TI-83: Simple 3 equation 3 variable

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 missing and other varaibles.
The next example page covers a Simple 4 equation 4 variable situation, with an integer 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 here to only show the key face, as in

. In addition, the

key will be shown as

and the

key will be shown as

, again to save space.

The problem we will use on this page is 2x - 6y + z = 18
-4x + 7y + 5z = 7
3x - y + 8z = 29

We are looking for the values of the variables that make all three equations true. [Remember that these are linear equations in three variables. In Cartesian space, each equation represents a plane. There are an infinite number of points on each plane, and an infinite number of solutions to the individual equations. Two planes can cross and their intersection is a line. Any point on that line is a solution to for both equations. Three planes can intersect in a point. That point of intersection is a solution to all three equations. It is the only point that solves all three equations. We need to find that point, that x, y, and z value that solve all three equations.]

Before we start using the calculator, note that these equations are given in standard form. That is, they appear as Ax + By + Cy = D, where A, B, C, and D are numeric values. For three equations, in three unknowns (variables) x, y and z, we could write the equations in a general standard form as:

Ax + By Cz = D
Ex + Fy + Gz = H
Ix + Jy + Kz = L As we will see, the calculator uses a more general standard form for these equations, namely: a_1,1 x₁ + a_1,2 x₂ + a_1,3 x₃= b₁
a_2,1 x₁ + a_2,2 x₂ + a_2,3 x₃= b₂
a_3,1 x₁ + a_3,2 x₂ + a_3,3 x₃= b₃ This new form can be more confusing in simple cases, such as two variables and two equations, but it is more useful in complex situations, such as 7 variables and 7 equations. The key to understanding this general form is that the numbers after the a's indicate first the equation (row) and second the variable to which the numeric coefficient is attached. Thus, a_2,1 indicates that this is the number in the second equation attached to the first variable. The variables are numbered by the subscript of x, so x₁ represents the first variable and x₃ represents the third variable. The constants on the right side of the equations are numbered by the equation (row) in which they appear. Therefore, b₁ is the constant for the first equation and b₃ is the constant for the second equation.

The problem that we were given was:

2x - 6y + z = 18
-4x + 7y + 5z = 7
3x - y + 8z = 29 and we remember that the general standard form on the calculator is: a_1,1 x₁ + a_1,2 x₂ + a_1,3 x₃= b₁
a_2,1 x₁ + a_2,2 x₂ + a_2,3 x₃= b₂
a_3,1 x₁ + a_3,2 x₂ + a_3,3 x₃= b₃ so, for this problem

a_1,1 is 2	x₁ is x	a_1,2 is -6	x₂ is y	a_1,3 is 1	x₃ is z	b₁ is 18
a_2,1 is -4	x₁ is x	a_2,2 is 7	x₂ is y	a_2,3 is 5	x₃ is z	b₂ is 7
a_3,1 is 3	x₁ is x	a_3,2 is -1	x₂ is y	a_3,3 is 8	x₃ is z	b₃ is 29

Special note should be given to a_1,3 above. In the first equation, 2x - 6y + z = 18 the coefficient of the variable z is understood to be 1 and is not written. However, in the table above we specifically remember and indicate that the coefficient is the value "1". A similar situation appears in the third equation where the coefficient of the variable y is understood to be -1, accounting for the implied 1 and the subtraction. Thus, in the table above, a_3,2 is -1.

The calculator uses a matrix to hold a rectangular array of numbers. There are two ways to enter a matrix into the calculator. First, you can use the [ and ] characters to type the matrix directly into the calcualtor. That method is demonstrated in Figure 1 below. Second, the calculator has a Matrix Editor that you can use to enter values into a matrix. That method was demonstrated in the 2 variable, 2 equation page. The other Figures on this page are used to demonstrate how the calculator produces the reduced row echelon form of the given matrix.

With all of that out of the way, we are finally ready to start using the calculator. The steps shown before assume that the calculator is turned on, that we are not in any menu, and that the screen is clear.

Figure 1 In Figure 1 we enter the desired matrix directly from the calculator keyboard. To do this we use the to generate a [ to signal the start of the matrix. A second produces a [ to signal the start of a row. Then we give the coeeficients and the constant of the row, separated by commas. The keys for our example are . Then we use and to produce ][ to signal the end of the first row and the start of the second row, respectively. The values in the second row are . We use and to produce ][ to signal the end of the second row and the start of the third row, respectively. The values in the third row are . Then, we need to end the third row with a ], and end the matrix with a ]. We do this with the keys and . Finally, we want to store this matrix on the calculator. We start to store it by pressing the key.

Figure 2 On a TI-83 or a TI-83 Plus we need to store a matrix into one of the 10 predefined matrices, [A], [B], [C], [D], [E], [F], [G], [H], [I], or [J]. To do this we need to find the name of the desired matrix. The names are available in the MATRIX menu. On the TI-83 we press the key to open the MATRIX menu. (As noted at the start of this page, on the TI-83 Plus we use the sequence keys to accomplish the same thing.) Figure 2 shows the MATRIX menu. The calculator used here already has a matrix stored in [A] and another matrix stored in [C], in fact, both matrices have 2 rows and 3 columns. We will store our matrix in place of the matrix already in [A]. To do this, we can press the key to select the [A] matrix name.

