) + 9 \cdot 2 \\ \amp = -4 - 2 + 18 \\ \amp = 12 \\ \amp \neq \textrm{RHS} \text{.} \end{align*}
As this collection of variable values does not satisfy the equation, it does not constitute a solution of the equation.

3. Testing system solutions.

For each collection of variable values, determine whether it constitutes a solution of the system of linear equations
\begin{equation*} \begin{sysofeqns}{rcrcrcr} -4 x \amp + \amp y \amp + \amp 9 z \amp = \amp 3, \\ 3 x \amp - \amp 4 y \amp - \amp 4 z \amp = \amp -6, \\ x \amp - \amp y \amp + \amp 2 z \amp = \amp 2 \end{sysofeqns} \end{equation*}

(a)

\(x = 2, y = 2, z = 1\text{.}\)
Solution.
Substitute the variable values into each of the left-hand side of the equations and compute
\begin{align*} \textrm{LHS}_1 \amp = -4 x + y + 9 z \\ \amp = -4 \cdot 2 + 2 + 9 \cdot 1 \\ \amp = -8 + 2 + 9 \\ \amp = 3 \\ \amp = \textrm{RHS}_1 \text{,} \end{align*}
\begin{align*} \textrm{LHS}_2 \amp = 3 x - 4 y - 4 z \\ \amp = 3 \cdot 2 - 4 \cdot 2 - 4 \cdot 1 \\ \amp = 6 - 8 - 4 \\ \amp = -6 \\ \amp = \textrm{RHS}_2 \text{,} \end{align*}
\begin{align*} \textrm{LHS}_3 \amp = x - y + 2 z \\ \amp = 2 - 2 + 2 \cdot 1 \\ \amp = 0 + 2 \\ \amp = 2 \\ \amp = \textrm{RHS}_3 \text{.} \end{align*}
As this collection of variable values simultaneously satisfies all equations in the system, it does constitute a solution of the system.

(b)

\(x = -2, y = 4, z = -1\text{.}\)
Solution.
Substitute the variable values into each of the left-hand side of the equations and compute
\begin{align*} \textrm{LHS}_1 \amp = -4 x + y + 9 z \\ \amp = -4 \cdot (-2) + 4 + 9 \cdot (-1) \\ \amp = 8 + 4 - 9 \\ \amp = 3 \\ \amp = \textrm{RHS}_1 \text{,} \end{align*}
\begin{align*} \textrm{LHS}_2 \amp = 3 x - 4 y - 4 z \\ \amp = 3 \cdot (-2) - 4 \cdot 4 - 4 \cdot (-1) \\ \amp = -6 - 16 + 4 \\ \amp = -18 \\ \amp \neq \textrm{RHS}_2 \text{.} \end{align*}
There is no need to continue on to the third equation, as we may already conclude that this collection of variable values does not simultaneously satisfy all equations in the system. Therefore, it does not constitute a solution of the system.

Guessing equation solutions.

For each equation, determine four specific solutions.

4.

\(2 x + 3 y = 5 \)
Solution.
There are no restrictions on the value of the \(y\) variable, so we may choose its value to be anything we like and then determine the corresponding \(x\)-value. For example, when \(y = -1\) we have
\begin{align*} 2 x + 3 \cdot (-1) \amp = 5 \\ 2 x - 3 \amp = 5 \\ 2 x \amp = 5 + 3 \\ 2 x \amp = 8 \\ x \amp = 4 \text{.} \end{align*}
Thus \(x = 4, y = -1\) is a specific solution to the equation. Here is a list of that solution and three others obtained by choosing other values for \(y\) and then solving for the corresponding \(x\) value, just as above.
  1. \(\displaystyle x = 4, y = -1\)
  2. \(\displaystyle x = 1, y = 1\)
  3. \(\displaystyle x = 5 / 2, y = 0\)
  4. \(\displaystyle x = 2, y = 1/3\)

5.

