Example 8 from Algebra and Trigonometry, 11.6 Solving Systems with Gaussian Elimination

Solving a Dependent System of Linear Equations Using Matrices

Solve the following system of linear equations using matrices.

\begin{array}{r} - x −2 y + z = −1 \\ 2 x + 3 y = 2 \\ y −2 z = 0 \end{array}

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Write the augmented matrix.

$[\begin{array}{r} −1 & −2 & 1 \\ 2 & 3 & 0 \\ 0 & 1 & −2 \end{array} | \begin{array}{r} −1 \\ 2 \\ 0 \end{array}]$
First, multiply row 1 by $−1$ to get a 1 in row 1, column 1. Then, perform row operations to obtain row-echelon form.
$- R_{1} \to [\begin{array}{r} 1 & 2 & −1 \\ 2 & 3 & 0 \\ 0 & 1 & −2 \end{array} | \begin{array}{r} 1 \\ 2 \\ 0 \end{array}]$
$R_{2} \leftrightarrow R_{3} \to [\begin{array}{r} 1 & 2 & - 1 \\ 0 & 1 & - 2 \\ 2 & 3 & 0 \end{array} | \begin{array}{r} 1 \\ 0 \\ 2 \end{array}]$
$−2 R_{1} + R_{3} = R_{3} \to [\begin{array}{r} 1 & 2 & −1 \\ 0 & 1 & −2 \\ 0 & −1 & 2 \end{array} | \begin{array}{r} 1 \\ 0 \\ 0 \end{array}]$
$R_{2} + R_{3} = R_{3} \to [\begin{array}{r} 1 & 2 & −1 \\ 0 & 1 & −2 \\ 0 & 0 & 0 \end{array} | \begin{array}{r} 2 \\ 1 \\ 0 \end{array}]$
The last matrix represents the following system.

$\begin{array}{l} x+2y-z=1 \\ y - 2 z = 0 \\ 0 = 0 \end{array}$
We see by the identity $0 = 0$ that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the second equation for $y$ and substituting it into the first equation we can solve for $z$ in terms of $x .$
$\begin{array}{l} x + 2 y - z = 1 \\ y = 2 z \\ x + 2 (2 z) - z = 1 \\ x + 3 z = 1 \\ z = \frac{1 - x}{3} \end{array}$
Now we substitute the expression for $z$ into the second equation to solve for $y$ in terms of $x .$
$\begin{array}{l} y - 2 z = 0 \\ z = \frac{1 - x}{3} \\ y - 2 (\frac{1 - x}{3}) = 0 \\ y = \frac{2 - 2 x}{3} \end{array}$
The generic solution is $(x, \frac{2 −2 x}{3}, \frac{1 - x}{3}) .$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solve the system using matrices.

\begin{matrix} x + 4 y - z = 4 \\ 2 x + 5 y + 8 z = 15 \\ x + 3 y −3 z = 1 \end{matrix}

How did it go?

Solution

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Solution

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