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

Step by step

The question

Question: find the inverse of A.

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

So D = 5 × 1 − 2 × 2.

D = 5 × 1 − 2 × 2

5 times 1 is 5.

2 times 2 is 4.

D = 5 − 4

5 minus 4 is 1.

So D = 1.

D = 1

2 Swap and change signs

The top left and bottom right swap places: 5 and 1.

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

Then we divide by D.

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

A⁻¹ = 11 × [1, −2; −2, 5]

Careful!

Only the other two change sign.

5 and 1 just swap places.

Careful: 5 and 1 swap, keep their signs

3 Divide by the determinant

Dividing by 1 changes nothing.

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

A⁻¹ = [1, −2; −2, 5]

4 Check

Let's check: A times the swapped matrix.

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

= [5, 2; 2, 1] × [1, −2; −2, 5]

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

We work out every times.

= [5 + (−4), −10 + 10; 2 + (−2), −4 + 5]

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