+ +  3 1 0  A =  1 1 2   0 2 4  + +Solution:
+ +  3 1 0  1 0 0  write down the augmented matrix  1 1 2  0 1 0  [ A  I ]  0 2 4  0 0 1  + + + +  1 1 2  0 1 0  Swap rows 1 and 2  3 1 0  1 0 0  (to bring a 1 to the upper left)  0 2 4  0 0 1  + + + +  1 1 2  0 1 0  Add 3 times row 1 to row 2  0 4 6  1 3 0  (to get a 0 in the first column)  0 2 4  0 0 1  + + + +  1 1 2  0 1 0  Swap second and third rows  0 2 4  0 0 1  (this will make the computation easier)  0 4 6  1 3 0  + + + +  1 1 2  0 1 0  Divide second row by 2  0 1 2  0 0 1/2  (to get a leading 1)  0 4 6  1 3 0  + + + +  1 0 0  0 1 1/2  Add 1 times row 2 to row 1  0 1 2  0 0 1/2  Add 4 times row 2 to row 3  0 0 2  1 3 2  (to get zeros in column 2) + + + +  1 0 0  0 1 1/2  Divide third row by 2  0 1 2  0 0 1/2  (to get a leading 1)  0 0 1  1/2 3/2 1  + + + +  1 0 0  0 1 1/2  Add 2 times row 3 to row 2  0 1 0  1 3 3/2  (to get a zero in column 3)  0 0 1  1/2 3/2 1  + +This has reduced the lefthand side to the identity matrix, so the righthand side is the inverse of A.
+ + 1  0 1 1/2  A =  1 3 3/2   1/2 3/2 1  + +To check the answer, we compute the product A^{1}A and see if it equals I. If it does, then we have the inverse.
+ + + + 1  0 1 1/2   3 1 0  A A =  1 3 3/2   1 1 2   1/2 3/2 1/2   0 2 4  + + + + + +  0(3)+1(1)+(1/2)(0) 0(1)+1(1)+(1/2)(2) 0(0)+1(2)+(1/2)(4)  =  1(3)+3(1)+(3/2)(0) 1(1)+3(1)+(3/2)(2) 1(0)+3(2)+(3/2)(4)   (1/2)(3)+(3/2)(1)+1(0) (1/2)(1)+(3/2)(1)+1(2) (1/2)(0)+(3/2)(2)+1(4)  + + + +  0+1+0 0+11 0+22  =  3+3+0 1+33 0+66   3/23/2+0 (1/2)(3/2)+1 03+4  + + + +  1 0 0  =  0 1 0   0 0 1  + + = ISince

