Tutorial: Using matrices to solve systems of equations

Go to Part A: Matrix of a linear system and row operations

Go to Part B: Solving a System by Gauss-Jordan

This tutorial: Part C: Unique solutions, infinitely many solutions,and inconsistent systems

(This topic is also in Section 3.2 in Finite Mathematics or Section 4.2 in Finite Mathematics and Applied Calculus)

Resources

Pivot and Gauss-Jordan tool

Excel pivot and Gauss-Jordan tool

Pivot program for graphing calculator (TI-82/83/84/85/86)

Chapter true/false quiz

#[Matrices in row-reduced echelon form][Matrices en forma escalonada reducida por filas]#

Good! You now know how to pivot and reduce a matrix in order to solve many systems of linear equations. Also, at the end of %%partBtut we briefly considered what it means for a matrix to "reduced" and we begin by reviewing this important concept:

First, recall that the leading entry of a row is the first non-zero entry in that row. Thus, the only rows that don't have leading entries rows of zeros.

#[Reduced row-echelon form][Rorma escalonada reducida por filas]#

A matrix is in row-reduced echelon form (or reduced for short) if:

The leading entry of each non-zero row is a 1.
The column of every leading entry is clear; that is, the other entries in its column are all zero. (Note: Columns that do not contain leading entries do not need to be clear.)
The rows are arranged so that the leading entries go from left to right as you go down the rows. Rows of zeros (if any) are at the bottom.

%%Examples

#[The following matrices are reduced:][Las siguientes matrices son redicidas:]#

#[The last of these, with $1$s down the diagonal and zeros everywhere else, is the $4 \times 4$ identity matrix $I$ (see the %%matrixAlgTut)][La última de estas, con $1$ en la diagonal y ceros en el resto del espacio, es la $4 \times 4$ matriz identidad $I$ (ver el %%matrixAlgTut)]#

%%Q: #[How do we reduce a matrix to row-reduced echelon form?][¿Cómo reducimos una matriz a la forma escalonada reducida por filas?]#
%%A: #[You already know from the %%partBtut how to reduce any matrix to row-reduced echelon form! (If you don\'t recall how to do that, go back to the that tutorial to review it.)][¡Ya sabes por el %%partBtut cómo reducir cualquier matriz a la forma escalonada reducida por filas! (Si no recuerdas cómo hacerlo, vuelve a ese tutorial para revisarlo).]#

#[In brief the procedure was the following:][En breve, el procedimiento fue lo siguiente:]#

#[In each row, use pivoting to clear the column of its leading entry.][En cada fila, utiliza la técnica de pivotar para despejar la columna de su entrada principal.]#
#[Convert all leading entries to 1s][Convierta todas las entradas principales a 1]#
#[Rearrange the rows so that the leading entries go from left to right, and rows of zeros are at the bottom. (As we noted in %%partAtut, it is traditional to allow only row swaps as "elementary" row operations, even though it is a mathematical fact that any rearrangement or the rows can be accomplshed by a sequence of row swaps. Thus, we permit arbitrary rearrangements of the rows, meaning that the row operations we permit are not really "elementary".) ][Reorganiza las filas de modo que las entradas principales vayan de izquierda a derecha y las filas de ceros queden en la parte inferior. (Como señalamos en la %%partAtut, es tradicional permitir solo intercambios de filas como operaciones de filas "elementales", aunque también es un hecho matemático que cualquier reordenamiento de filas se puede lograr mediante una secuencia de intercambios de fila. De esta forma, permitimos reordenamientos arbitrarios de las filas, lo que significa que las operaciones de filas que permitimos no son realmente "elementales".)]#

#[Obtaining the solution of a system from the reduced augmented matrix][Obtener la solución de un sistema a partir de la matriz aumentada reducida]#

%%Q: #[Ok, so now that we know how to reduce a matrix to reduced row echelon form, what does it do for us?][Bien, ahora que sabemos cómo reducir una matriz a la forma escalonada reducida, ¿qué hace por nosotros?]#
%%A: Let's look again at the some examples of a reduced matrix corresponding to systems of linear equations; some of these are examples of reduced matrices we saw near the end of %%partBtut:

#[Two unknowns in diagonal form: The reduced matrix][Dos incógnitas en la forma diagonal: La matriz reducida]#

represents the two equations

$x =9,\ \ y = -\dfrac{5}{2}$,

and therefore a unique solution: $x$ has to be equal to $9$, %%and $y$ has to be equal to $-\frac{5}{2}$.

