TI-83: Simple 3 equation 3 variable situation, with fractional coefficients, but integer 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, but with no solution, non-parallel.
The next example page covers a Simple 3 equation 3 variable situation, but with fractional answers.

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 (2/3)x - (4/5)y - (1/2)z = -133/30
(5/3)x - (9/2)y + (8/5)z = -257/6
(6/7)x - (5/8)y + (4/3)z = -2143/168

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. In Figure 2 we have moved to enter a new value into row 1, column 1 of the matrix. The new values, 2/3 is the coefficient of the first variable, x, in the first equation. In Figure 2 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 3.

Figure 3 In Figure 3 we can see that the first coefficient has been entered into the matrix. We might note that the display of that value has just a few digits. This is for display only. Were we to return to that first row, first column spot we would see, at the bottom of the screen, the full precision of the value. A similar situation is discussed below in Figure 4. Meanwhile, Figure 3 has us entering the next coefficient, ^–4/5 at the bottom of the screen. Press the key to move to accept that value and move forward with the problem.

Figure 4 To get here we have entered all of the coefficients and constants from the original problem, for all three equations. The last value that we entered was ^–2143/168. It went into row 3, column 4, the currently highlighted cell in the matrix. Note that the matrix display of this is now ^–12.76, but this is just a short version of the value. The longer version is given at the bottom of the screen as ^–12.7559523... but even that has more digits on the calculator than are displayed three. [You could press to move the cursor to the number at the bottom of the screen and then use the key to move over to display more digits if need be.]

Figure 5 For Figure 5 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 4. Note that all the values in the display are in decimal format. At this point, having entered all the values from the problem into the matrix, we can quit the matrix editor by pressing .

Figure 6 Once out of the matrix editor, we return to the matrix menu so that we can select the name of the matrix for display. To leave this screen, since [A] is the current selection, press the key.

Figure 7 The action of Figure 6 just pasted the nameof the matrix onto our main screen. We would like to see the matrix, but in fractional form. Therefore, press the key to open the MATH menu as shown in Figure 8.

Figure 8 We want the first item in the menu, . Therefore, just press the key to paste that command onto the main screen, as shown in Figure 9.

Figure 9 Here our command is set up. Press to tell the calculator to perform the command.

Figure 10 The calculator has displayed the matrix, or at least the left side of the matrix, using fractional notation for the entries. As usual, we can use this display to verify the entry of the values of the current problem.

Figure 11 We can move to the right side of the matrix using the key.

Figure 12 We can recall the command rref([A]) and then use the key to perform the command, resulting in Figure 12. From this we could reconstruct the transformed equations as 1x + 0y + 0z = ^–2
0x + 1y + 0z = 7
0x + 0y + 1z = ^–5 or x=^–2, y=7, and z=^–5
or we could give the solution as a point (^–2,7,^–5).

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. In Figure 2 we have moved to enter a new value into row 1, column 1 of the matrix. The new values, 2/3 is the coefficient of the first variable, x, in the first equation. In Figure 2 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 3.
Figure 3	In Figure 3 we can see that the first coefficient has been entered into the matrix. We might note that the display of that value has just a few digits. This is for display only. Were we to return to that first row, first column spot we would see, at the bottom of the screen, the full precision of the value. A similar situation is discussed below in Figure 4. Meanwhile, Figure 3 has us entering the next coefficient, ^–4/5 at the bottom of the screen. Press the key to move to accept that value and move forward with the problem.
Figure 4	To get here we have entered all of the coefficients and constants from the original problem, for all three equations. The last value that we entered was ^–2143/168. It went into row 3, column 4, the currently highlighted cell in the matrix. Note that the matrix display of this is now ^–12.76, but this is just a short version of the value. The longer version is given at the bottom of the screen as ^–12.7559523... but even that has more digits on the calculator than are displayed three. [You could press to move the cursor to the number at the bottom of the screen and then use the key to move over to display more digits if need be.]
Figure 5	For Figure 5 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 4. Note that all the values in the display are in decimal format. At this point, having entered all the values from the problem into the matrix, we can quit the matrix editor by pressing .
Figure 6	Once out of the matrix editor, we return to the matrix menu so that we can select the name of the matrix for display. To leave this screen, since [A] is the current selection, press the key.
Figure 7	The action of Figure 6 just pasted the nameof the matrix onto our main screen. We would like to see the matrix, but in fractional form. Therefore, press the key to open the MATH menu as shown in Figure 8.
Figure 8	We want the first item in the menu, . Therefore, just press the key to paste that command onto the main screen, as shown in Figure 9.
Figure 9	Here our command is set up. Press to tell the calculator to perform the command.
Figure 10	The calculator has displayed the matrix, or at least the left side of the matrix, using fractional notation for the entries. As usual, we can use this display to verify the entry of the values of the current problem.
Figure 11	We can move to the right side of the matrix using the key.
Figure 12	We can recall the command rref([A]) and then use the key to perform the command, resulting in Figure 12. From this we could reconstruct the transformed equations as 1x + 0y + 0z = ^–2 0x + 1y + 0z = 7 0x + 0y + 1z = ^–5 or x=^–2, y=7, and z=^–5 or we could give the solution as a point (^–2,7,^–5).