y1 = x1 + x2 + x3
y2 = x1 + 2x2 + 3x3
y3 = x1 + 4x2 + 9x3You can easily check that this map is invertible and find the inverse by row reducing the augmented matrix
[ 1 1 1 | y1 ] [ 1 0 0 | 3y1 - 5y2/2 + y3/2 ]
(1) [ 1 2 3 | y2 ] ----> [ 0 1 0 | -3y1 + 4y2 - y3 ]
[ 1 4 9 | y3 ] [ 0 0 1 | y1 - 3y2/2 + y3/2 ]The usual method for computing matrix inverses looks a bit different, but amounts to the same thing. In this method we augment with the identity matrix I (instead of with y1,y2 and y3) and row reduce as before:
[ 1 1 1 | 1 0 0 ] [ 1 0 0 | 3 -5/2 1/2 ]
(2) [ 1 2 3 | 0 1 0 ] ----> [ 0 1 0 | -3 4 -1 ]
[ 1 4 9 | 0 0 1 ] [ 0 0 1 | 1 -3/2 1/2 ]Notice that the matrix that appears on the far right in (2) is the matrix that transforms (y1,y2,y3) to the corresponding (x1,x2,x3) as we found in (1) above. These two methods are completely equivalent. Let me explain why.
In (2) we augment with three vectors simultaneously and row reduce. When we augment with (1,0,0) it gets transformed to (3,-3,1). Similarly, augmenting with (0,1,0) transforms to (-5/2, 4, -3/2) and augmenting with (0,0,1) gives (1/2,-1,1/2). This is the same as saying that
[ 1 1 1] [ 3 ] [1] [ 1 1 1 ] [-5/2] [0] [ 1 1 1 ] [1/2] [0]
[ 1 2 3] [-3 ] = [0] and [ 1 2 3 ] [ 4 ] = [1] and [ 1 2 3 ] [ -1] = [0]
[ 1 4 9] [ 1 ] [0] [ 1 4 9 ] [-3/2] [0] [ 1 4 9 ] [1/2] [1]This means that we can given (y1,y2,y3) we can always find (x1,x2,x3) so that
[ 1 1 1 ] [x1] [y1]
[ 1 2 3 ]*[x2] = [y2]
[ 1 4 9 ] [x3] [y3]That is, given (y1,y2,y3) there is exactly one such (x1,x2,x3), namely
[x1] [ 3 ] [-5/2] [1/2] [ 3 -5/2 1/2 ] [y1]
[x2] = y1*[-3 ] + y2*[ 4 ] + y3* [-1 ] = [ -3 4 -1 ]*[y2]
[x3] [ 1 ] [-3/2] [1/2] [ 1 -3/2 1/2 ]*[y3]