Inverse of [3, 2, 1; 0, 1, 2; 1, 0, 0] Step by Step

Step by step

The question

Question: find the inverse of A.

A = [3, 2, 1; 0, 1, 2; 1, 0, 0]

The inverse A⁻¹ is the matrix that times A gives I: 1s down the diagonal, 0s elsewhere.

A × A⁻¹ = I

1 Find the determinant

First we need D, the determinant of A, by the diagonals.

D = |3, 2, 1; 0, 1, 2; 1, 0, 0|3, 2; 0, 1; 1, 0

We write the first two columns again, on the right.

Then every diagonal has three numbers.

Down: 3 × 1 × 0, 2 × 2 × 1, 1 × 0 × 0.

down = 3 × 1 × 0 + 2 × 2 × 1 + 1 × 0 × 0

0, 4, 0.

Together: 4.

= 0 + 4 + 0 = 4

Up: 1 × 1 × 1, 0 × 2 × 3, 0 × 0 × 2.

up = 1 × 1 × 1 + 0 × 2 × 3 + 0 × 0 × 2

1, 0, 0.

Together: 1.

= 1 + 0 + 0 = 1

Down minus up: 4 minus 1 is 3.

So D = 3.

D = 4 − 1 = 3

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.

minors = [1×0 − 0×2, 0×0 − 1×2, 0×0 − 1×1; 2×0 − 0×1, 3×0 − 1×1, 3×0 − 1×2; 2×2 − 1×1, 3×2 − 0×1, 3×1 − 0×2]

Work out each one: 0, −2, −1; 0, −1, −2; 3, 6, 3.

= [0, −2, −1; 0, −1, −2; 3, 6, 3]

3 Change the signs

The signs go plus, minus, plus, like a checkerboard.

Where there is a minus, the minor changes sign.

signs: [+, −, +; −, +, −; +, −, +]

So the cofactors are rows 0, 2, −1, 0, −1, 2 and 3, −6, 3.

The purple ones changed sign.

C = [0, 2, −1; 0, −1, 2; 3, −6, 3]

Careful!

Only the minus places change sign: where the row number plus the column number is odd.

Careful: the − places: row + column 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 1/3.

A⁻¹ = 13 × [0, 0, 3; 2, −1, −6; −1, 2, 3]

5 Divide by the determinant

Every entry over 3, made simpler: rows 0, 0, 1, 2/3, −1/3, −2 and −1/3, 2/3, 1.

A⁻¹ = [0, 0, 1; 2 over 3, −1 over 3, −2; −1 over 3, 2 over 3, 1]

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 × column 1: 3 × 0 + 2 × 23 + 1 × (−13) = 1

Row 1 times column 2: 0.

row 1 × column 2: 3 × 0 + 2 × (−13) + 1 × 23 = 0

The other entries work the same way: 1s down the diagonal, 0s elsewhere.

So the answer is right!

A × A⁻¹ = I

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.

Answer