TI-83: Simple 4 equation 4 variable

We are looking for the values of the variables that make all four equations true. [Remember that these are linear equations in four variables. In the earlier example of two variables in two equations, we could associate the problem with lines on in a Cartesian plane where each equation represents a line. In the 3 equation 3 variable situation we could associate each equation with a plane in Cartesian space. For 4 equations and 4 variables we no longer have the luxury of a physical model. Nonetheless the observations of the earlier situations hold true. There are an infinite number of 4 values (one for each of x, y, z, and w) that make each equation true. In the problem given above, there is exactly one set of 4 values that make all four equations true. We need to find that set of 4 values, for x, y, z, and w, that solve all four equations.]

Before we actually start using the calculator, remember that the calculator will be using a general form for each of the equations, expecting the equations to have the variables is the same order. The earlier pages had much longer explanations of this. Here we will just point out that the calculator will use

ai,j	for the coefficient of the jth variable in the ith equation. Thus, a2,3 is the coefficient for the 3rd variable in the 2nd equation.
xj	for the jth variable. Thus, x4 is the fourth variable (in our case w).
bi	is the constant value in the ith equation. Thus b3 is the constant value in the third equation.

Now, onto the problem on the calculator.

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 was demonstrated in the 3 variable, 3 equation page. Second, you can use the Matrix Editor. That is the method presented here, in Figures 1 through 11. The other Figures on this page are used to demonstrate how the calculator produces the reduced row echelon form of the given matrix.

Figure 1	Press the key to openthe matrix menu, and then to shift the menu selection to EDIT. The calculator used here shows that two matrices have been defined previously, [A] and [C]. Matrix [A] is a 3x4 matrix (3 rows and 4 columns) and [C] is a 2x3 matrix. For our problem we will need to use a 4x5 matrix to hold the coefficients and the constants. We will use [C] for our work on this page. We select [C] by pressing the key. This shifts the calcualtor to Figure 2.
Figure 2	The blinking cursor is on the "2" in the 2x3 designation of the size of the matrix. We want a matrix with 4 rows. Therefore, we press keys to change the number of of rows. The cursor will jump to the "3" in what was 2x3, representing the number of columns. We will need 5 columns. Therefore, press the key to set the number of of columns to the desired 5. The result is shown in Figure 3.
Figure 3	Figure 3 shows the effect of the change made so far. In particular, the matrix now has four rows. The two new rows are filled with 0's. At the same time, the matrix still has 3 columns. Note that we have not pressed the ENTER key yet. The cursor is still in the column field of the matrix size at the top of the screen. Once we press the calculator will add the extra two columns to the right side of the matrix, as shown in Figure 4.
Figure 4	At this point the matrix has been resized. Now we need to enter the desired values into the matrix in the desired places. The cursor is at row 1 column 1 where the old value is 4. However, we want the calculator to hold the matrix 2 5 – 9 3 151 5 6 – 4 2 103 3 – 4 2 7 16 11 7 4 – 8 – 32
Figure 5	Therefore, press to change the value stored in row 1 column 1. The display changes to that shown in Figure 5. Note that we have not finished entering the new value. As such, the value in the actual matrix is still 4. We can press the key to accept the value.
Figure 6	Figure 6 shows the screen after we have entered for row 1 column 2 and to start the process to change the element at row 1 column 3. We can press to move to Figure 7.
Figure 7	Our value for row 1 column 3 has been accepted. The calculator has shifted to the element at row 1 column 4. The current value is 0.
Figure 8	The next keystrokes are . Press to accept the value 151 for row 1 column 5 and move to the next screen.
Figure 9	The calculator shifts to the element at row 2 column 1. the number in that cell is — 5, left over from the earlier matrix [C]. We continue to enter values for the rest of the cells of the matrix.
Figure 10	Figure 10 shows the screen after we have entered all of the values for our desired matrix. Note that the calculator is still at row 4 column 5. We can take this opportunity to review the elements of the matrix and to compare them to the desired matrix. In Figure 10 we can see only the final three columns of the matrix. After reviwing those twelvve values, we can press the key four times to move the highlight back to row 5 column 1, as shown in Figure 11.
Figure 11	Now we can compare the values in the leftmost three columns with the desired matrix. If we find an incorrect value then we can use the cursor keys to move the highlight to that element. Then we can enter a new value for that element. Once we are satisfied that the values are correct, we can get out of the matrix editor by pressing the key sequence . This will take us back to the main screen, shown, for this calcualtor, in Figure 12.
Figure 12	The calculator used in this example had a blank main screen. In order to try to solve our problem we want to create the command rref([C]). We can find the rref( command in the matrix menu.
Figure 13	Press to open the matrix menu, and to shift the highlight to the MATH menu. Then use the to move down the menu to find the rref( item.
Figure 14	Press to select that item and paste it onto the screen, as shown in Figure 14.
Figure 15	Return to the matrix menu with the key in order to select the name of the matrix, [C] via the key.
Figure 16	Then complete the command with . The result is shown in Figure 16. We read this display by converting it back to equations 1x + 0y + 0z + 0w = 3 0x + 1y + 0z + 0w = 5 0x + 0y + 1z + 0w = – 11 0x + 0y + 0z + 1w = 7 These give us x=3, y=5, z=– 11, and w=7.

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