Inverse of [0, 1; -1, 2] Step by Step | Inverse of a 2 × 2 matrix

Step by step

The question

Question: find the inverse of A.

A = [0, 1; −1, 2]

The inverse of A is written A⁻¹.

A times its inverse gives I: 1s down the diagonal, 0s elsewhere.

Times I changes nothing, like times 1.

A × A⁻¹ = I, I = [1, 0; 0, 1]

1 Find the determinant

First we need D, the determinant of A.

D = |0, 1; −1, 2|

So D = 0 × 2 − (−1 × 1).

D = 0 × 2 − (−1 × 1)

0 times 2 is 0.

−1 times 1 is −1.

D = 0 − (−1)

Taking away a negative is adding: 0 + 1.

0 − (−1) = 0 + 1

0 minus (−1) is 1.

So D = 1.

D = 1

2 Swap and change signs

The top left and bottom right swap places: 0 and 2.

The other two stay where they are, but change their signs: 1 becomes −1, −1 becomes 1.

Then we divide by D.

Dividing by 1 is the same as times 1/1.

A⁻¹ = 11 × [2, −1; 1, 0]

Careful!

Only the other two change sign.

0 and 2 just swap places.

Careful: 0 and 2 swap, keep their signs

3 Divide by the determinant

Dividing by 1 changes nothing.

So the answer is rows 2, −1 and 1, 0.

A⁻¹ = [2, −1; 1, 0]

4 Check

Let's check: A times the swapped matrix.

It must give 1s down the diagonal, 0s elsewhere.

= [0, 1; −1, 2] × [2, −1; 1, 0]

= [0 × 2 + 1 × 1, 0 × (−1) + 1 × 0; −1 × 2 + 2 × 1, −1 × (−1) + 2 × 0]

We work out every times.

= [0 + 1, 0 + 0; −2 + 2, 1 + 0]

Then we add: 1 down the diagonal, 0 everywhere else.

= [1, 0; 0, 1]

Divided by 1, that is I.

So A times our answer is I: the answer is right!

A × A⁻¹ = I

Let's look back.

First: find the determinant.

Then: swap and change signs.

Then: divide by the determinant.

Last: check.

Answer