TI-83: Complex 4 equation 3 variable situation, where there is a solution

The main page for solving systems of linear equations on the TI-83 and TI-83 Plus.
The previous example page covers a Complex 3 equation 2 variable situation, where two lines are identical and there is a solution.
The next example page covers a Complex 4 equation 3 variable situation, without a solution.

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 is 3x + 4y - z = 1
5x - 2y + 7z = 19
2x - y - z = -12
7x + 3y + 2z = 5 Similar to the previous examples, we have more equations than we have variables. In this example there are three variables, but we have 4 equations. Our use of the rref function has been in situations where the number of equations is equal to the number of variables. In a real sense, we have an extra equation here. If we only had three equations then we could use rref to find their point of intersection. Let us move ahead in that direction. At first we will restrict our analysis to the first three equations: 3x + 4y - z = 1
5x - 2y + 7z = 19
2x - y - z = -12

Figure 1	Here we are in the MATRIX menu and we have moved to highlight the EDIT tab. We will use matrix [A] for this problem. The current dimensions for [A], 2 rows and 3 columns, will need to be changed.
Figure 2	In Figure 2 the dimensions of [A] have been changed. Note the added row of values at the bottom of the matrix.
Figure 3	For FIgure 3 we have entered all of the coefficients and constants for the three equations that we are using. Note that we see the right side of the matrix in Figure 3.
Figure 4	We can use the key to move the display so that we can verify the values in the other column of the matrix.
Figure 5	We quit the matrix editor, recall the rref command, and execute it. The result, shown in Figure 5 gives us the transformed equations that indicate that the solution is x=^–2, y=3, and z=5. Knowing that we have a solution for the first three of our equations means that we can test those values on the fourth equation: 7x + 3y + 2z = 5 7^–2 + 33 + 25 ?= 5 ^–14 + 9 + 10 ?= 5 5 = 5 Therefore, the values x=^–2, y=3, and z=5* not only satisfy the first three equations they also satisfy the fourth. That makes those values be the solution of the system of four linear equations.
Figure 6	Of course, we could throw the fourth equation into the matrix along with two of the other equations and use rref to solve that system. That is what we have done in Figure 6, replacing the first equation with the values of the fourth.
Figure 7	Returning to the main screen, recalling and running the comand, produces the matrix shown in Figure 7. Again, we have the same solution, but this time for the fourth, second, and third equations.
Figure 8	Returning to the main screen and performing the command again gives us the same unique solution.