\(3 x - 4 y + z = -6 \)
Solution.
There are no restrictions on the value of either the \(y\) or \(z\) variable — we may choose any combination of values for these two variables and then determine the corresponding \(x\)-value. For example, when \(y = 1\) and \(z = 0\) we have
\begin{align*} 3 x - 4 \cdot 1 + 0 \amp = -6 \\ 3 x - 4 \amp = -6 \\ 3 x \amp = -6 + 4 \\ 3 x \amp = -2 \\ x \amp = - 2 / 3 \text{.} \end{align*}
Thus, \(x = - 2 / 3, y = 1, z = 0\) is a specific solution to the equation. Here is a list of that solution and three others obtained by choosing other combinations of values for \(y\) and \(z\) and then solving for the corresponding \(x\) value, just as above.
  1. \(\displaystyle x = - 2 / 3, y = 1, z = 0\)
  2. \(\displaystyle x = -7 / 3, y = 0, z = 1\)
  3. \(\displaystyle x = -2, y = 0, z = 0\)
  4. \(\displaystyle x = -1, y = 1, z = 1\)

Solutions as geometric points of intersection.

For each system of equations, use a graphing tool to graph the corresponding lines on the set same of axes. Use the plot to determine if the system is consistent or inconsistent.

6.

\(\displaystyle \begin{sysofeqns}{rcrcr} x \amp + \amp 4 y \amp = \amp 6 \\ x \amp - \amp 2 y \amp = \amp 0 \\ 3 x \amp + \amp y \amp = \amp 7 \end{sysofeqns}\)
Solution.
Three lines intersecting at a single point in common.
The three lines intersect at a common point, and so the system is consistent. Your graphing tool may tell you that this common point is \((2,1)\text{,}\) and you may verify that this is correct algebraically by confirming that \(x = 2, y = 1\) constitutes a solution to the system.

7.

\(\displaystyle \begin{sysofeqns}{rcrcr} x \amp - \amp y \amp = \amp -2 \\ 2 x \amp + \amp y \amp = \amp 5 \\ x \amp + \amp 2 y \amp = \amp 1 \end{sysofeqns}\)
Solution.
Three intersecting lines that do not intersect at a single point in common.
While these lines intersect in pairs, there is no common point where all three intersect. Therefore, the system is inconsistent.

Creating augmented matrices.

Convert each system into an augmented matrix.

8.

\(\displaystyle \begin{sysofeqns}{rcrcr} x \amp + \amp 3 y \amp = \amp 1 \\ 2 x \amp + \amp 7 y \amp = \amp 2 \end{sysofeqns}\)
Answer.
\(\displaystyle \begin{abmatrix}{rr|r} 1 \amp 3 \amp 1 \\ 2 \amp 7 \amp 2 \end{abmatrix}\)

9.

\(\displaystyle \begin{sysofeqns}{rcrcr} 5 a \amp - \amp b \amp = \amp 2 \\ 2 a \amp + \amp 6 b \amp = \amp 5 \\ 4 a \amp + \amp 3 b \amp = \amp -7 \end{sysofeqns}\)
Answer.
\(\displaystyle \begin{abmatrix}{rr|r} 5 \amp -1 \amp 2 \\ 2 \amp 6 \amp 5 \\ 4 \amp 3 \amp -7 \end{abmatrix}\)

10.

\(\displaystyle \begin{sysofeqns}{rcrcrcr} x \amp - \amp 2 y \amp + \amp 2 z \amp = \amp 3 \\ 3 x \amp + \amp 2 y \amp - \amp 4 z \amp = \amp 4 \end{sysofeqns}\)
Answer.
\(\displaystyle \begin{abmatrix}{rrr|r} 1 \amp -2 \amp 2 \amp 3 \\ 3 \amp 2 \amp -4 \amp 4 \end{abmatrix}\)

11.

\(\displaystyle \begin{sysofeqns}{rcrcrcr} 5 x_1 \amp - \amp 3 x_2 \amp + \amp 2 x_3 \amp = \amp 2 \\ x_1 \amp - \amp 2 x_2 \amp \amp \amp = \amp 4 \\ -3 x_1 \amp - \amp 3 x_2 \amp + \amp 2 x_3 \amp = \amp 1 \end{sysofeqns}\)
Answer.
\(\displaystyle \begin{abmatrix}{rrr|r} 5 \amp -3 \amp 2 \amp 2 \\ 1 \amp -2 \amp 0 \amp 4 \\ -3 \amp -3 \amp 2 \amp 1 \end{abmatrix}\)

12.

