# elementary matrix - single element change [closed]

I have a question on how to get matrix B from matrix A.

A =   1 2 3 4 5
6 7 8 9 10
0 0 1 0 0

B is given, too

B = 6 7 0 9 10
1 2 0 4 5
0 0 1 0 0


the tools are swapping, adding and multiplying individual lines to one another. main problem here: how to get these 0s in line 1 and 2 is not obvious for me in B.

• Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Commented Oct 13, 2019 at 12:19

First swap the first and second rows to get $$\begin{pmatrix}6 & 7 & 8 & 9 & 10 \\ 1 & 2 & 3 & 4 & 5 \\ 0 & 0 & 1 & 0 & 0 \end{pmatrix}$$

Then subtract 8 times the third row from the first row and subtract 3 times the third row from the second row.

• They probably wrote the question wrong, but note that you wouldn't be able to subtract with the operations that they wrote. Commented Oct 13, 2019 at 12:29

Use the 3rd row to clear the 3rd column except the bottom entry then swap the 1st & 2nd row, period.

If the valid operations are 'swapping, adding and multiplying individual lines to one another', then it is impossible to get the matrix $$B$$ from matrix $$A$$. note that these operations can only increase the total sum of all the elements.

However, these are not the only operations allowed by multiplication by elementary matrices. Elementary transformation allow you to add to a row, another row multiplied by a number. This operation would allow you to add to the first row, the last row multiplied by $$-1$$. This is, subtract the last row from the first row. Likewise, you can subtract the last row from the second row. Finally swap the first two rows to get $$B$$.

• multiplying 3rd row with -1. okay. I undertand the approach. But I need a -8 for cancellig out the (2,3) and a -3 cancelling out (1,3). So I would have to multiply the last row by -3 and by -8. This does not work out, once the last row is changed, I guess. Commented Oct 13, 2019 at 12:56
• @HiDuEi The operation doesn't change the row being subtracted. It only changes the row to which it is being subtracted. In other words if the rows are $R_1,R_3$ then new rows would be $R_1-8R_2, R_3$. Note how $R_3$ stays the same. Commented Oct 13, 2019 at 13:43