# Adding a Multiple of One Row to Another Doesn't Change Determinant ... Geometric?

• January 3rd 2013, 05:57 PM
zhandele
Adding a Multiple of One Row to Another Doesn't Change Determinant ... Geometric?
If I add a multiple of one row of a square matrix to another row, which is the most common elementary row operation, I don't change the determinant thereby. I know two proofs of this and I have no problem with it. But what does it mean geometrically? I'm going for a more intuitive understanding.

I generated a random 2x2 matrix

$\left(\begin{array}{cc}-4&7\\8&10\end{array}\right)$

then added the first row to the second to get

$\left(\begin{array}{cc}-4&7\\4&17\end{array}\right)$

Then I modeled these two pairs of vectors, and the determinant-parallelograms they generate, on a Geogebra screen (see attachments). I know the two parallelograms should have the same area, and they do, but it's not obvious why. The change in one vector is "undone" by the change in the other, it seems.

I feel I must've missed something. I hope somebody can suggest something.
• January 3rd 2013, 11:14 PM
Deveno
Re: Adding a Multiple of One Row to Another Doesn't Change Determinant ... Geometric?
to see why this is, let's look at a matrix that effects this change. suppose you add row 1 to row 2. this matrix that does this (via left-multiplication) is:

$P = \begin{bmatrix}1&0\\1&1 \end{bmatrix}$

this is a "shear matrix", it sends squares in the plane to rhombuses (which have the same area as the squares).

if we replace the 1 in the 2,1-position with an arbitrary scaling factor r, we just change the "slant" of the resulting rhombus, which doesn't affect the area (as it still has the same height perpendicular to its base).

surely you can see that det(PA) = det(P)det(A) = (1)(det(A)) = det(A).