##### 33A Practice Midterm #3
emilywang
To get the inverse matrix, compute rref [A|I] rref [A|I] = [I|\$A^{-1}] Row reduce: 1 1 3 | 1 0 0 1 0 0 | 2/3 -1/2 1/3 2 2 4 | 0 1 0 --> 0 1 0 | -8/3 2 -1/3 4 1 0 | 0 0 1 0 0 1 | 1 -1/2 0
emilywang: May 4, 2015, 9:44 p.m.
Sorry about the weird formatting! // rref [A|I] = [I|B] where B is the inverse of A. // Let I = row 1: 1 0 0, row 2: 0 1 0, row 3: 0 0 1. Then the inverse of A becomes: row 1: 2/3, -1/2, 1/3. row 2: -8/3, 2, -1/3. row 3: 1, -1/2, 0
syp881122: May 6, 2015, 4:33 p.m.
I use the fomula: A inverse= 1/deat(A)*adj(A) det(A)=1*(0-4)-1*(0-16)+3*(2-8)=-6 then A transfer ={ 1 2 4} { 1 2 1} { 3 4 0} A= -4 3 -2 16 -12 2 -6 3 0 Then use A divide by det(A) I got same answer.