### TI-83: Simple 2 equation 2 variable

The main page for solving systems of linear equations on the TI-83 and TI-83 Plus.
The next example page covers a Simple 3 equation 3 variable situation.

WARNING: The TI-83 and TI-83 Plus are almost identical in terms of the material presented here. The major difference is the labels that are on certain keys. The TI-83 has a key, whereas on the TI-83 Plus requires 2 keys to achieve the same result, namely, the key. The text below will be done from the perspective of the TI-83. That is, all reference to the MATRIX key will be demonstrated via the key. If the user has a TI-83 Plus then the key strokes should be . To save some space, and to ignore this difference, the numeric keys (the gray ones) have been changed here to only show the key face, as in . In addition, the key will be shown as and the key will be shown as , again to save space.

3x + 4y = 14
5x - 7y = -45
We are looking for the values of the variables that make both equations true. [Remember that these are linear equations in two variables. On the Cartesian plane, each equation represents a straight line. There are an infinite number of points on each line, and an infinite number of solutions to the individual equations. However, these particular lines cross at a point. That point of intersection is a solution to both equations. It is the only point that solves both equations. We need to find that point, that x and y value that solve both equations.]

Before we start using the calculator, note that these equations are given in standard form. That is, they appear as Ax + By = C, where A, B, and C are numeric values. For two equations, in two unknowns (variables) x and y, we could write the equations in a general standard form as:

Ax + By = C
Dx + Ey = F
Once the equations are in standard form we might notice that much of what we have written is extraneous. The whole problem is really characterized by the six values, A, B, C, D, E, and F, arranged in a rectangular array:
 A B C D E F
Looking forward to more complex problems, we can see that if we have more variables and more equations then we will run out of letters to represent the coefficients and the variables. There is a more general form for the equations, namely:
a1,1 x1 + a1,2 x2 = b1
a2,1 x1 + a2,2 x2 = b2
This new form can be more confusing in simple cases, such as two variables and two equations, but it is more useful in complex situations, such as 7 variables and 7 equations. The key to understanding this general form is that the numbers after the a's indicate first the equation (row) and second the variable to which the numeric coefficient is attached. Thus, a2,1 indicates that this is the number in the second equation attached to the first variable. The variables are numbered by the subscript of x, so x1 represents the first variable and x2 represents the second variable. The constants on the right side of the equations are numbered by the equation (row) in which they appear. Therefore, b1 is the constant for the first equation and b2 is the constant for the second equation. Using this form, our simple two variable, two equation general problem can be represented completely by the rectangular array
 a1,1 a1,2 b1 a2,1 a2,2 b2

The problem that we were given was:

3x + 4y = 14
5x - 7y = -45
and we remember that the more general standard form is:
a1,1 x1 + a1,2 x2 = b1
a2,1 x1 + a2,2 x2 = b2
so, for this problem
 a1,1 is 3 x1 is x a1,2 is 4 x2 is y b1 is 14 a2,1 is 5 x1 is x a2,2 is -7 x2 is y b2 is -45
The calculator uses a matrix to hold a rectangular array of numbers. There are two ways to enter a matrix into the calculator. First, you can use the [ and ] characters to type the matrix directly into the calcualtor. That method is demonstrated in Figure 1 below. Second, the calculator has a Matrix Editor that you can use to enter values into a matrix. That method is demonstrated in Figures 12 through 22 below. The other Figures are used to demonstrate how the calculator produces the reduced row echelon form of the given matrix.

With all of that out of the way, we are finally ready to start using the calculator. The steps shown before assume that the calculator is turned on, that we are not in any menu, and that the screen is clear.
 Figure 1 In Figure 1 we enter the desired matrix directly from the calculator keyboard. To do this we use the to generate a [ to signal the start of the matrix. A second produces a [ to signal the start of a row. Then we give the coeeficients and the constant of the row, separated by commas. The keys for our example are . Then we use and to produce ][ to signal the end of the first row and the start of the second row, respectively. The values in the second row are . Then, we need to end the second row with a ], and end the matrix with a ]. We do this with the keys and . Finally, we want to store this matrix on the calculator. We start to store it by pressing the key. Figure 2 On a TI-83 or a TI-83 Plus we need to store a matrix into one of the 10 predefined matrices, [A], [B], [C], [D], [E], [F], [G], [H], [I], or [J]. To do this we need to find the name of the desired matrix. The names are available in the MATRIX menu. On the TI-83 we press the key to open the MATRIX menu. (As noted at the start of this page, on the TI-83 Plus we use the sequence keys to accomplish the same thing.) Figure 2 shows the MATRIX menu. The calculator used here already has a matrix stored in [A], in fact it has a matrix with 10 rows and 7 columns stored in [A]. We will store our matrix in place of that one. To do this, we can press the key to select the [A] matrix name. Figure 3 Figure 3 shows the result of our work in Figure 2. The name [A] has been appended to the store command in our screen. The action has not been performed at this point. We have merely created the command that defines the matrix and assigns it to [A]. To perform that command we press the key. Figure 4 Having performed the command to leave Figure 3 the calculator stores the matrix into [A] and then displays it on the screen as This is the form that the TI-83 uses to display a matrix. Figure 5 Our next step is to find the reduced row echelon form of this matrix. We could do this ourselves, using what are known as elementary row operations. However, the TI-83 has a function that will try to produce the reduced row echelon form of the original matrix. That command is rref(). We can find this command in the MATRIX menu. Therefore, we open that menu again, via the key. The result is shown in Figure 5. Note that [A] shows up in the list of matrix names as having 2 rows and 3 columns. Figure 6 Our rref() function is not in the list of names. Rather, we will find it under the MATH option. Therefore, we press the key to move the highlight at the top onto the MATH option, as shown in Figure 6. Now we have a new list of options on the screen. Unfortunately, rref() is not one of them. Therefore, use the key to move the selection highlight down the screen until it the screen appears as shown in Figure 7. Figure 7 In Figure 7 we have located the rref( function. Press to select that option and paste rref( onto the main screen. Figure 8 The full command that we want to construct is rref([A]). In Figure 8 we have the start of this command. Now we need to append the name of our matrix. To do this we will need to open the MATRIX menu again. Press to move to Figure 9. Figure 9 Once again, the list of matrix names is presented. [A] is already highlighted. Therefore, we can press to select [A] and move to Figure 10. Figure 10 Once [A] has been pasted into the command, we press to cpomplete the comamnd, as shown in Figure 10. All that remains is to press to get the calculator to perform the function and to produce the reduced row echelon form of the original matrix [A]. Figure 11 Finally, we have the reduced row echolon form of the original matrix. This new form appears as remember that this is merely a shorthand version of the two linear equations 1X + 0Y = – 2 0X + 1Y = 5 and those equations tell us that X=– 2 while Y=5. That point, the point (– 2,5), is the solution to the original problem.