#[We refer to this reduced matrix as being in diagonal form because its leading entries go down the leading diagonal (first entry in Row 2, second entry in Row 2) of the matrix.][Nos referimos a esta matriz reducida como si estuviera en forma diagonal porque sus entradas principales van hacia abajo por la diagonal principal (primera entrada en la fila 2, segunda entrada en la fila 2) de la matriz.]#

#[Three unknowns in diagonal form: The reduced matrix][Tres incógnitas en la forma diagonal: La matriz reducida]#

represents the three equations

$x = \dfrac{3}{2},\ \ y = \dfrac{1}{2},\ \ z = -2$,

and therefore a unique solution ($x$ has to be equal to $\frac{3}{2}$, $y$ has to be equal to $\frac{1}{2}$, %%and $z$ has to be equal to $-2$.)

#[This reduced matrix is also in diagonal form as its leading entries also go down the leading diagonal (first entry in Row 2, second entry in Row 2, third entry in Row 3) of the matrix.][Esta matriz reducida es también en la forma diagonal porque sus entradas principales van hacia abajo por la diagonal principal (primera entrada en la fila 2, segunda entrada en la fila 2, tercera entrada en la fila 3) de la matriz.]#

#[Two unknowns not in diagonal form: ][Dos incógnitas no en la forma diagonal: ]# On the other hand, the reduced matrix

is not in diagonal form. This matrix represents the two equations

$x - 2y = \dfrac{1}{2},\ \ 0 = 0$.

Ignoring the second equation (which tells us nothing interesting), we get, solving for the first unknown in the equation that remains,

$x = 2y + \dfrac{1}{2}$.

#[This equation tells us what $x$ needs to be (once we have a value for $y$), but not what $y$ needs to be, so $y$ can have any value whatsoever, and so the solution is as follows:][Estas ecuaciones nos dicen lo que $x$ debe ser (una vez que tenemos un valor para $y$), pero no lo que $y$ debe ser, por lo que $y$ puede tener cualquier valor, y la solución es la siguiente:]#

#[General solution][Solución general]#
#[Form of every possible solution][Forma de cada posible solución]#

#[A partiular solution][Una solución particular]#
#[Choose $y$ to be ][Elija $y$ con el valor ]# $\tfrac{3}{2}$

$x = 2y + \dfrac{1}{2}$
$y$ #[arbitrary][arbitraria]#

$x = 2\left(\color{darkred}{\dfrac{3}{2}}\right) + \dfrac{1}{2} = \dfrac{7}{2}$
$y = \color{darkred}{\dfrac{3}{2}}$

#[The general solution gives us a formula for every possible solution, whereas choosing different values for $y$ gives us different particular solutions.][La solución general nos da una fórmula para cada solución posible, mientras que elegir diferentes valores para $y$ nos da diferentes soluciones particulares.]# #[ Choosing any other value for $y$ will similarly result in another particular solution, so we can conclude that there are, in fact infinitely many solutions.][ Elegir cualquier otro valor para $y$ resultará de forma similar en otra solución particular, por lo que podemos concluir que, de hecho, existen infinitas soluciones.]# #[The set of all solutions is therefore][El conjunto de todas las soluciones es por tanto]#

\t $\text{Solution set}{}=\Big\{\Big(2y + \dfrac{1}{2},\ y \Big) \Big\mid y \text{ arbitrary} \Big\}$

#[As the variable $y$ can be chosen to have any value we like, we call $y$ a parameter and refer to the solution set as a one-parameter set of solutions.][Como la variable $y$ puede tener cualquier valor, llamamos a $y$ un parámetro y nos referimos al conjunto de soluciones como un conjunto de soluciones de un solo parámetro.]#

#[Three unknowns not in diagonal form: ][Tres incógnitas no en la forma diagonal: ]# Similarly, the reduced matrix

represents the three equations

$x - 3y = \dfrac{1}{2},\ \ z = \dfrac{5}{6},\ \ 0 = 0$.

Ignoring the third equation, we get, solving for the first unknown in each remaining equation,

$x = 3y + \dfrac{1}{2}, \ \ z = \dfrac{5}{6}$.

