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

Step by step

The question

Question: find the inverse of A.

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

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

Then every diagonal has three numbers.

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

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

0, 0, 1.

Together: 1.

= 0 + 0 + 1 = 1

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

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

0, 0, 0.

Together: 0.

= 0 + 0 + 0 = 0

Down minus up: 1 minus 0 is 1.

So D = 1.

D = 1 − 0 = 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 − −1×0, −1×0 − 0×0, −1×(−1) − 0×1; −2×0 − −1×1, 3×0 − 0×1, 3×(−1) − 0×(−2); −2×0 − 1×1, 3×0 − −1×1, 3×1 − −1×(−2)]

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

= [0, 0, 1; 1, 0, −3; −1, 1, 1]

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, 0, 1, −1, 0, 3 and −1, −1, 1.

The purple ones changed sign.

C = [0, 0, 1; −1, 0, 3; −1, −1, 1]

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 1/1.

A⁻¹ = 11 × [0, −1, −1; 0, 0, −1; 1, 3, 1]

5 Divide by the determinant

Dividing by 1 changes nothing.

So the answer is rows 0, −1, −1, 0, 0, −1 and 1, 3, 1.

A⁻¹ = [0, −1, −1; 0, 0, −1; 1, 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 × 0 + 1 × 1 = 1

Row 1 times column 2: 0.

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