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

Step by step

The question

Question: find the inverse of A.

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

So D = 5 × 2 − 2 × 3.

D = 5 × 2 − 2 × 3

5 times 2 is 10.

2 times 3 is 6.

D = 10 − 6

10 minus 6 is 4.

So D = 4.

D = 4

2 Swap and change signs

The top left and bottom right swap places: 5 and 2.

The other two stay where they are, but change their signs: 3 becomes −3, 2 becomes −2.

Then we divide by D.

Dividing by 4 is the same as times 1/4.

A⁻¹ = 14 × [2, −3; −2, 5]

Careful!

Only the other two change sign.

5 and 2 just swap places.

Careful: 5 and 2 swap, keep their signs

3 Divide by the determinant

Every entry over 4.

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

Then we make each one simpler.

The answer: rows 1/2, −3/4 and −1/2, 5/4.

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

4 Check

Let's check: A times the swapped matrix.

It must give 4s down the diagonal, 0s elsewhere.

= [5, 3; 2, 2] × [2, −3; −2, 5]

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

We work out every times.

= [10 + (−6), −15 + 15; 4 + (−4), −6 + 10]

Then we add: 4 down the diagonal, 0 everywhere else.

= [4, 0; 0, 4]

Divided by 4, 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