Inverse of [3, 5; 1, 2] Step by Step | Inverse of a 2 × 2 matrix

Step by step

The question

Question: find the inverse of A.

A = [3, 5; 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 = |3, 5; 1, 2|

So D = 3 × 2 − 1 × 5.

D = 3 × 2 − 1 × 5

3 times 2 is 6.

1 times 5 is 5.

D = 6 − 5

6 minus 5 is 1.

So D = 1.

D = 1

2 Swap and change signs

The top left and bottom right swap places: 3 and 2.

The other two stay where they are, but change their signs: 5 becomes −5, 1 becomes −1.

Then we divide by D.

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

A⁻¹ = 11 × [2, −5; −1, 3]

Careful!

Only the other two change sign.

3 and 2 just swap places.

Careful: 3 and 2 swap, keep their signs

3 Divide by the determinant

Dividing by 1 changes nothing.

So the answer is rows 2, −5 and −1, 3.

A⁻¹ = [2, −5; −1, 3]

4 Check

Let's check: A times the swapped matrix.

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

= [3, 5; 1, 2] × [2, −5; −1, 3]

= [3 × 2 + 5 × (−1), 3 × (−5) + 5 × 3; 1 × 2 + 2 × (−1), 1 × (−5) + 2 × 3]

We work out every times.

= [6 + (−5), −15 + 15; 2 + (−2), −5 + 6]

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