\(\displaystyle \begin{sysofeqns}{rcrcrcrcr} 2 w \amp + \amp x \amp + \amp y \amp + \amp 9 z \amp = \amp 5 \\ w \amp + \amp x \amp \amp 2 y \amp - \amp 5 z \amp = \amp -4 \\ \amp + \amp x \amp + \amp 4 y \amp + \amp 4 z \amp = \amp -5 \end{sysofeqns}\)
Answer.
\(\displaystyle \begin{abmatrix}{rrrr|r} 2 \amp 1 \amp 1 \amp 9 \amp 5 \\ 1 \amp 1 \amp 2 \amp -5 \amp -4 \\ 0 \amp 1 \amp 4 \amp 4 \amp -5 \end{abmatrix}\)

Interpreting augmented matrices.

For each matrix, write out a system of equations whose augmented matrix is the one given. Then determine the solution(s) to the system.

13.

\(\displaystyle \begin{abmatrix}{rr|r} 1 \amp 2 \amp 5 \\ 0 \amp 1 \amp 2 \end{abmatrix}\)
Solution.
Using variables \(x\) and \(y\text{,}\) the corresponding system is
\begin{equation*} \begin{sysofeqns}{rcrcr} x \amp + \amp 2 y \amp = \amp 5 \\ \amp \amp y \amp = \amp 2 \end{sysofeqns}\text{.} \end{equation*}
From the second equation, we immediately have \(y = 2\text{,}\) and we can substitute that value into the first equation to obtain
\begin{gather*} x + 2 y = 5 \\ x + 2 \cdot 2 = 5 \\ x + 4 = 5 \\ x = 5 - 4 \\ x = 1 \text{.} \end{gather*}
So the solution to the system is \(x = 1, y = 2\text{.}\)

14.

\(\displaystyle \begin{abmatrix}{rr|r} 1 \amp -2 \amp 0 \\ 0 \amp 1 \amp 3 \\ 0 \amp 0 \amp 0 \end{abmatrix}\)
Solution.
Using variables \(x\) and \(y\text{,}\) the corresponding system is
\begin{equation*} \begin{sysofeqns}{rcrcr} x \amp - \amp 2 y \amp = \amp 0 \\ \amp \amp y \amp = \amp 3 \\ 0 x \amp + \amp 0 y \amp = \amp 0 \end{sysofeqns}\text{.} \end{equation*}
Notice that the third equation essentially says \(0 = 0\text{,}\) which is true but contains no useful information. From the second equation, we immediately have \(y = 3\text{,}\) and we can substitute that value into the first equation to obtain
\begin{gather*} x - 2 y = 0 \\ x - 2 \cdot 3 = 0 \\ x - 6 = 0 \\ x = 6 \text{.} \end{gather*}
So the solution to the system is \(x = 6, y = 3\text{.}\)

15.

\(\displaystyle \begin{abmatrix}{rrr|r} 1 \amp -3 \amp -2 \amp 5 \\ 0 \amp 1 \amp 3 \amp -3 \\ 0 \amp 0 \amp 1 \amp 2 \end{abmatrix}\)
Solution.
Using variables \(x,y,z\) the corresponding system is
\begin{equation*} \begin{sysofeqns}{rcrcrcr} x \amp - \amp 3 y \amp - \amp 2 z \amp = \amp 5 \\ \amp \amp y \amp + \amp 3 z \amp = \amp -3 \\ \amp \amp \amp \amp z \amp = \amp 2 \end{sysofeqns}\text{.} \end{equation*}
From the third equation, we immediately have \(z = 2\text{,}\) and we can substitute that value into the second equation to obtain
\begin{gather*} y + 3 z = -3 \\ y + 3 \cdot 2 = -3 \\ y + 6 = -3 \\ y = -9 \text{.} \end{gather*}
Knowing values for both \(y\) and \(z\text{,}\) we can substitute them into the first equation to obtain
\begin{gather*} x - 3 y - 2 z = 5 \\ x - 3 \cdot (-9) - 2 \cdot 2 = 5 \\ x + 23 = 5 \\ x = - 18 \text{.} \end{gather*}
So the solution to the system is \(x = -18, y = -9, z = 2\text{.}\)

16.

