TI-83: Simple 3 equation 3 variable, with integer coefficients, fractional answers

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 fractional coefficients, but integer answers.
The next example page covers a Simple 4 equation 4 variable situation, with no 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 84x + 147y - 105z = -22
42x - 147y + 168z = 94
126x - 84y + 42z = 97

Figure 1 Open the matrix menu and move to the editor. We will use [A] for this problem.

Figure 2 [A] already has the correct dimensions for the current problem, namely 3 rows and 4 columns.

Figure 3 In Figure 3 we have moved to enter a new value into row 1, column 1 of the matrix. The new values, 84 is the coefficient of the first variable, x, in the first equation. In Figure 3 we have entered the value but have not yet hit the ENTER key to really submit the new value.
Press the key to move to Figure 4.

Figure 4 In Figure 4 we can see that the first coefficient has been entered into the matrix. Figure 4 has us entering the next coefficient, 147 at the bottom of the screen. Press the key to move to accept that value and move forward with the problem.

Figure 5 To get here we have entered all of the coefficients and constants from the original problem, for all three equations. The calculator display, in Figure 5, only shows the final three columns. This is a good time to verify the accuracy of our entry. It is often helpful to read down the columns to check the values against corresponding values in the original equation.

Figure 6 For Figure 6 we have merely used the key to move the display so that we can verify the values in the first column, which was not shown in Figure 5. Once the values have been verified, press to exit the matrix editor.

Figure 7 Having left the matrix editor we are back at the main screen, filled with information from previous work. The important information on this screen is that our next desired command, rref([A]) is also the last command that we used. Therefore, we can recall that command via the sequence.

Figure 8 The command has been recalled, we press to have the calculator perform the command.

Figure 9 The result, shown in Figure 9, gives decimal values for the values in the last column. We want to see these as fractional values. Press the key to open the MATH menu.

Figure 10 In the MATH menu, the first item, , is the one we want. Press the key to paste that onto the main screen.

Figure 11 Because there was nothing before the pasted , the calculator assumes that we want to convert the previous answer to fractional form. Indeed, we do, so press the key to perform the command.

Figure 12 Figure 12 shows the fractional form of the display. We could reconstruct the transformed equations as 1x + 0y + 0z = 1/2
0x + 1y + 0z = ^–1/3
0x + 0y + 1z = 1/7 and read from this the solution, x=1/2, y=1/3, and z=1/7.

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, with fractional coefficients, but integer answers.
The next example page covers a Simple 4 equation 4 variable situation, with no solution.

Figure 1	Open the matrix menu and move to the editor. We will use [A] for this problem.
Figure 2	[A] already has the correct dimensions for the current problem, namely 3 rows and 4 columns.
Figure 3	In Figure 3 we have moved to enter a new value into row 1, column 1 of the matrix. The new values, 84 is the coefficient of the first variable, x, in the first equation. In Figure 3 we have entered the value but have not yet hit the ENTER key to really submit the new value. Press the key to move to Figure 4.
Figure 4	In Figure 4 we can see that the first coefficient has been entered into the matrix. Figure 4 has us entering the next coefficient, 147 at the bottom of the screen. Press the key to move to accept that value and move forward with the problem.
Figure 5	To get here we have entered all of the coefficients and constants from the original problem, for all three equations. The calculator display, in Figure 5, only shows the final three columns. This is a good time to verify the accuracy of our entry. It is often helpful to read down the columns to check the values against corresponding values in the original equation.
Figure 6	For Figure 6 we have merely used the key to move the display so that we can verify the values in the first column, which was not shown in Figure 5. Once the values have been verified, press to exit the matrix editor.
Figure 7	Having left the matrix editor we are back at the main screen, filled with information from previous work. The important information on this screen is that our next desired command, rref([A]) is also the last command that we used. Therefore, we can recall that command via the sequence.
Figure 8	The command has been recalled, we press to have the calculator perform the command.
Figure 9	The result, shown in Figure 9, gives decimal values for the values in the last column. We want to see these as fractional values. Press the key to open the MATH menu.
Figure 10	In the MATH menu, the first item, , is the one we want. Press the key to paste that onto the main screen.
Figure 11	Because there was nothing before the pasted , the calculator assumes that we want to convert the previous answer to fractional form. Indeed, we do, so press the key to perform the command.
Figure 12	Figure 12 shows the fractional form of the display. We could reconstruct the transformed equations as 1x + 0y + 0z = 1/2 0x + 1y + 0z = ^–1/3 0x + 0y + 1z = 1/7 and read from this the solution, x=1/2, y=1/3, and z=1/7.