Systems of Equations in Three Variables - Revision history

Khanh: /* Cramer's rule */

2022-01-15T18:16:35Z

Cramer's rule

Khanh: /* Row reduction */

2022-01-15T18:16:08Z

Row reduction

Lila at 18:35, 21 October 2021

2021-10-21T18:35:30Z

Lila: /* Resources */

2021-10-21T18:32:57Z

Resources

Lila at 18:32, 21 October 2021

2021-10-21T18:32:01Z

Show changes

Lila: /* Resources */

2021-09-15T17:04:37Z

Resources

Lila: /* Examples */

2021-09-15T17:02:14Z

Examples

Lila: /* Resources */

2021-09-15T17:01:01Z

Resources

Lila: /* Examples */

2021-09-15T16:56:17Z

Examples

Lila: /* Examples */

2021-09-15T16:54:30Z

Examples

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 Though Cramer's rule is important theoretically, it has little practical value for large matrices, since the computation of large determinants is somewhat cumbersome. (Indeed, large determinants are most easily computed using row reduction.)
-Further, Cramer's rule has very poor numerical properties, making it unsuitable for solving even small systems reliably, unless the operations are performed in rational arithmetic with unbounded precision.{{Citation needed|date=March 2017}}
+Further, Cramer's rule has very poor numerical properties, making it unsuitable for solving even small systems reliably, unless the operations are performed in rational arithmetic with unbounded precision.
 ===Matrix solution===

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 ===Row reduction===
 In '''row reduction''' (also known as '''Gaussian elimination'''), the linear system is represented as an augmented matrix:
 :<math>\left[\begin{array}{rrr|r}

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 x &&\; + \;&& 4y &&\; + \;&& 3z &&\; = \;&& 8 &
 \end{alignat}</math>
-Solving the first equation for ''x'' gives {{nowrap|''x'' {{=}} 5 + 2''z'' &minus; 3''y''}}, and plugging this into the second and third equation yields
+Solving the first equation for ''x'' gives ''x'' = 5 + 2''z'' - 3''y'', and plugging this into the second and third equation yields
 :<math>\begin{alignat}{5}
 -4y &&\; + \;&& 12z &&\; = \;&& -8 & \\
 -2y &&\; + \;&& 7z &&\; = \;&& -2 &
 \end{alignat}</math>
-Solving the first of these equations for ''y'' yields {{nowrap|''y'' {{=}} 2 + 3''z''}}, and plugging this into the second equation yields {{nowrap|''z'' {{=}} 2}}. We now have:
+Solving the first of these equations for ''y'' yields ''y'' = 2 + 3''z'', and plugging this into the second equation yields ''z'' = 2. We now have:
 :<math>\begin{alignat}{7}
   x &&\; = \;&& 5 &&\; + \;&& 2z &&\; - \;&& 3y & \\
@@ Line 63: / Line 63: @@
   z &&\; = \;&& 2 && && && && &
 \end{alignat}</math>
-Substituting {{nowrap|''z'' {{=}} 2}} into the second equation gives {{nowrap|''y'' {{=}} 8}}, and substituting {{nowrap|''z'' {{=}} 2}} and {{nowrap|''y'' {{=}} 8}} into the first equation yields {{nowrap|''x'' {{=}} &minus;15}}. Therefore, the solution set is the single point {{nowrap|(''x'', ''y'', ''z'') {{=}} (&minus;15, 8, 2)}}.
+Substituting ''z'' = 2 into the second equation gives ''y'' = 8, and substituting ''z'' = 2 and ''y'' = 8 into the first equation yields ''x'' = -15. Therefore, the solution set is the single point (''x'', ''y'', ''z'') = (-15, 8, 2).
 ===Row reduction===
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 & 0 & 1 & 2
 \end{array}\right].\end{align}</math>
-The last matrix is in reduced row echelon form, and represents the system {{nowrap|''x'' {{=}} &minus;15}}, {{nowrap|''y'' {{=}} 8}}, {{nowrap|''z'' {{=}} 2}}. A comparison with the example in the previous section on the algebraic elimination of variables shows that these two methods are in fact the same; the difference lies in how the computations are written down.
+The last matrix is in reduced row echelon form, and represents the system ''x'' = -15, ''y'' = 8, ''z'' = 2. A comparison with the example in the previous section on the algebraic elimination of variables shows that these two methods are in fact the same; the difference lies in how the computations are written down.
 ===Cramer's rule===

