How Can I Use WolframAlpha To Get Reduced Row Echelon Form?

A matrix is entered between curly brackets in WolframAlpha (http://www.wolframalpha.com). Additionally, each row of the matrix is entered in curly brackets too.

You can also put a matrix in reduced row echelon form. We could put the augmented matrix

\displaystyle \left[ \begin{array}{*{35}{l}} 1 & 2 & -1 \\ 4 & 3 & 1 \\ \end{array} \right]

Use the text “row reduce” and then enter the matrix. The solution is x = 1 and y = -1.

Let’s try this with another system of linear equations

Convert this system into a 3 x 4 augmented matrix:

WolframAlpha understands several commands for putting an augmented matrix into reduced row echelon form. You can use the command rref { }or the command row reduce { }. The matrix goes inside the curly brackets. However, the matrix must be put in carefully. Each row needs to be typed in inside of curly brackets with the entries separated by a commas. In this case, you would type

on the command line in WolframAlpha.

After you press Enter, the reduced row echelon form is computed,

This indicates that the solution to the system is

x = 65,000, y = 45,000, z = 40,000.

How Do I Solve A System of Linear Equations By Graphing?

To solve a system of linear equations in two variables by graphing, you must first solve each equation for the dependent variable. Once this is done, we can use the equations to find an appropriate window for the graph. It is often useful to also solve the system algebraically…this helps us to establish the horizontal extent of the window.

Problem The annual number of cars produced t years after 2000 by a small car manufacturer is y=6.032t+34.543 thousand cars. A larger producer has annual production y=-10.564t+100.340 thousand cars. In what year will the annual production be equal? What will the production be then?

Solution Examine the two equations. The vertical intercepts are 34.543 and 100.340. From 34.543, the first line rises. From 100.340, the second line decreases. Based on this, we can deduce that a vertical window from 0 to 110 is appropriate. If we solve the system by substitution, we see that the point of intersection should be t = 3.965. This suggests a horizontal window of 0 to 5 or 0 to 10. In this window, we can find the point of intersection at (3.965, 58.460).

The value 3.965 corresponds to a time late in the year 2003. The number of cars produced at that time is 58.460 thousand cars or 58,460 cars. Be careful in rounding the t value. Although t = 3.965 rounds to 4 (the year 2004), this time is in the year 2003. Rounding to the nearest integer would put the point of intersection in the next year…a mistake when the problem asked in what year.

How Do I Solve A System Of Equations That Results In Two Equations And Three Variables?

When using Left to Right Elimination or Gauss’s Methods, a system of three equation in three variables often results in a unique solution. In other words, a single value for each of the three variables solves each of the three equations in the system. However when some systems are put into row echelon form, all of the constants and variables in one of the equations drop out. In a sense, one of the equations has disappeared and now you are left with a system of two equations in three variables.

Don’t confuse this with an inconsistent system. In an inconsistent system, the row operations lead to one of the equations being false…something like 0 = 6. This type of system has no solutions.

Let’s examine a system of three equations in three variables in which one of the equations drops out. Take the system

\displaystyle \begin{matrix} x&+2y&-4z &=5 \\ 2x&+3y&-z &=3 \\ 3x&+5y&-5z &=8 \\ \end{matrix}

If we use left to right elimination to solve this system, we get the equivalent system

\displaystyle \begin{matrix} &x&+2y&-4z &=5 \\ &&y&-7z &=7 \\ & & &0 &=0 \\ \end{matrix}

This new system contains only two equations since 0 = 0 does not describe a relationship between the variables. This new system has many solutions. We can show this by solving the second equation for y to give

\displaystyle y=7z+7

and substituting it into the first equation:

\displaystyle x+2\left( 7z+7 \right)-4z=5

Now solve this equation for x,

\displaystyle \begin{matrix} x+14z+14-4z=5 \\ x+10z+14=5 \\ x=-10z-9 \\ \end{matrix}

The solution to the system is

\displaystyle \begin{matrix} x&=&-10z-9 \\ y&=&7z+7 \\ \end{matrix}

This may also be written as an ordered triple \left( -10z-9,7z+7,z \right).  We have solved for x and y, but z is not specified. We can pick it to be anything. For instance, if  z = 0, then x = -0 and y = 7. This means that the ordered triple (-9, 7, 0) solves the original system of equations. If z = -1, then x = 1 and y = 0, So (1, 0, -1) also solves the system.

How Do You Solve a Linear System with WolframAlpha?

Many of you may already be familiar with using a graphing calculator to put a matrix in reduced row echelon form. Did you know that you can do the same thing with WolframAlpha?

To see how this is done, let’s start from the system of linear equations

wolframrref_03

Convert this system into a 3 x 4 augmented matrix:

wolframrref_04

WolframAlpha understands several commands for putting an augmented matrix into reduced row echelon form. You can use the command rref { }or the command row reduce { }. The matrix goes inside the curly brackets. However, the matrix must be put in carefully. Each row needs to be typed in inside of curly brackets with the entries separated by a commas. In this case, you would type

wolframrref_01

on the command line in WolframAlpha.

After you press Enter, the reduced row echelon form is computed,

wolframrref_02

This indicates that the solution to the system is

x = 65,000, y = 45,000, z = 40,000.