The first 11 Figures on this page demonstated a calculator solution to the given problem. This solution required us to create a matrix to hold the coeeficients and the constants of the two equations. We created that matrix by typing it into the calculator, back in Figure 1, and then storing it into a matrix variable in Figure 2 and 3. There is an alternate method to enter or change a matrix, namely, we can use the matrix editor. Figures 12 through 22 demonstrate using the matrix editor to enter a new matrix into the matrix variable [C]. The new matrix will represent the two equations

4x – 3y = 27
– 5x + 7y = – 50

 Figure 12 We open the matrix menu by pressing the key. Figure 13 Next we press to move the highlight to the EDIT menu item. In Figure 13 we note that the highlight has moved, and that [A] is selected. We want to select [C]. Figure 14 The key sequence moves the selection highlight to our desired [C]. Press to move to Figure 15. Figure 15 Here the calculator has opened the matrix [C] in the matrix editor. At the moment this is a 1 row and 1 column matrix, and the one element in it is 0. However, the flashing cursor is on the number of rows in the matrix. Figure 16 In Figure 16 we have changed the number of rows in the matrix to 2 by typing and we have started to change the number of columns to 3 by typing . The screen already shows the two rows, but it has not changed to chow 3 columns. We will need to press again to do this and move to Figure 17. Figure 17 Both rows and all three columns are visible in the display. The cursor is on item 1,1 which currently holds 1. We want to have that item be 4. Therefore, we press the key to move to Figure 18. Figure 18 Note that the new value is being formed at the bottom of the screen, but that it has not yet been assigned to the element of the matrix. To signal the end of the number and to have the value placed into the matrix, we press , and move on to Figure 19. Figure 19 Our 4 is in the correct place. THe highlight is on the element in the first row, second column. At the bottom of the screen we see that the current value is 0. We want it to be – 3. We press to enter that value at the bottom of the screen, shown in Figure 20. Figure 20 The – 3 value is shown at the bottom. Press top accept that value and move on to the next matrix element. Thereafter, we continue to enter array elements until we have the values shown in Figure 21. Figure 21 Figure 21 reflects the steps taken to enter the matrix, up to but not inlcuding the entry of the final value, – 50, into the element at row 2 column 3. Once we press to complete the entry of this value, our matrix will be complete. Figure 22 Here we have the completed matrix. Note that the highlight is still on the element in the lst row, last column. We could use the cursor keys to move around within the matrix. Right now we will verify that all of the values in the matrix are correct. Figure 23 Once we are satisfied that hte matrix holds the correct values, we press the sequence to exit the matrix editor. This retuns us to the main screen, which is unchanged since we left after Figure 11. We need to formulate the rref([C]) command. To do this we press to open the matrix menu, shown in Figure 24. Figure 24 We use the key to move the highlight to the MATH menu option. Figure 25 Use over and over until we have the selector highlight on the rref( function. Press to paste that selection onto the main screen, shown in Figure 26. Figure 26 The command has been started. We need to generate the name of our matrix, C, so we press to open the matrix menu again, as shown in Figure 27. Figure 27 We can select the [C] item by pressing the key. That will paste [C] onto the main screen. Figure 28 We complete the command with and the calculator responds as in Figure 28. The matrix that we put into [C] represented the two equations 4x – 3y = 27 – 5x + 7y = – 50 The reduced row echelon form represents the two equations 1x + 0y = 3 0x + 1y = – 5 That is, x=3 and y=– 5.

The main page for solving systems of linear equations on the TI-83 and TI-83 Plus.
The next example page covers a Simple 3 equation 3 variable situation.