← Older revision		Revision as of 18:32, 21 October 2021
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	* [https://www.youtube.com/watch?v=GjbRnAjVlXM Solving a System of 3 Variables With Substitution], patrickJMT		* [https://www.youtube.com/watch?v=GjbRnAjVlXM Solving a System of 3 Variables With Substitution], patrickJMT
	* [https://www.youtube.com/watch?v=tGPSEXVYw_o Solving a System of Two Equations with Three Variables (Infinite Solutions)], patrickJMT		* [https://www.youtube.com/watch?v=tGPSEXVYw_o Solving a System of Two Equations with Three Variables (Infinite Solutions)], patrickJMT
		+
		+	== Licensing ==
		+	Content obtained and/or adapted from:
		+	* [https://en.wikipedia.org/wiki/System_of_linear_equations System of linear equations, Wikipedia] under a CC BY-SA license

@@ Line 8: / Line 8: @@
 * [https://tutorial.math.lamar.edu/classes/alg/systemsthreevrble.aspx Linear Systems with Three Variables], Paul's Online Notes (Lamar Math)
 * [https://courses.lumenlearning.com/wmopen-collegealgebra/chapter/introduction-systems-of-linear-equations-three-variables/ Systems of Equations: Three Variables], Lumen Learning
-* [https://www.youtube.com/watch?v=CdpFu7t0dJ4 Using Elimination to Solve a 3-Variable System], patrickJMT
+* [https://www.youtube.com/watch?v=CdpFu7t0dJ4 Solving a System of 3 Variables With Elimination], patrickJMT
 * [https://www.youtube.com/watch?v=tGPSEXVYw_o Solving a System of Two Equations with Three Variables (Infinite Solutions)], patrickJMT

@@ Line 2: / Line 2: @@
 ==Examples==
 * One solution: <math> x + y + z = 3 </math>, <math> 2x - 3y + z = 0 </math>, and <math> -5x - 5y + 23z = 13 </math>. <math> x + y + z = 3 \implies 5x + 5y + 5z = 15</math>. We can add this to the third equation to get <math> 28z = 28 </math>, which means z = 1. So, the first two equations can be rewritten as <math> x + y = 2 </math> and <math> 2x - 3y = -1 </math>. Using substitution, elimination, or graphing, we can calculate that x = 1 and y = 1 with these two equations. Thus, the solution to the system is (x, y, z) = (1, 1, 1).
-* No solutions: <math> x + y + z = 1 </math>, <math> x + y + z = 2 </math>, <math> x + y + z = 3 </math>. These equations represent three parallel planes, and there is no x, y, and z that satisfy all three equations simultaneously. So, this system has no solutions.
+* No solutions: <math> x + y + z = 1 </math> and <math> x + y + z = 5 </math>. These equations represent two parallel planes, and there is no x, y, and z that satisfy both equations simultaneously. So, this system has no solutions.
-* Infinite solutions: <math> x + y = z </math>, <math> x + y = 2z </math>, <math> x + y = 4z </math>. x + y = 0 for all x and y such that y = -x. Since <math> z = 2z = 4z </math> when z = 0, this system has an infinite number of solutions of the form (x, -x, 0) where x can be any real number (for example, <math> (3, -3, 0), (-0.5, 0.5, 0), </math> and <math> (\pi, -\pi, 0)</math> are solutions of this system of equations).
+* Infinite solutions: <math> x + y = z </math> and <math> x + y = 2z </math>. x + y = 0 for all x and y such that y = -x. Since <math> z = 2z </math> when z = 0, this system has an infinite number of solutions of the form (x, -x, 0) where x can be any real number (for example, <math> (3, -3, 0), (-0.5, 0.5, 0), </math> and <math> (\pi, -\pi, 0)</math> are solutions of this system of equations).
 ==Resources==

← Older revision		Revision as of 17:01, 15 September 2021
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	==Resources==		==Resources==
	* [https://tutorial.math.lamar.edu/classes/alg/systemsthreevrble.aspx Linear Systems with Three Variables], Paul's Online Notes (Lamar Math)		* [https://tutorial.math.lamar.edu/classes/alg/systemsthreevrble.aspx Linear Systems with Three Variables], Paul's Online Notes (Lamar Math)
		+	* [https://courses.lumenlearning.com/wmopen-collegealgebra/chapter/introduction-systems-of-linear-equations-three-variables/ Systems of Equations: Three Variables], Lumen Learning
		+	* [https://www.youtube.com/watch?v=CdpFu7t0dJ4 Using Elimination to Solve a 3-Variable System], patrickJMT
		+	* [https://www.youtube.com/watch?v=tGPSEXVYw_o Solving a System of Two Equations with Three Variables (Infinite Solutions)], patrickJMT