Row operations and matrix determinants

Question

I am stuck on a particular point about using row operations to simplify calculating the determinant of a matrix. In particular, that adding some scalar multiple of the other rows to a particualr row leaves the determinant unchanged, and multiplying any row by some scalar multiplied the determinant by that same scalar.

This seems to lead me to a bit of a problem. Consider the matrix below (apologies that this is actually determinant with the straight lines. I am struggling with matrices in LaTeX):

$$ \begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\\ \end{vmatrix} $$

Using row operations we can say

$ \begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\\ \end{vmatrix} \equiv \begin{vmatrix} a_{11}+a_{21}&a_{12}+a_{22}&a_{13}+a_{33}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\\ \end{vmatrix}$

But then considering the second matrix as a completely new case, and ignoring the fact that it was derived from the first matrix, I cannot see anything wrong with now subtracting twice the first row from the second

$\begin{vmatrix} a_{11}+a_{21}&a_{12}+a_{22}&a_{13}+a_{33}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\\ \end{vmatrix} \equiv \begin{vmatrix} a_{11}+a_{21}&a_{12}+a_{22}&a_{13}+a_{33}\\ -2a_{11}-a_{21}&-2a_{12}-a_{22}&-2a_{13}-a_{33}\\ a_{31}&a_{32}&a_{33}\\ \end{vmatrix}$

Except now comparing this to the first matrix, we have multiplied the second row by a scalar (-1) which should change the determinant as well as adding just a scalar multiple of row 1 to row 2, but then we also added a scalar multiple of row 2 to row 1? So i'm not sure if the effect cancels out somewhere...

Would appreciate a pointer as to why the above is ok (or not!)

Answer 1

Your calculation is correct, but you have a small misunderstanding about the effect of a general row operation.

Call the original matrix $A$. Then its determinant is $det(A)$

Multiplying a row by $k$ makes the determinant $k*det(A)$. Switching adjacent rows makes it $-det(A)$. The determinant of the identity matrix is $1$. These three rules completely determine the function that maps $n \times n$ matrices into the real numbers. We call that function the determinant.

You can use these three properties to show that adding two rows does not change the determinant. But doubling the first row and adding in to the second row doubles the determinant.

Bottom line: adding rows does nothing to the determinant. But general row operations do change the determinant.

Answer 2

Every elementary row operation on a matrix is equivalent to left-multiplying the matrix by a particular nonsingular matrix. The determinant of a product of matrices is equal to the product of their determinants, so the effect of an elementary row operation on the determinant of a matrix is to multiply it by some number.

When you multiply a row by some scalar $\lambda$, that’s the same as multiplying the matrix by a diagonal matrix with $\lambda$ in the corresponding row and $1$s everywhere else along the diagonal. It should be obvious that the determinant of this diagonal matrix is $\lambda$, and so this operation multiplies the determinant of the original matrix by $\lambda$.

When you add a $\lambda$ times row $j$ to row $i$, the corresponding elementary matrix is the identity matrix with its $(i,j)$-th element set to $\lambda$. (Verify for yourself that this is true.) The determinant of such a matrix is $1$, so this operation leaves the determinant unchanged.

The two operations you’ve performed can be written in matrix form as $$\pmatrix{1&0&0\\-2&1&0\\0&0&1}\pmatrix{1&1&0\\0&1&0\\0&0&1}\pmatrix{a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}}.$$ You can easily verify that the first two matrices have determinant $1$, so the entire product has the same determinant as the original matrix. Now, if you’d multiplied the first row in place, however, that would be a different story.

There’s a geometric explanation for these effects of row operations that might be helpful in understanding what’s going on. The determinant of a matrix can be viewed as the (signed) volume of a paralleletope with sides defined by the rows of the matrix interpreted as vectors. Multiplying a row by a scalar is equivalent to changing the length of one set of parallel sides, so it’s natural that this changes the volume proportionally. On the other hand, adding a scalar multiple of one row to another shifts a vertex in a direction parallel to another edge. Now, what matters for computing the volume is the perpendicular distance of the vertex to that edge, but that isn’t changed by a parallel shift, so the volume remains unchanged. Try playing around with some parallelograms to see this in action.