\(\displaystyle \begin{abmatrix}{rrr|r} 1 \amp -3 \amp -2 \amp 5 \\ 0 \amp 1 \amp 3 \amp -3 \\ 0 \amp 0 \amp 1 \amp 2 \\ 0 \amp 0 \amp 0 \amp 4 \end{abmatrix}\)
Solution.
Using variables \(x,y,z\) the corresponding system is
\begin{equation*} \begin{sysofeqns}{rcrcrcr} x \amp - \amp 3 y \amp - \amp 2 z \amp = \amp 5 \\ \amp \amp y \amp + \amp 3 z \amp = \amp -3 \\ \amp \amp \amp \amp z \amp = \amp 2 \\ 0 x \amp + \amp 0 y \amp + \amp 0 z \amp = \amp 4 \end{sysofeqns}\text{.} \end{equation*}
At first glance this seems very similar to the system from Exercise 15. However, here we have a fourth equation that essentially says \(0 = 4\text{,}\) which is false, and there is no combination of values for \(x,y,z\) that will make it true. So while the values \(x = -18, y = -9, z = 2\) will simultaneously satisfy the first three equations in the system, there is no combination of variable values that will simultaneously satisfy all four equations, and therefore the system is inconsistent.

Performing row operations.

In each case, perform the stated row operation on the given matrix. Then explore how the row operation represents an algebraic manipulation/combination of equations in the underlying system.

17.

Interchange the first and second rows.
\begin{equation*} \begin{abmatrix}{rrr|r} 4 \amp 8 \amp -7 \amp 3 \\ 1 \amp 2 \amp -3 \amp -1 \\ -2 \amp 0 \amp -4 \amp 13 \end{abmatrix} \end{equation*}
Answer.
\begin{equation*} \begin{abmatrix}{rrr|r} 4 \amp 8 \amp -7 \amp 3 \\ 1 \amp 2 \amp -3 \amp -1 \\ -2 \amp 0 \amp -4 \amp 13 \end{abmatrix} \xrightarrow{R_1 \leftrightarrow R_2} \begin{abmatrix}{rrr|r} 1 \amp 2 \amp -3 \amp -1 \\ 4 \amp 8 \amp -7 \amp 3 \\ -2 \amp 0 \amp -4 \amp 13 \end{abmatrix} \end{equation*}
Using variables \(x,y,z\text{,}\) the original system of equations is
\begin{equation*} \begin{sysofeqns}{rcrcrcr} 4 x \amp + \amp 8 y \amp - \amp 7 z \amp = \amp 3 \\ 1 x \amp + \amp 2 y \amp - \amp 3 z \amp = \amp -1 \\ -2 x \amp \amp \amp - \amp 4 z \amp = \amp 13 \end{sysofeqns}\text{.} \end{equation*}
Interchanging the order of the first two equations leads to system
\begin{equation*} \begin{sysofeqns}{rcrcrcr} 1 x \amp + \amp 2 y \amp - \amp 3 z \amp = \amp -1 \\ 4 x \amp + \amp 8 y \amp - \amp 7 z \amp = \amp 3 \\ -2 x \amp \amp \amp - \amp 4 z \amp = \amp 13 \end{sysofeqns}\text{,} \end{equation*}
and the coefficients/constants of this re-ordered system correspond exactly to the entries in the result matrix above.

18.