#[These equations tell us what $x$ and $z$ need to be (once we have a value for $y$), but not what $y$ needs to be, so $y$ can have any value whatsoever, and the solution is as follows:][Estas ecuaciones nos dicen lo que $x$ y $z$ deben ser (una vez que tenemos un valor para $y$), pero no lo que $y$ debe ser, por lo que $z$ puede tener cualquier valor, y la solución es la siguiente:]#

#[General solution][Solución general]#
#[Form of every possible solution][Forma de cada posible solución]#

#[A partiular solution][Una solución particular]#
#[Choose $y$ to be ][Elija $y$ con el valor ]# $\tfrac{3}{2}$

$x = 3y + \dfrac{1}{2}$
$y$ #[arbitrary][arbitraria]#
$z = \dfrac{5}{6}$

$x = 3\left(\color{darkred}{\dfrac{3}{2}}\right) + \dfrac{1}{2} = 5$
$y = \color{darkred}{\dfrac{3}{2}}$
$z = \dfrac{5}{6}$

#[As the variable $y$ can be chosen to have any value we like, we obtain a one-parameter set of solutions:][Como la variable $y$ puede tener cualquier valor, obtenemos un conjunto de soluciones como un conjunto de soluciones de un solo parámetro:]#

\t $\text{Solution set}{}=\Big\{\Big(3y + \dfrac{1}{2},\ y,\ \dfrac{5}{6}\Big) \Big\mid y \text{ arbitrary} \Big\}$

#[Four unknowns not in diagonal form][Cuatro incógnitas no en la forma diagonal]# #[The reduced matrix][La matriz reducida]#

in the unknowns $x, y, z, t$ represents the three equations $y - 3z = 4,\ t = -\dfrac{5}{6},\ 0 = 0$. Solving for the first unknown in each equation,

$y = 3z + 4, \quad t = -\dfrac{5}{6}$.

#[These equations tell us what $y$ and $t$ need to be (once we have values for $x$ and $z$), but not what $x$ and $z$ need to be, so the solution is][Como no se menciona $x$, puede ser arbitrario y obtenemos la solución.]#

#[General solution][Solución general]#

#[A partiular solution][Una solución particular]#

$x$ #[arbitrary][arbitraria]#
$y = 3z+4$
$z$ #[arbitrary][arbitraria]#
$t = -\dfrac{5}{6}$

$x = \color{darkred}{5}$
$y = 3(\color{darkred}{5})+4 = 19$
$z = \color{green}{\dfrac{1}{7}}$
$t = -\dfrac{5}{6}$

Choosing different values for x and z gives us different particular solutions.

\t $\text{Solution set}{}=\Big\{\Big(x,\ 3z+4,\ z,\ -\dfrac{5}{6} \Big) \Big\mid x, z \text{ arbitrary} \Big\}$

#[To summarize:][Para resumir:]#

Summary: How to obtain the general solution of a system of linear equations:

Reduce the associated augmented matrix to row-reduces echelon form.
#[Translate each row into an equation, and solve for the unknown that occurs first (and corresponds to the leading entry for that row). ][Traduzca cada fila en una ecuación, y resuelve la incógnita que ocurre primero (y corresponde a la entrada destacada de aquella fila). ]#
All unknowns not solved for are arbitrary.

#[Spotting an inconsistent system][Reconocer un sistema inconsistente]#

#[Up to this point we never mentioned what happens if the leading entry of some row in your reduced augmented matrix is in the last (answer) column, as in, say][Hasta este punto nunca consideramos qué sucedería si la entrada principal de cualquiera fila en la matriz reducida es en la última (respuesta) columna, como, por ejemplo en]#

#[The next-to-last row translates into the equation][La penúltima fila se traduce en la ecuación]#

$0x + 0y + 0z + 0s = 1$,

#[or just][o simplemente]#

$0 = 1$,

#[which is a false statement. If we remember that the rows of the augmented matrix (reduced or not) tell us what equations any solution of the system needs to satisfy, we must conclude that there can be no solution, as any solution would need to satisfy a false equation! In other words, i ][que es una declaración falsa. Si recordamos que las filas de la matriz aumentada (reducida o no) nos indican qué ecuaciones debe satisfacer cualquier solución del sistema, debemos concluir que no puede haber solución, ya que cualquier solución tendría que satisfacer una ecuación falsa. En otras palabras, el sistema de ecuaciones original es inconsistente y no tiene soluciones.]#

#[The moral of the story: If, during the process of row-reduction, you get any row of the form][La moraleja de la historia: si durante el proceso de reducción de filas obtienes cualquier fila de la forma]#

$[0\ \ 0\ \ 0\ \ 0\ \ ...\ \ 0\ \ k]$

#[where $k$ is nonzero, then you can stop working because the original system is inconsistent.][donde $k$ es distinto de cero, entonces puedes dejar de trabajar porque el sistema original es inconsistente.]#

#[You now have some options: ][Ahora tienes algunas opciones: ]#

#[Try some of the online exercises on systems of linear equations][Pruseba algunos de los ejercicios enlínea sobre sistemas de ecuaciones lineales]#.
#[Alternatively, try some of the exercises in ][Alternativamente, prueba los ejercicios en ]# Section 3.2 in Finite Mathematics or Section 4.2 in Finite Mathematics and Applied Calculus
Move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.