TI-83: Simple 4 equation 4 variable situation, with no solution

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, with integer coefficients, fractional answers but with missing and other varaibles.
The next example page covers a Complex 3 equation 2 variable situation, where there is no unique 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 + 5y + 2z - 7w = 10
4x - 2y + 11z - 3w = 71
5x - 6y - 3z - 5w = 43
x + 16y + 7z - 9w = -23

We are looking for the values of the variables that make all four equations true.

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

Figure 2 [A] does not have the correct dimensions for the current problem. We need to change those dimensions to be 4x5 to hold the four equations each with four coefficients and a constant. Not seen in Figure 2 is the blinking cursor, although it was over the initial 3 at the top of the screen.

Figure 3 Pressing changes the dimensions and moves us to the first element of the matrix. Note that there is a new row at the bottom of the matrix, prefilled with 0 values.

Figure 4 In Figure 4 all of the coefficient and constant values have been entered into the matrix. Only the last three columns of the matrix are visible. We can use these to verify the entry of correct values from the problem.

Figure 5 To check the values in the left columns of the matrix we use the key to move the display, as shown in Figure 5.

Figure 6 Use to exit the matrix editor and return to the main screen.

Figure 7 Use to recall the previous command. That is not the one we want, so we use again to recall the command before that.

Figure 8 Now that we have the right command, we might as well append the command to convert the answer to fractional form just in case we need it. Press to get to the MATH menu.

Figure 9 Press to select the highlighted command.

Figure 10 With the command in place, press to perform the command.

Figure 11 The answer, or at least the left part of the answer matrix, appears in Figure 11. The bottom row gives us cause for concern. All four of the values there are 0. We need to inspect the right portion of the answer to get the resst of the information.

Figure 12 Use the key to see the right portion, as in FIgure 12. Here we see that we have 0 values all the way across the last row. If we translate that last row back into the equation form we would have 0x + 0y + 0z +0w = 0 Therefore, we conclude that there are an infinite number of solutions to this system of linear equations.

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, with integer coefficients, fractional answers.
The next example page covers a Complex 3 equation 2 variable situation, wihere there is no unique solution.

Figure 1	Open the matrix menu and move to the editor. We will use [A] for this problem.
Figure 2	[A] does not have the correct dimensions for the current problem. We need to change those dimensions to be 4x5 to hold the four equations each with four coefficients and a constant. Not seen in Figure 2 is the blinking cursor, although it was over the initial 3 at the top of the screen.
Figure 3	Pressing changes the dimensions and moves us to the first element of the matrix. Note that there is a new row at the bottom of the matrix, prefilled with 0 values.
Figure 4	In Figure 4 all of the coefficient and constant values have been entered into the matrix. Only the last three columns of the matrix are visible. We can use these to verify the entry of correct values from the problem.
Figure 5	To check the values in the left columns of the matrix we use the key to move the display, as shown in Figure 5.
Figure 6	Use to exit the matrix editor and return to the main screen.
Figure 7	Use to recall the previous command. That is not the one we want, so we use again to recall the command before that.
Figure 8	Now that we have the right command, we might as well append the command to convert the answer to fractional form just in case we need it. Press to get to the MATH menu.
Figure 9	Press to select the highlighted command.
Figure 10	With the command in place, press to perform the command.
Figure 11	The answer, or at least the left part of the answer matrix, appears in Figure 11. The bottom row gives us cause for concern. All four of the values there are 0. We need to inspect the right portion of the answer to get the resst of the information.
Figure 12	Use the key to see the right portion, as in FIgure 12. Here we see that we have 0 values all the way across the last row. If we translate that last row back into the equation form we would have 0x + 0y + 0z +0w = 0 Therefore, we conclude that there are an infinite number of solutions to this system of linear equations.