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Exercises Systems of equations

Example 6 from Algebra and Trigonometry, 11.8 Solving Systems with Cramer's Rule

Use Cramer’s Rule to Solve a Dependent System

Solve the system with an infinite number of solutions.

\begin{array}{r} x - 2 y + 3 z = 0 & (1) \\ 3 x + y - 2 z = 0 & (2) \\ 2 x - 4 y + 6 z = 0 & (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
Let’s find the determinant first. Set up a matrix augmented by the first two columns.

$| \begin{array}{r} 1 & −2 & 3 \\ 3 & 1 & −2 \\ 2 & −4 & 6 \end{array} | \begin{array}{r} 1 & −2 \\ 3 & 1 \\ 2 & −4 \end{array} |$
Then,

$1 (1) (6) + (- 2) (- 2) (2) + 3 (3) (- 4) - 2 (1) (3) - (- 4) (- 2) (1) - 6 (3) (- 2) = 0$

As the determinant equals zero, there is either no solution or an infinite number of solutions. We have to perform elimination to find out.
1. Multiply equation (1) by $−2$ and add the result to equation (3):
  $\frac{\begin{array}{r} - 2 x + 4 y - 6 z = 0 \\ 2 x - 4 y + 6 z = 0 \end{array}}{0 = 0}$
2. Obtaining an answer of $0 = 0,$ a statement that is always true, means that the system has an infinite number of solutions. Graphing the system, we can see that two of the planes are the same and they both intersect the third plane on a line. See Figure 2.
Figure 2

How did it go?

PreviousUsing Cramer’s Rule to Solve an Inconsistent System Next Illustrating 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.