Inverse of [2, -1; 3, 1] Step by Step | Inverse of a 2 × 2 matrix

Step by step

The question

Question: find the inverse of A.

A = [2, −1; 3, 1]

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 = |2, −1; 3, 1|

So D = 2 × 1 − 3 × (−1).

D = 2 × 1 − 3 × (−1)

2 times 1 is 2.

3 times (−1) is −3.

D = 2 − (−3)

Taking away a negative is adding: 2 + 3.

2 − (−3) = 2 + 3

2 minus (−3) is 5.

So D = 5.

D = 5

2 Swap and change signs

The top left and bottom right swap places: 2 and 1.

The other two stay where they are, but change their signs: −1 becomes 1, 3 becomes −3.

Then we divide by D.

Dividing by 5 is the same as times 1/5.

A⁻¹ = 15 × [1, 1; −3, 2]

Careful!

Only the other two change sign.

2 and 1 just swap places.

Careful: 2 and 1 swap, keep their signs

3 Divide by the determinant

Every entry over 5.

They are as simple as they can be.

The answer: rows 1/5, 1/5 and −3/5, 2/5.

A⁻¹ = [1 over 5, 1 over 5; −3 over 5, 2 over 5]

4 Check

Let's check: A times the swapped matrix.

It must give 5s down the diagonal, 0s elsewhere.

= [2, −1; 3, 1] × [1, 1; −3, 2]

= [2 × 1 + −1 × (−3), 2 × 1 + −1 × 2; 3 × 1 + 1 × (−3), 3 × 1 + 1 × 2]

We work out every times.

= [2 + 3, 2 + (−2); 3 + (−3), 3 + 2]

Then we add: 5 down the diagonal, 0 everywhere else.

= [5, 0; 0, 5]

Divided by 5, 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