Interchange the third and fifth rows.
\begin{equation*} \begin{abmatrix}{rrrr|r} 1 \amp 1 \amp -6 \amp 3 \amp 3 \\ 0 \amp 1 \amp 6 \amp -1 \amp 2 \\ 0 \amp 0 \amp 0 \amp 4 \amp -1 \\ 0 \amp 0 \amp 0 \amp -4 \amp -7 \\ 0 \amp 0 \amp 0 \amp 1 \amp 3 \end{abmatrix} \end{equation*}
Answer.
\begin{equation*} \begin{abmatrix}{rrrrr|r} 1 \amp 1 \amp -6 \amp 3 \amp 3 \amp 7 \\ 0 \amp 1 \amp 6 \amp -1 \amp 2 \amp 3 \\ 0 \amp 0 \amp 0 \amp 4 \amp -1 \amp -8 \\ 0 \amp 0 \amp 0 \amp -4 \amp -7 \amp 7 \\ 0 \amp 0 \amp 0 \amp 1 \amp 3 \amp 3 \end{abmatrix} \xrightarrow{R_3 \leftrightarrow R_5} \begin{abmatrix}{rrrrr|r} 1 \amp 1 \amp -6 \amp 3 \amp 3 \amp 7 \\ 0 \amp 1 \amp 6 \amp -1 \amp 2 \amp 3 \\ 0 \amp 0 \amp 0 \amp 1 \amp 3 \amp 3 \\ 0 \amp 0 \amp 0 \amp -4 \amp -7 \amp 7 \\ 0 \amp 0 \amp 0 \amp 4 \amp -1 \amp -8 \end{abmatrix} \end{equation*}
Using variables \(x_1,x_2,x_3,x_4,x_5\text{,}\) the original system of equations is
\begin{equation*} \begin{sysofeqns}{rcrcrcrcrcr} x_1 \amp + \amp x_2 \amp - \amp 6 x_3 \amp + \amp 3 x_4 \amp + \amp 3 x_5 \amp = \amp 7 \\ \amp \amp x_2 \amp + \amp 6 x_3 \amp - \amp x_4 \amp + \amp 2 x_5 \amp = \amp 3 \\ \amp \amp \amp \amp \amp \amp 4 x_4 \amp - \amp x_5 \amp = \amp -8 \\ \amp \amp \amp \amp \amp \amp -4 x_4 \amp - \amp 7 x_5 \amp = \amp 7 \\ \amp \amp \amp \amp \amp \amp x_4 \amp + \amp 3 x_5 \amp = \amp 3 \end{sysofeqns}\text{.} \end{equation*}
Interchanging the order of the third and fifth equations leads to system
\begin{equation*} \begin{sysofeqns}{rcrcrcrcrcr} x_1 \amp + \amp x_2 \amp - \amp 6 x_3 \amp + \amp 3 x_4 \amp + \amp 3 x_5 \amp = \amp 7 \\ \amp \amp x_2 \amp + \amp 6 x_3 \amp - \amp x_4 \amp + \amp 2 x_5 \amp = \amp 3 \\ \amp \amp \amp \amp \amp \amp x_4 \amp + \amp 3 x_5 \amp = \amp 3 \\ \amp \amp \amp \amp \amp \amp -4 x_4 \amp - \amp 7 x_5 \amp = \amp 7 \\ \amp \amp \amp \amp \amp \amp 4 x_4 \amp - \amp x_5 \amp = \amp -8 \end{sysofeqns}\text{,} \end{equation*}
and the coefficients/constants of this re-ordered system correspond exactly to the entries in the result matrix above.

19.

Multiply the second row by \(1/3\text{.}\)
\begin{equation*} \begin{abmatrix}{rrr|r} 2 \amp 4 \amp 3 \amp -7 \\ 3 \amp 6 \amp 6 \amp 9 \\ 0 \amp 4 \amp 13 \amp 2 \end{abmatrix} \end{equation*}
Answer.
\begin{equation*} \begin{abmatrix}{rrr|r} 2 \amp 4 \amp 3 \amp -7 \\ 3 \amp 6 \amp 6 \amp 9 \\ 0 \amp 4 \amp 13 \amp 2 \end{abmatrix} \xrightarrow{\frac{1}{3} R_2} \begin{abmatrix}{rrr|r} 2 \amp 4 \amp 3 \amp -7 \\ 1 \amp 2 \amp 2 \amp 3 \\ 0 \amp 4 \amp 13 \amp 2 \end{abmatrix} \end{equation*}
Using variables \(x,y,z\text{,}\) the original second row represents the equation
\begin{equation*} 3 x + 6 y + 6 z = 9 \text{.} \end{equation*}
The coefficients and constant in this equation have a factor of \(3\) in common, and multiplying the equation by \(1/3\) represents “cancelling out” this common factor on both sides:
\begin{equation*} 3 x + 6 y + 6 z = 9 \quad \longrightarrow \quad x + 2 y + 2 z = 3 \text{.} \end{equation*}
The coefficients/constant in this new equation correspond exactly to the new second row in our manipulated matrix above.

20.

