Why use elementary matrices?

Question

I just started learning linear algebra and I'm having a hard time figuring out why creating an elementary matrix to perform row operations on another matrix is necessary if we could just perform the row operations on the matrix itself. Also, which of the two methods would be more efficient?

Answer 1

It's obviously easier to just perform the elementary row operation on the matrix instead of creating a whole new matrix to represent the elementary row operation and performing a matrix-matrix multiplication. The reason linear algebra courses define elementary matrices is to help prove things about elementary row operations.

For instance, when solving a system of equations $Ax = b$ using Gauss-Jordan elimination, how do you know that performing elementary row operations won't change the set of solutions to the system of equations?

If we let $E$ be the elementary matrix corresponding to a row operation, then trivially, if $Ax = b$, then $EAx = Eb$. So any solution to the original system is still a solution after we've performed an elementary row operation. But how do we know that we haven't added any solutions? We know because the elementary matrix $E$ corresponding to each elementary row operation is invertible (check this), so $EAx = Eb$ implies that $E^{-1}EAx = E^{-1}Eb$, i.e. $Ax = b$. Hence, any solution to $EAx = Eb$ is also a solution to $Ax = b$.

Also, suppose we want to calculate the determinant of a square matrix $A$. Let's suppose we perform elementary row operations $E_1,E_2,\ldots,E_k$ that transform the matrix into an upper-triangular matrix $U$, so $E_kE_{k-1}\cdots E_2E_1A = U$. Then, since the determinant of a product of square matrices is the product of determinants, we have $\det(E_k)\det(E_{k-1})\cdots\det(E_2)\det(E_1)\det(A) = \det(U)$. The determinants of elementary row matrices are easy to compute, and since $U$ is upper-triangular, $\det(U)$ is just the product of the diagonal entries. So we know we can find the determinant of $A$ easily by row reducing it into upper-triangular form.

EDIT: While I was typing this, Qiaochu Yuan posted a comment which is a good summary of this answer.