+- -+ | 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 left-hand side to the identity matrix, so the right-hand 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-1A 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+1-1 0+2-2 | = | -3+3+0 1+3-3 0+6-6 | | 3/2-3/2+0 -(1/2)-(3/2)+1 0-3+4 | +- -+ +- -+ | 1 0 0 | = | 0 1 0 | | 0 0 1 | +- -+ = ISince