We will keep our notations in part (1). We now generalize the result proved in Problem 2.

**Problem**. Let with Prove that is a product of elementary matrices.

**Solution**. Suppose that i.e. the -entry of is The first column of cannot be all zero because then which is not true. So if then choosing we will have the matrix whose -entry is Then we can replace by and so we may assume that Now let and Then the first row and the first column entries of are all zero except for the -entry which is We can now do the same with the second row and the second column and continue this process until we get matrices and each of which is a product of elementary matrices and

is a diagonal matrix. Note that because for all (see the first part of Problem 1) and Thus by Problem 2, is a product of elementary matrices. Hence

is also a product of elementary matrices, by the second part of Problem 1.

**Definition 1**. The **general linear group** of order over a field is

**Definition 2**. The **special linear group** of order over a field is

**Remark 1**. It is clear that is a group with matrix multiplication and that is a subgroup of If, as usual, we let be the multiplicative group of the field then we can define the group homomorphism by Obviously is onto and So is a normal subgroup of and

**Remark 2**. By the first part of Problem 1, every elementary matrix is in Thus any product of elementary matrices is also in On the other hand, by the above problem, every element of is a product of elementary matrices. So we get the following important result:

as a group, is generated by all elementary matrices.