Can we multiply a column in a matrix by an integer?

Question

Sorry if this question is too basic.

We can suppose that we have a matrix that reduces to the identity matrix in reduced row-echelon form. My question is fairly simple: Can we multiply one or more columns by a constant and still be able to reduce the matrix to the identity?

I'd like a proof of this, but for some reason the proof eludes me.

My ideas for a proof I am thinking that we can start by examining the determinant of the matrix. Since we are multiplying a column by a constant, the determinant is simply multiplied by a constant. Then we can use Cramer's rule to finish off the proof.

Answer 1

So long is the integer is non-zero, you certainly can. This corresponds to right-multiplication by a diagonal matrix with ones along the diagonal, except in the column that you want to multiply by an integer, which should have that integer in it instead of a one. Such a matrix is invertible (so long as the integer is non-zero), and given two square matrices $A,B$ of the same size with $B$ invertible, we have that $A$ is invertible (row-reducible to the identity matrix) if and only if $AB$ is.

Added: You can also use a determinant approach with my proof sketch, since if $k$ is the integer and $B$ is a diagonal matrix as described above, then $\det(B)=k$ and $$\det(AB)=\det(A)\det(B)=k\det(A).$$ Clearly, if $k\neq0,$ then $\det(AB)\neq0$ if and only if $\det(A)\neq0.$

Answer 2

The number of non-zero rows in reduced row-echolon form is the row rank of the matrix. Now the row rank equals the column rank, and multiplication of a column by a non-zero number does not change the column rank.