Given an n-by-n matrix A, form the augmented matrix A bar I (i.e. you have the n-by-n identity matrix on the right hand side of the bar). Apply row operations to this augmented matrix until you obtain an augmented matrix B bar C with B in reduced echelon form.
If B = I then A is invertible with inverse C.
If B is not the identity then A is not invertible.
So if A is invertible, you start with A bar I and you end up with I bar A inverse. We're going to postpone the proof for now, and work out some examples before developing the theory necessary for the proof.
Let A be the 2-by -2 matrix 1, minus 1; 1, 1. The augmented matrix is 1, minus 1, bar, 1, 0; 1, 1, bar, 0, 1. To put the left-hand block into reduced echelon form, we do R_2 maps to R_2 minus R_1: 1, minus 1, bar, 1, 0; 0, 2, bar, minus 1, 1 then we do R_2 maps to a half R_2 to get: 1, 1, bar, 1, 0; 0, 1, bar, minus a half, a half. Next we do R_1 maps to R_1 + R_2, which gives: 1, 0, bar, a half, a half' 0, 1, bar, minus a half, a half Now the left-hand block is in reduced echelon form; it is the identity matrix, so A is invertible and its inverse is the right-hand block: a half times the matrix 1, 1; minus 1, 1 You can check this agrees with the 2-by-2 inverse we talked about last time.
Here's a messy example. Let's invert the 3-by-3 matrix A = 1, 2, 3; 0, 1, 4; 1, 5, 0. We have A bar I equals 1, 2, 3, bar, 1, 0, 0; 0, 1, 4, bar, 0, 1, 0; 1, 5, 0, bar, 0, 0, 1. Perform R_3 maps to R_3 minus R_1 to clear the first column: 1 , 2 , 3 , bar , 1 , 0 , 0 ; 0 , 1 , 4 , bar , 0 , 1 , 0 ; 0 , 3 , minus 3 , bar , minus 1 , 0 , 1 Next, R_3 maps to R_3 minus 3 R_2 gets us into echelon form: 1 , 2 , 3 , bar , 1 , 0 , 0 ; 0 , 1 , 4 , bar , 0 , 1 , 0 ; 0 , 0 , minus 15 , bar , minus 1 , minus 3 , 1. Now R_3 maps to minus R_3 over 15 gives: 1 , 2 , 3 , bar , 1 , 0 , 0 ; 0 , 1 , 4 , bar , 0 , 1 , 0 ; 0 , 0 , 1 , bar , 1 over 15 , 1 over 5 , minus 1 over 15. Clear the last column with R_2 maps to R_2 minus 4 R_3 and R_1 maps to R_1 minus 3 R_3 to get: 1 , 2 , 0 , bar , 4 over 5 , minus 3 over 5 , 1 over 5 ; 0 , 1 , 0 , bar , minus 4 over 15 , 1 over 5 , 4 over 15 ; 0 , 0 , 1 , bar , 1 over 15 , 1 over 5 , minus 1 over 15. Finally, R_1 maps to R_1 minus 2 R_2 yields: 1 , 0 , 0 , bar , 4 over 3 , minus 1 , minus 1 over 3 ; 0 , 1 , 0 , bar , minus 4 over 15 , 1 over 5 , 4 over 15 ; 0 , 0 , 1 , bar , 1 over 15 , 1 over 5 , minus 1 over 15. The left-hand block is the identity matrix, so A is invertible and the right-hand block is A inverse.
This means A inverse equals 4 over 3 , minus 1 , minus 1 over 3 ; minus 4 over 15 , 1 over 5 , 4 over 15 ; 1 over 15 , 1 over 5 , minus 1 over 15. We can check this by computing A inverse A: we get I.