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

Mar 26, 2018

You can see it in the picture!

#### Explanation: I guess you can follow.

Mar 26, 2018

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.

${R}_{1} \leftrightarrow {R}_{3}$

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

${R}_{2} - 3 {R}_{1} \to {R}_{3}$

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

${R}_{3} - 4 {R}_{1} \to {R}_{3}$

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

${R}_{3} - 2 {R}_{2} \to {R}_{3}$

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

${R}_{2} + 4 {R}_{3} \to {R}_{2}$

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

${R}_{1} - 2 {R}_{3} \to {R}_{1}$

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

$- \frac{1}{5} {R}_{2} \to {R}_{2}$

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

${R}_{1} - 2 {R}_{2} \to {R}_{1}$

[ (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) ]