TI-83: Simple 3 equation 3 variable, parallel planes

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 one previously defined matrix, namely, [A]. For this problem we will re-use matrix [A]. Because [A] is already highlighted, press to select that matrix.
Figure 2	The existing [A] has two rows and three columns. We want the matrix to hold three rows and four columns. Before we place new values into the matrix, the editor allows us to change the size of the matrix.
Figure 3	In order to size and fill the matrix, press to specify 3 rows and to specify 4 columns. Then fill the matrix with the desired values: . The result should be identical to that shown in Figure 3. The final three columns of the matrix are visible in Figure 3. We can verify the values in those three columns.
Figure 4	We use the key to move the highlight back to the first column so that we can verify the values in that column.
Figure 5	Leave the matrix editor by pressing . This returns us to the main screen. For the calculator used here, the main screen shows us the results from the computation of the previous example. In particular, the previous command, rref([A]), is exactly the command that we want to give again.
Figure 6	Recall the previous command by pressing the keys . This produces the image shown in Figure 6. We could tell the calculator to perform the command, but we will take this opportunity to add the "change it to a fraction" command to the end of the current command. We will do this so that the eventual resulting matrix will display fractions instead of decimal approximation.
Figure 7	Use the key to open the MATH menu which is shown in Figure 7. Then, we can select the "change it to a fraction" command by pressing .
Figure 8	The top three lines in Figure 8 are left over from an earlier example. THe fourth line in Figure 8 contains the command that we recalled and modified. To execute this new command we press . The result is the matrix shown in lines 5 through 7 of the display. Again, as was the case in the previous example, this is is not the matrix that we expect to find. We expect to have 1's down the main diagonal with 0's above and below that diagonal. Instead, the third row, third column position holds a 0, and both the third column first and second rows do not hold 0's. If we translate the matrix back into a system of equations we get 1x + 0y – (17/76)z = 0 0x + 1y – (-99/76)z = 0 0x + 0y + 0z = 1 Although the first and second of these equations are possible, the third can not have any solution. No matter what values we choose for x, y, and z, if we multiply them by 0 the products will be 0, and the sum of the three 0's will be 0, not 1. Therefore, this matrix indicates that there is no solution to the original system of linear equations.

If we return to the original problem,

The main page for solving systems of linear equations on the TI-83 and TI-83 Plus.
The previous example page covers a Simple 2 equation 2 variable situation but with parallel lines.
The next example page covers a Simple 3 equation 3 variable situation, but with no solution since the planes do not intersect, but are not parallel.