Today I uploaded a slightly updated version of MNToolkit. To spark interest in that application, I will present how to solve systems of linear equations both stepwise and automatically using this application. Herefore we consider the following linear system:
The matrix form of that system is:
In the beginning we start a new calculation. The matrix generator pops up automatically (see the screenshot below). Since MNToolkit can only work on single matrices, we have to perform the following trick: We consider the vector on the right side of the equation as another matrix column. Doing this we get a matrix consisting of 3 rows and 4 columns. This trick is possible because all steps of the Gauss algorithm are performed on the vector, too.
Therefore we have to pick 3 rows and 4 columns within the matrix generator. After that we edit the entries of the matrix on the left side of the window by clicking on them. When all entries are done, we have the following situation:
Now we click ‘OK’. The subsequent situation is shown in the next image. I also marked the button we will use next:
Each click on that button performs a single step of the Gauss algorithm. Hence you can convert the matrix into reduced row echelon form step by step. Alternatively you can use the button directly to the left of the marked button to perform the complete conversion with one click (you will still see the intermediate steps). In the next image you see the result of clicking the (marked) single step button twice:
As shown the application also displays the matrices of the intermediate steps. Even the performed operations are documented. Herewith it should be easy to reconstruct the actions.
After a finite number of clicks on the single step button (alternatively after one click on the previously mentioned complete conversion button) the matrix is transfered into reduced row echelon form. The next picture shows the result:
Now the 3×3-submatrix on the left site is the unit matrix. Therefore the linear system has a unique solution. You can read this solution directly out of this matrix. You only have to review how we constructed the 3×4-matrix before. The first row of the resulting matrix corresponds to the equation:
When we perform the same on the two other rows, we get the only solution of the linear system as:
This matches the last column of the 3×4-matrix (hence we could have seen this directly). Inserting this solution into the equation of the initial linear system verifies this solution.
In case you would like to use this tool, here is the download link:
Download-Link: MNToolkit 0.83