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Exercises Systems of equations

Example 8 from Algebra and Trigonometry, 11.8 Solving Systems with Cramer's Rule

Using Cramer’s Rule and Determinant Properties to Solve a System

Find the solution to the given 3 × 3 system.

\begin{array}{l} 2 x + 4 y + 4 z = 2 & (1) \\ 3 x + 7 y + 7 z = −5 & (2) \\ x + 2 y + 2 z = 4 & (3) \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.

1
Solution
Using Cramer’s Rule, we have

$D = | \begin{matrix} 2 & 4 & 4 \\ 3 & 7 & 7 \\ 1 & 2 & 2 \end{matrix} |$
Notice that the second and third columns are identical. According to Property 3, the determinant will be zero, so there is either no solution or an infinite number of solutions. We have to perform elimination to find out.
1. Multiply equation (3) by –2 and add the result to equation (1).
  $\frac{\begin{array}{l} - 2 x - 4 y - 4 x = - 8 \\ 2 x + 4 y + 4 z = 2 \end{array}}{0 = - 6}$
Obtaining a statement that is a contradiction means that the system has no solution.

How did it go?

PreviousIllustrating Properties of Determinants

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 11.8, 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.