Figure 3 Figure 3 shows the result of our work in Figure 2. The name [A] has been appended to the store command in our screen. We created the command that defines the matrix and assigns it to [A]. To perform that command we press the key. Performing the command stores the matrix into [A] and then displays it on the screen as This is the form that the TI-83 uses to display a matrix.

Figure 4 Our next step is to find the reduced row echelon form of this matrix. We could do this ourselves, using what are known as elementary row operations. However, the TI-83 has a function that will try to produce the reduced row echelon form of the original matrix. That command is rref(). We can find this command in the MATRIX menu. Therefore, we open that menu again, via the key, and then move to the MATH submenu via the key. Unfortunately, the rref( option does not appear on this screen. We will need to use the key to move the selector highlight down the list of options until we find the desired option, shown in Figure 4a.

Figure 4a In Figure 4a we have located the rref( function. Press to select that option and paste rref( onto the main screen.

Figure 5 The full command that we want to construct is rref([A]). In Figure 5 we have the start of this command. Now we need to append the name of our matrix. To do this we will need to open the MATRIX menu again. Press to move to Figure 6.

Figure 6 Once again, the list of matrix names is presented. Note that the dimensions for [A] have changed. [A] is already highlighted. Therefore, we can press to select [A] and move to Figure 7.

Figure 7 Once [A] has been pasted into the command, we press to complete the comamnd, as shown in Figure 7. All that remains is to press to get the calculator to perform the function and to produce the reduced row echelon form of the original matrix [A].

Figure 8 Finally, we have the reduced row echolon form of the original matrix. This new form appears as remember that this is merely a shorthand version of the three linear equations 1x + 0y + 0z = ^–2
0x + 1y + 0z = ^–3
0x + 0y + 1z = 4 and those equations tell us that x=^–2, y=^–3, and z=4. That point, the point (^–2,^–3,4), is the solution to the original problem.

Figure 1	In Figure 1 we enter the desired matrix directly from the calculator keyboard. To do this we use the to generate a [ to signal the start of the matrix. A second produces a [ to signal the start of a row. Then we give the coeeficients and the constant of the row, separated by commas. The keys for our example are . Then we use and to produce ][ to signal the end of the first row and the start of the second row, respectively. The values in the second row are . We use and to produce ][ to signal the end of the second row and the start of the third row, respectively. The values in the third row are . Then, we need to end the third row with a ], and end the matrix with a ]. We do this with the keys and . Finally, we want to store this matrix on the calculator. We start to store it by pressing the key.
Figure 2	On a TI-83 or a TI-83 Plus we need to store a matrix into one of the 10 predefined matrices, [A], [B], [C], [D], [E], [F], [G], [H], [I], or [J]. To do this we need to find the name of the desired matrix. The names are available in the MATRIX menu. On the TI-83 we press the key to open the MATRIX menu. (As noted at the start of this page, on the TI-83 Plus we use the sequence keys to accomplish the same thing.) Figure 2 shows the MATRIX menu. The calculator used here already has a matrix stored in [A] and another matrix stored in [C], in fact, both matrices have 2 rows and 3 columns. We will store our matrix in place of the matrix already in [A]. To do this, we can press the key to select the [A] matrix name.
Figure 3	Figure 3 shows the result of our work in Figure 2. The name [A] has been appended to the store command in our screen. We created the command that defines the matrix and assigns it to [A]. To perform that command we press the key. Performing the command stores the matrix into [A] and then displays it on the screen as This is the form that the TI-83 uses to display a matrix.
Figure 4	Our next step is to find the reduced row echelon form of this matrix. We could do this ourselves, using what are known as elementary row operations. However, the TI-83 has a function that will try to produce the reduced row echelon form of the original matrix. That command is rref(). We can find this command in the MATRIX menu. Therefore, we open that menu again, via the key, and then move to the MATH submenu via the key. Unfortunately, the rref( option does not appear on this screen. We will need to use the key to move the selector highlight down the list of options until we find the desired option, shown in Figure 4a.
Figure 4a	In Figure 4a we have located the rref( function. Press to select that option and paste rref( onto the main screen.
Figure 5	The full command that we want to construct is rref([A]). In Figure 5 we have the start of this command. Now we need to append the name of our matrix. To do this we will need to open the MATRIX menu again. Press to move to Figure 6.
Figure 6	Once again, the list of matrix names is presented. Note that the dimensions for [A] have changed. [A] is already highlighted. Therefore, we can press to select [A] and move to Figure 7.
Figure 7	Once [A] has been pasted into the command, we press to complete the comamnd, as shown in Figure 7. All that remains is to press to get the calculator to perform the function and to produce the reduced row echelon form of the original matrix [A].
Figure 8	Finally, we have the reduced row echolon form of the original matrix. This new form appears as remember that this is merely a shorthand version of the three linear equations 1x + 0y + 0z = ^–2 0x + 1y + 0z = ^–3 0x + 0y + 1z = 4 and those equations tell us that x=^–2, y=^–3, and z=4. That point, the point (^–2,^–3,4), is the solution to the original problem.