![]() | Open the matrix menu and move to the editor. We will use [A] for this problem. |
![]() | [A] already has the correct dimensions for the current problem, namely 3 rows and 4 columns. |
![]() | 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 |
![]() | 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 ![]() |
![]() | 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. |
![]() | For Figure 6 we have merely used the ![]() ![]() ![]() |
![]() | 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 ![]() ![]() |
![]() | The command has been recalled, we press ![]() |
![]() | 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 ![]() |
![]() | In the MATH menu, the first item, ![]() ![]() |
![]() | Because there was nothing before the pasted ![]() ![]() |
![]() | Figure 12 shows the fractional form of the display. We could reconstruct the transformed
equations as 0x + 1y + 0z = –1/3 0x + 0y + 1z = 1/7 |
©Roger M. Palay
Saline, MI 48176
November, 2010