Multiply the first row by \(1/7\text{.}\)
\begin{equation*} \begin{abmatrix}{rrrr|r} 7 \amp -21 \amp 15 \amp 0 \amp 7 \\ 12 \amp - 5 \amp -4 \amp 1 \amp 8 \\ 11 \amp - 1 \amp -1 \amp 3 \amp -2 \end{abmatrix} \end{equation*}
Answer.
\begin{equation*} \begin{abmatrix}{rrrr|r} 7 \amp -21 \amp 15 \amp 0 \amp 7 \\ 12 \amp - 5 \amp -4 \amp 1 \amp 8 \\ 11 \amp - 1 \amp -1 \amp 3 \amp -2 \end{abmatrix} \xrightarrow{\frac{1}{7} R_1} \begin{abmatrix}{rrrr|r} 1 \amp -3 \amp \frac{15}{7} \amp 0 \amp 1 \\ 12 \amp -5 \amp -4 \amp 1 \amp 8 \\ 11 \amp -1 \amp -1 \amp 3 \amp -2 \end{abmatrix} \end{equation*}
Using variables \(x_1,x_2,x_3,x_4\text{,}\) the original first row represents the equation
\begin{equation*} 7 x_1 - 21 x_2 + 15 x_3 = 7 \text{.} \end{equation*}
Most of the coefficients and constant in this equation have a factor of \(7\) in common. Usually, in purely algebraic manipulations we would not divide this equation through by that common factor. But when we develop an algorithm for systematically simplifying systems of equations through matrix operations in Chapter 2, we will find that it is advantageous to perform operations to reduce the “leading” coefficient in some of the equations in the system to \(1\text{,}\) which is precisely what our chosen operation achieves:
\begin{equation*} 7 x_1 - 21 x_2 + 15 x_3 = 7 \quad \longrightarrow \quad x_1 - 3 x_2 + \tfrac{15}{7} x_3 = 1 \text{.} \end{equation*}
The coefficients/constant in this new equation correspond exactly to the new first row in our manipulated matrix above. (While we often try to avoid fractions in our matrices, there is no reason a matrix cannot contain fractional entries because there is no reason the coefficients in an equation cannot be fractions.)

21.

Add \(2\)-times the first row to the third row.
\begin{equation*} \begin{abmatrix}{rrrr|r} 1 \amp 3 \amp 1 \amp -2 \amp 1 \\ 0 \amp 1 \amp 1 \amp -2 \amp 3 \\ -2 \amp 1 \amp -6 \amp 8 \amp 5 \end{abmatrix} \end{equation*}
Answer.
\begin{equation*} \begin{abmatrix}{rrrr|r} 1 \amp 3 \amp 1 \amp -2 \amp 1 \\ 0 \amp 1 \amp 1 \amp -2 \amp 3 \\ -2 \amp 1 \amp -6 \amp 8 \amp 5 \end{abmatrix} \xrightarrow{R_3 + 2 R_1} \begin{abmatrix}{rrrr|r} 1 \amp 3 \amp 1 \amp -2 \amp 1 \\ 0 \amp 1 \amp 1 \amp -2 \amp 3 \\ 0 \amp 7 \amp -4 \amp 4 \amp 3 \end{abmatrix} \end{equation*}
Notice that the first row has not changed. We wished to modify the third row using a multiple of the first, but the first row itself is not modified by that multiple.
Using variables \(w,x,y,z\text{,}\) the original first and third rows represent the equations
\begin{align*} w + 3 x + y - 2 z \amp = 1 \text{,} \\ -2 w + x - 6 y + 8 z \amp = 5 \text{.} \end{align*}
Combine left- and right-hand sides identically. Since each left-hand side is assumed to be equal to its corresponding right-hand side, any specific combination of left-hand sides will be equal to the identical combination of right-hand sides. In the combination of left-hand sides, we simplify by collecting like terms.
\begin{gather*} {\text{LHS}}_3 + 2 ({\text{LHS}}_1) = {\text{RHS}}_3 + 2 ({\text{RHS}}_1) \\ (-2 w + x - 6 y + 8 z) + 2 (w + 3 x + y - 2 z) = 5 - 2 \cdot 1 \\ (-2 w + x - 6 y + 8 z) + (2 w + 6 x + 2 y - 4 z) = 5 - 2 \\ 7 x - 4 y + 4 z = 3 \end{gather*}
The coefficients/constant in this new equation correspond exactly to the new third row in our manipulated matrix above. (Recall that we have assigned the variable \(w\) to the first column in the original matrix. The fact that \(w\) no longer appears in the new equation corresponds to the zero value in the first entry of the new third row.)

