1. Prove Cases 2 and 3 in the Theorem about the product EA

Case 2.

Suppose that E is obtained by multiplying the i-th row of Im by a non-zero number, x. Then each row ra except ri has a leading 1 on the (a,a) place and zeros everywhere else. The row ri has the non-zero number x on the (i,i)-place and zeros everywhere else. If a is not equal to i then the element EA(a,b), the product of ra and cb, will be equal to the a-th entry of cb, which is A(a,b). Therefore EA(a,b)=A(a,b) whenever a is not equal to i. Thus all rows of EA except for the i-th row coincide with the corresponding rows of A. The i-th row is the only one to change. The entries of this row are products of ri and cb (b=1,...,n). The i-th row of E has x on the i-th place and zeros everywhere else So the product of ri and cb is x times the i-th element of cb. Thus EA(i,b)=xA(i,b) for every b=1,2,...,n. Therefore the i-th row of EA is obtained by multiplying the i-th row of A by x. This is the same operation we used to obtain E from i. The proof is complete.

Case 3.

Suppose that E is obtained by swapping the i-th and j-th rows of Im. Then each row ra except ri and rj has a 1 on the (a,a) place and zeros everywhere else. ri has the 1 on the (i,j) place and zeros everywhere else. rj has the 1 on the (j,i) place and zeros everywhere else. If a is not equal to i or j, then the element EA(a,b), the product of ra and cb will be equal to the a-th entry of cb, that is A(a,b). Therefore EA(a,b) = A(a,b) whenever a is not equal to i or j and all rows but the i-th and j-th rows coincide with the corresponding rows of A. The i-th and j-th rows are the only ones to change. The entries of ri are the products of ri and cb (b = 1...n). ri has a 1 on the j-th element and zeros everywhere else. So the product of ri and cb is simply the j-th element of cb. EA(i,b) = A(j,b) for every b = 1...n. Therefore, the i-th row of EA is the j-th row of A. We can prove in the same way that the j-th row of EA is the i-th row of A. The i-th and j-th rows of A are swapped. This is the same row operation that performed to obtain E from i. The proof is complete.

2. Express the matrix
 A := [ 1 0 1 ] [ 0 1 1 ] [ 1 1 0 ]

as a product of elemetary matrices.

To do this, we first work backwards. We reduce A to the identity matrix. This takes five row operations. To get matrix A from the identity matrix, we perform the inverse operations in opposite order. These five row operations are
• add row 3 to row 1 2) add -1*row 3 to row 2
• multiply row 3 by 2
• add row 2 to row 3
• add row 1 to row 2

So we have five elementary matrices (E1, E2, E3, E4, and E5) and B, the identity matrix I3.

> B:=matrix(3,3,[[1,0,0],[0,1,0],[0,0,1]]);
 B := [ 1 0 0 ] [ 0 1 0 ] [ 0 0 1 ]

 E1 := [ 1 0 1 ] [ 0 1 0 ] [ 0 0 1 ]

 E2 := [ 1 0 0 ] [ 0 1 -1 ] [ 0 0 1 ]

> E3:=mulrow(B,3,2);
 E3 := [ 1 0 0 ] [ 0 1 0 ] [ 0 0 2 ]