Step by step
The question
Question: find the inverse of A.
The inverse A⁻¹ is the matrix that times A gives I: 1s down the diagonal, 0s elsewhere.
1 Find the determinant
First we need D, the determinant of A, by the diagonals.
We write the first two columns again, on the right.
Then every diagonal has three numbers.
Down: 1 × 3 × 1, 0 × (−2) × 0, −1 × (−2) × (−1).
3, 0, −2.
Together: 1.
Up: 0 × 3 × (−1), −1 × (−2) × 1, 1 × (−2) × 0.
0, 2, 0.
Together: 2.
Down minus up: 1 minus 2 is −1.
So D = −1.
2 Find the minors
Each entry has a minor: cover its row and its column.
A 2 by 2 is left.
Its determinant, down minus up, is the minor.
Work out each one: 1, −2, 2; −1, 1, −1; 3, −4, 3.
3 Change the signs
The signs go plus, minus, plus, like a checkerboard.
Where there is a minus, the minor changes sign.
So the cofactors are rows 1, 2, 2, 1, 1, 1 and 3, 4, 3.
The purple ones changed sign.
Careful!
Only the minus places change sign: where the row number plus the column number is odd.
4 Swap rows and columns
Row 1 of C becomes column 1, row 2 becomes column 2, row 3 becomes column 3.
Then we divide by D: times 1 over −1.
5 Divide by the determinant
Dividing by −1 changes every sign.
So the answer is rows −1, −1, −3, −2, −1, −4 and −2, −1, −3.
6 Check
Let's check: A times our answer must be I.
Row 1 of A times column 1 of the answer: 1.
Row 1 times column 2: 0.
The other entries work the same way: 1s down the diagonal, 0s elsewhere.
So the answer is right!
Let's look back.
First: find the determinant.
Then: find the minors.
Then: change the signs.
Then: swap rows and columns.
Then: divide by the determinant.
Last: check.