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Exercises Systems of equations

Example 6 from Algebra and Trigonometry, 11.6 Solving Systems with Gaussian Elimination

Performing Row Operations on a 3×3 Augmented Matrix to Obtain Row-Echelon Form

Perform row operations on the given matrix to obtain row-echelon form.

[\begin{array}{r} 1 & −3 & 4 \\ 2 & −5 & 6 \\ −3 & 3 & 4 \end{array} | \begin{array}{r} 3 \\ 6 \\ 6 \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
The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by $−2$ and add it to row 2. Then replace row 2 with the result.

$−2 R_{1} + R_{2} = R_{2} \to [\begin{array}{r} 1 & −3 & 4 \\ 0 & 1 & −2 \\ −3 & 3 & 4 \end{array} | \begin{array}{r} 3 \\ 0 \\ 6 \end{array}]$
Next, obtain a zero in row 3, column 1.

$3 R_{1} + R_{3} = R_{3} \to [\begin{array}{r} 1 & −3 & 4 \\ 0 & 1 & −2 \\ 0 & −6 & 16 \end{array} | \begin{array}{r} 3 \\ 0 \\ 15 \end{array}]$
Next, obtain a zero in row 3, column 2.

$6 R_{2} + R_{3} = R_{3} \to [\begin{array}{r} 1 & −3 & 4 \\ 0 & 1 & −2 \\ 0 & 0 & 4 \end{array} | \begin{array}{r} 3 \\ 0 \\ 15 \end{array}]$
The last step is to obtain a 1 in row 3, column 3.

$\frac{1}{4} R_{3} = R_{3} \to [\begin{array}{r} 1 & −3 & 4 \\ 0 & 1 & −2 \\ 0 & 0 & 1 \end{array} | \begin{array}{r} 3 \\ −6 \\ \frac{15}{4} \end{array}]$

Your turn

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

Write the system of equations in row-echelon form.

\begin{array}{l} x - 2 y + 3 z = 9 \\ - x + 3 y = - 4 \\ 2 x - 5 y + 5 z = 17 \end{array}

How did it go?

PreviousSolving a Dependent System Next Solving a System of Linear Equations Using Matrices

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 11.6, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.