Question

How do you find the inverse of `A=``((4, -2, 1), (3, 1, 2), (1, 2, 2))`?

Answer 1

You can see it in the picture!

Explanation:

I guess you can follow.

Answer 2

I prefer to append an identity matrix to the right, then use elementary row operations to reduce the original matrix to an identity matrix; when this done, the matrix on the is the inverse.

Explanation:

Given:

#[ (4, -2, 1), (3, 1, 2), (1, 2, 2) ]#

Append an identity matrix:

#[ (4, -2, 1,|,1,0,0), (3, 1, 2,|,0,1,0), (1, 2, 2,|,0,0,1) ]#

Perform elementary row operations until the matrix on the left is an identity matrix.

`R_1harrR_3`

#[ (1, 2, 2,|,0,0,1), (3, 1, 2,|,0,1,0), (4, -2, 1,|,1,0,0) ]#

`R_2-3R_1toR_3`

#[ (1, 2, 2,|,0,0,1), (0, -5, -4,|,0,1,-3), (4, -2, 1,|,1,0,0) ]#

`R_3-4R_1toR_3`

#[ (1, 2, 2,|,0,0,1), (0, -5, -4,|,0,1,-3), (0, -10, -7,|,1,0,-4) ]#

`R_3-2R_2toR_3`

#[ (1, 2, 2,|,0,0,1), (0, -5, -4,|,0,1,-3), (0, 0, 1,|,1,-2,2) ]#

`R_2+4R_3 to R_2`

#[ (1, 2, 2,|,0,0,1), (0, -5, 0,|,4,-7,5), (0, 0, 1,|,1,-2,2) ]#

`R_1-2R_3toR_1`

#[ (1, 2, 0,|,-2,4,-3), (0, -5, 0,|,4,-7,5), (0, 0, 1,|,1,-2,2) ]#

`-1/5R_2toR_2`

#[ (1, 2, 0,|,-2,4,-3), (0, 1, 0,|,-4/5,7/5,-1), (0, 0, 1,|,1,-2,2) ]#

`R_1 - 2R_2 to R_1`

#[ (1, 0, 0,|,-2/5,6/5,-1), (0, 1, 0,|,-4/5,7/5,-1), (0, 0, 1,|,1,-2,2) ]#

The inverse matrix is on the right:

#[ (-2/5,6/5,-1), (-4/5,7/5,-1), (1,-2,2) ]#