22.

Subtract the second row from the first. Note: This operation should modify the first row only.
\begin{equation*} \begin{abmatrix}{rrr|r} 3 \amp 9 \amp -2 \amp 3 \\ 2 \amp 6 \amp 1 \amp 1 \end{abmatrix} \end{equation*}
Answer.
\begin{equation*} \begin{abmatrix}{rrr|r} 3 \amp 9 \amp -2 \amp 3 \\ 2 \amp 6 \amp 1 \amp 1 \end{abmatrix} \xrightarrow{R_1 - R_2} \begin{abmatrix}{rrr|r} 1 \amp 3 \amp -3 \amp 2 \\ 2 \amp 6 \amp 1 \amp 1 \end{abmatrix} \end{equation*}
Using variables \(x,y,z\text{,}\) the original equations are
\begin{align*} 3 x + 9 y - 2 z \amp = 3 \text{,} \\ 2 x + 6 y + z \amp = 1 \text{.} \end{align*}
Combine left- and right-hand sides identically. Since each left-hand side is assumed to be equal to its corresponding right-hand side, any specific combination of left-hand sides will be equal to the identical combination of right-hand sides. In the combination of left-hand sides, we simplify by collecting like terms.
\begin{gather*} {\text{LHS}}_1 - {\text{LHS}}_2 = {\text{RHS}}_1 - {\text{RHS}}_2 \\ (3 x + 9 y - 2 z) - (2 x + 6 y + z) = 3 - 1 \\ x + 3 y - 3 z = 2 \end{gather*}
The coefficients/constant in this new equation correspond exactly to the new first row in our manipulated matrix above.
In Exercise 20, our choice of operation had a purpose: to reduce the leading coefficient in that equation to \(1\text{.}\) But the operation had the side-effect of introducing fractions into our matrix. Here we were able to reduce the leading coefficient in the first equation to \(1\) without introducing fractions, as would have happened if we had instead multiplied the first row by \(1/3\text{.}\)

23.

Subtract \(3\)-times the third row from the second. Take care when subtracting a negative! Note: This operation should modify the second row only.
\begin{equation*} \begin{abmatrix}{rrrr|r} 1 \amp -3 \amp 0 \amp -1 \amp 1 \\ 0 \amp 0 \amp 16 \amp 4 \amp 1 \\ 0 \amp 0 \amp 5 \amp -2 \amp -1 \end{abmatrix} \end{equation*}
Answer.
\begin{equation*} \begin{abmatrix}{rrrr|r} 1 \amp -3 \amp 0 \amp -1 \amp 1 \\ 0 \amp 0 \amp 16 \amp 4 \amp 1 \\ 0 \amp 0 \amp 5 \amp -2 \amp -1 \end{abmatrix} \xrightarrow{R_2 - 3 R_3} \begin{abmatrix}{rrrr|r} 1 \amp -3 \amp 0 \amp -1 \amp 1 \\ 0 \amp 0 \amp 1 \amp 10 \amp 4 \\ 0 \amp 0 \amp 5 \amp -2 \amp -1 \end{abmatrix} \end{equation*}
Using variables \(x_1,x_2,x_3,x_4\text{,}\) the original equations are
\begin{align*} 16 x_3 + 4 x_4 \amp = 1 \text{,} \\ 5 x_3 - 2 x_4 \amp = -1 \text{.} \end{align*}
Combine left- and right-hand sides identically. Since each left-hand side is assumed to be equal to its corresponding right-hand side, any specific combination of left-hand sides will be equal to the identical combination of right-hand sides. In the combination of left-hand sides, we simplify by collecting like terms.
\begin{gather*} {\text{LHS}}_2 - 3 ({\text{LHS}}_3) = {\text{RHS}}_2 - 3 ({\text{RHS}}_3) \\ (16 x_3 + 4 x_4) - 3 (5 x_3 - 2 x_4) = 1 - 3 (-1) \\ 16 x_3 + 4 x_4 - 15 x_3 \mathbin{\color{red} \boldsymbol{+}} 6 x_4 = 1 \mathbin{\color{red} \boldsymbol{+}} 3 \\ x_3 + 10 x_4 = 4 \end{gather*}
The coefficients/constant in this new equation correspond exactly to the new second row in our manipulated matrix above.
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