PROBLEM 2
(a) To row-reduce the matrix
[ 0 0 1 | 1 0 0 ]
[ 0 2 1 | 0 1 0 ]
[ 3 2 1 | 0 0 1 ]
we switch rows I and III:
[ 3 2 1 | 0 0 1 ]
[ 0 2 1 | 0 1 0 ]
[ 0 0 1 | 1 0 0 ]
and multiply row I by (1/3):
[ 1 2/3 1/3 | 0 0 1/3 ]
[ 0 2 1 | 0 1 0 ]
[ 0 0 1 | 1 0 0 ]
This fixes the first column. Next multiply row II by (1/2):
[ 1 2/3 1/3 | 0 0 1/3 ]
[ 0 1 1/2 | 0 1/2 0 ]
[ 0 0 1 | 1 0 0 ]
Because the (3,3) entry happens to equal 1, we have already obtained
1's on the diagonal and 0's below, so now we just need to get 0's
above the leading 1's. Subtract (1/2) times row III from row II:
[ 1 2/3 1/3 | 0 0 1/3 ]
[ 0 1 0 | -1/2 1/2 0 ]
[ 0 0 1 | 1 0 0 ]
subtract (1/3) times row III from row I:
[ 1 2/3 0 | -1/3 0 1/3 ]
[ 0 1 0 | -1/2 1/2 0 ]
[ 0 0 1 | 1 0 0 ]
and finally subtract (2/3) times row II from row I to obtain the rref:
[ 1 0 0 | 0 -1/3 1/3 ]
[ 0 1 0 | -1/2 1/2 0 ]
[ 0 0 1 | 1 0 0 ]
(b) The matrix we started out with was [A | I_3] where A is the 3x3 matrix
[ 0 0 1 ]
[ 0 2 1 ]
[ 3 2 1 ]
We know that if the rref of [A | I_n] is [I_n | B] then B is
the inverse of A. Therefore in our case this inverse is
[ 0 -1/3 1/3 ]
[ -1/2 1/2 0 ]
[ 1 0 0 ]
(c) When A is invertible we solve the linear system represented by Ax=b
by multiplying from the left by the inverse of A. Here this gives
[ 0 -1/3 1/3 ] [ 3 ] [ 1 ]
x = [ -1/2 1/2 0 ] [ -1 ] = [ -2 ]
[ 1 0 0 ] [ 2 ] [ 3 ]
Checking our work:
0*1 + 0*(-2) + 1*3 = 3
0*1 + 2*(-2) + 1*3 = -1
3*1 + 2*(-2) + 1*3 = 2
as required.