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

Step by step

The question

Question: find the inverse of A.

A = [3, 0, 2; 2, 1, 3; −2, 1, 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, 0, 2; 2, 1, 3; −2, 1, 0|3, 0; 2, 1; −2, 1

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

Then every diagonal has three numbers.

Down: 3 × 1 × 0, 0 × 3 × (−2), 2 × 2 × 1.

down = 3 × 1 × 0 + 0 × 3 × (−2) + 2 × 2 × 1

0, 0, 4.

Together: 4.

= 0 + 0 + 4 = 4

Up: −2 × 1 × 2, 1 × 3 × 3, 0 × 2 × 0.

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

−4, 9, 0.

Together: 5.

= −4 + 9 + 0 = 5

Down minus up: 4 minus 5 is −1.

So D = −1.

D = 4 − 5 = −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.

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

Work out each one: −3, 6, 4; −2, 4, 3; −2, 5, 3.

= [−3, 6, 4; −2, 4, 3; −2, 5, 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 −3, −6, 4, 2, 4, −3 and −2, −5, 3.

The purple ones changed sign.

C = [−3, −6, 4; 2, 4, −3; −2, −5, 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 over −1.

A⁻¹ = 1−1 × [−3, 2, −2; −6, 4, −5; 4, −3, 3]

5 Divide by the determinant

Dividing by −1 changes every sign.

So the answer is rows 3, −2, 2, 6, −4, 5 and −4, 3, −3.

A⁻¹ = [3, −2, 2; 6, −4, 5; −4, 3, −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 × column 1: 3 × 3 + 0 × 6 + 2 × (−4) = 1

Row 1 times column 2: 0.

row 1 × column 2: 3 × (−2) + 0 × (−4) + 2 × 3 = 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