Honors Algebra 2: Systems of 3 Linear Equations: Dependent or Inconsistent Systems
- Class: Honors Algebra 2
- Author: Peter Atlas
- Text: Algebra and Trigonometry: Structure and Method, Brown
Calculator: You may use a calculator for help with the arithmetic, but you may not use a systems program or matrices.
Each of the systems below is either inconsistent, or dependent. If it is inconsistent, so state. If it is dependent, characterize the solution set.
- \( x + 2y + z = 1 \\
-x - y + 2z = 0 \\
y + 3z = 4\)
Solution
Inconsistent
- \(x + 2y + z = 1 \\
x - y + z = 1 \\
2x + y + 2z = 2\)
Solution
\( \displaystyle \left\{ \left(x, y, z \right) \in \mathbb{R} | \left( 1 - z, 0, z \right) \right\} \)
- \( x + 2y + z = 1 \\
3x + 3y + z = 2 \\
2x + y = 2\)
Solution
Inconsistent
- \( x + y + 2z = 1 \\
x - y + z = 1 \\
2x + 3z = 2\)
Solution
\( \displaystyle \left\{ \left(x, y, z \right) \in \mathbb{R} | \left( \frac{2 - 3z}{2}, \frac{z}{2}, z \right) \right\} \)
- \( x + y = 9 \\ y + z = 7 \\ x - z = 2\)
Solution
Dependent. A characterization of the solution would be \( \displaystyle \{ (x, y, z) \in \mathbb{R} | (2 + z, 7 - z, z) \} \)
- \( 2y + z = 3(-x + 1) \\
x - 3y + z = 4 \\
-2(3x + 2y + z) = 1\)
Solution
Inconsistent
- \( 3x + 2y + z = 3 \\
x - 3y + z = 4 \\
-6x - 4y - 2z = 1\)
Solution
Inconsistent
- \( x + 2y + 4z = 3 \\
4x - 2y - 6z = 2 \\
\displaystyle x - \frac{y}{2} - \frac{3z}{2} = \frac{1}{2}\)
Solution
Dependent. A characterization of the solution would be \( \displaystyle \left\{ \left(x, y, z \right) \in \mathbb{R} | \left( \frac{2}{5}z + \frac{5}{2}, \frac{7 - 7z}{5}, z \right) \right\} \)