### Inverse

##### 4. Computation of inverse matrix

Recall that an invertible matrix A must be a square matrix such that Ax = b has a unique solution for any b. In particular, the reduced row echelon form of A must be the identity matrix. In other words, we can use row operations to change A to I. Conversely, we may use the observation to compute the inverse matrix.

Row operations change [A I] to [I B] ⇒ x is the inverse of A

The justification of the method can be found later in the proof of criterion for invertibility.

Example To find the inverse of

 A = [ 3 1 -1 ] 1 -1 1 2 2 1

we form

 [A I] = [ 3 1 -1 1 0 0 ] 1 -1 1 0 1 0 2 2 1 0 0 1

The row operations corresponding to what we have done to A in this example and this example give us

 [ 1 -1 1 0 1 0 ] 0 4 -4 1 -3 0 0 0 3 -1 1 1

Further row operations give us

 [ 1 0 0 1/4 1/4 0 ] 0 1 0 -1/12 -5/12 1/3 0 0 1 -1/3 1/3 1/3

Thus we conclude

 A-1 = [ 1/4 1/4 0 ] -1/12 -5/12 1/3 -1/3 1/3 1/3

By the way, the vector

 [ 1/4 1/4 0 ] [ 2 ] = [ 1 ] -1/12 -5/12 1/3 2 1 -1/3 1/3 1/3 6 2

is exactly the solution of the system in this example.

Exercise To find the inverse of

 A = [ 1 0 1 0 ] 1 1 0 1 0 0 1 1 1 1 1 1

we form

 [A I] = [ 1 0 1 0 1 0 0 0 ] 1 1 0 1 0 1 0 0 0 0 1 1 0 0 1 0 1 1 1 1 0 0 0 1

Row operations on the matrix give us

 [ 1 0 0 0 1 1 0 -1 ] 0 1 0 0 -1 -1 -1 2 0 0 1 0 0 -1 0 1 0 0 0 1 0 1 1 -1

Thus we have

 A-1 = [ 1 0 0 0 1 1 0 -1 ] 0 1 0 0 -1 -1 -1 2 0 0 1 0 0 -1 0 1 0 0 0 1 0 1 1 -1

In terms of linear transformations, the inverse of

T(x1, x2, x3, x4) = (x1 + x3, x1 + x2 + x4, x3 + x4, x1 + x2 + x3 + x4): R4R4

is

T-1(x1, x2, x3, x4) = (x1 + x2 - x4, - x1 - x2 - x3 + 2x4, - x2 + x4, x2 + x3 - x4): R4R4