TI-83: Simple 4 equation 4 variable with missing and other variables

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, with missing variables.
The next example page covers a Simple 2 equation 2 variable situation, where the lines are parallel.

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 in some places to show the key face, as in

. In addition, the

key may be shown as

and the

key may be shown as

, again to save space.

The problem we will use on this page has both non-standard equations and missing variables. These topics have been covered in earlier pages. Therefore, we will move through the problem fairly quickly. The original statement of the problem is 8r + 37 = 3n + 5s
6m + 7s = 16
2n - 3r = 60 - 9m
2r - 6s = 38 -8m

Our first step is to identify the variables. In this case, the variables are r, n, s and m. Second we need to decide on the order of the varaibles, and the traditional m, then n, then r, then s seems reasonable. Third, we can rewrite the equations, with the variables in order, and not leaving any out. We will add terms with the coefficient set to 0 to make up for any missing terms. The rewrite of the problem produces

0m - 3n + 8r - 5s = -37
6m + 0n + 0r +7s = 16
9m + 2n - 3r + 0s = 60
8m + 0n + 2r - 6s = 38

By adding the extra terms (having 0 as the coefficient) we have not changed any equation but we have changed the form of the equation so that it fits the same pattern that we have been using in the earlier pages. We need to generate the matrix of the coefficients and the constants in the system of equations above. In particular, we need the matrix

0	^–3	8	^–5	^–37
6	30	0	7	16
9	2	^–3	0	60
8	0	2	^–6	38

We are ready to use the calculator.

Figure 1 Open the matrix menu with the key and then shift the highlight to the right with the keys. The calculator used here has two previously defined matrices, [A] and [C]. For this problem we will re-use matrix [A]. Because [A] is already highlighted, press to select that matrix.

Figure 2 Recall from Figure 1 that the matrix [A] had existed as a 3 row and 4 column matrix. We need a 4 row and 5 column matrix. For the calculator used here, we press to produce Figure 2. Note that the calculator has added the fourth row and that all of the valeus in that row have been set to 0.
Now we need to enter each of the values from the matrix given above.

Figure 3 Figure 3 shows the display after all of the valeus have been entered. We can compare the valeus on the display to the values in our matrix. [Note that our comparison is not very accurate at this point. There is an error here; an incorrect value has been entered. We will catch and fix the error later.]

Figure 4 Use the key to shift the display to show the first columns of the matrix. Now we can verify the contents of those columns.

Figure 5 Leave the matrix editor by pressing to return to the main screen, shown in Figure 5. This screen shows the results that were left on the calculator from a previous example. Fortunately, teh previous command is exactly the one that we want to issue again, but this time the command will use our revised matrix [A].

Figure 6 Press leave the matrix editor by pressing to recall the previous command.

Figure 7 Press to execute the command. The result is shown in Figure 7. The values for the variables appear in the final column. It is possible that these are the correct valeus, but we do not expect to get fractional answers. All of the previous examples have produced nice integer answers. Perhaps there was an error in our original matrix. Fortunately, the orginal matrix is still there in [A].

Figure 8 Let us examine the matrix [A]. To do this we return to the matrix menu via the key, and we select the name of our matrix, [A], by pressing the key. This will paste the name of the matrix onto the main screen, as shown in Figure 8. Then press the to display the va;ue of the matrix.
A comparison between the matrix shown in Figure 8 and the matrix

0 ^–3 8 ^–5 ^–37

6 30 0 7 16

9 2 ^–3 0 60

8 0 2 ^–6 38

reveals the error in [A] in the item at row 1 column 5. The value there should be ^–37.

Figure 9 We will correct the error by using the matrix editor. Press to open the matrix menu, to select the matrix editor, to edit [A], and then use the key to move the highlight to the item at row 1 coumn 5.

Figure 10 Figure 10 shows the result of pressing . Note that the new value is shown at the bottom of the screen. The new value will not be placed into the matrix until we press the key, which will move us to Figure 11.

Figure 11 The change has been made, but we cannot see it because it is now off of the screen. We can press the key a few times to shift the display to show the final column. This was done to produce Figure 12.

Figure 12 From Figure 12 we can verify that we have corrected the error.

Figure 13 Leave the matrix editor by pressing to return to the main screen, then press to recall the [A] command, and press to recall the rref[A] command. The result is shown in Figure 13.
Finally, press press to perform the rref[A] command.

Figure 14 The reduced row echelon form of the matrix is shown in Figure 14. From this matrix, we read the solution to the system of four linear equations in four variables. In particular, for the variables m, n, r, and s, we have m=5, n=^–3, r=^–7, and s=^–2