TI-83: Simple 3 equation 3 variable, missing values

Our first step is to identify the variables. In this case, the variables are x, y, and z, although the first equation does not have a z and the third equation is missing an x term. Second we need to decide on the order of the varaibles, and the traditional x, then y, then z seems reasonable. Third, we can rewrite the equations, with the variables in order, and not leaving any out. We will add a z term to the first, but with a coefficient of 0. We will add an x term to the third equation, again with a coefficent of 0. The rewrite of the problem produces

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 are ready to use the calculator.

Figure 1	To get the calculator to Figure 1, press to open the MATRIX menu, and to shift the display to the EDIT submenu. The calcualtor used here has two previously defined matrices, [A] with 7 rows and 8 columns, and [C] with 4 rows and 5 columns. We need to have a 3 row and 4 column matrix. Matrix [A] is highlighted. Press to move to Figure 2 where we can change the size of [A] and then change its contents.
Figure 2	Figure 2 shows the status of [A] before we make any changes. We change the size of the matrix by pressing to produce the 3 rows and 4 columns shown in Figure 3.
Figure 3	The size of [A] has changed, now we need to change the contents of the matrix to reflect the three equations that we have. 4x + 7y + 0z= 48 3x - 8y + 4z = -50 0x - 5y + 11z = 15 For such equations, we want then matrix to hold   4     7     0     48   3 – 8 4 – 50 0 – 5 11 15
Figure 4	Figure 4 shows the calculator display after all of the new values have been entered. Note that we can seen the final 3 columns of entries.
Figure 5	Use the key to move the highlight to the left until it is at the first row. Now the display shows the rest of the matrix. Once we have confirmed that the matrix holds thedesired values, we can leave the matrix editor by pressing the key sequence. That will return us to the main screen.
Figure 6	The calculator used here had been used to solve the problem presented on the previous page. Therefore, the main calculator page holds the solution matrix from that page. We need to construct the rref([A]) command on this main calculator page.
Figure 7	To construct the rref[A]) command we use the key to open the MATRIX menu, and then the key to move to the MATH sub-menu, shown in Figure 8.
Figure 8	Once on the MATH sub-menu, use the key to move the highlight down until it points to the rref( command. Then, press to paste that command onto the main calculator page.
Figure 9	We complete the command in Figure 9 by pressing to return to the MATRIX menu. The NAMES sub-menu will be displayed and the [A] matrix will be highlighted. Press to paste the highlighted value, [A], onto the main page. Add the final parentheis to the command with the key.
Figure 10	Press to have the calculator perform the command. The result is shown in Figure 10. This is a matrix representation for the equations 1x + 0y + 0z= – 2 0x + 1y + 0z = 8 0x + 0y + 1z = 5 Which means that the solution is x=– 2, y=8, and z=5.

The main page for solving systems of linear equations on the TI-83 and TI-83 Plus.
The previous example page covers a Simple 7 equation 7 variable situation.
The next example page covers a Simple 4 equation 4 variable situation, but with both missing terms and other variables.