Inverse of [4, 2; 3, 2] Step by Step | Inverse of a 2 × 2 matrix

Step by step

The question

Question: find the inverse of A.

A = [4, 2; 3, 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 = |4, 2; 3, 2|

So D = 4 × 2 − 3 × 2.

D = 4 × 2 − 3 × 2

4 times 2 is 8.

3 times 2 is 6.

D = 8 − 6

8 minus 6 is 2.

So D = 2.

D = 2

2 Swap and change signs

The top left and bottom right swap places: 4 and 2.

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

Then we divide by D.

Dividing by 2 is the same as times 1/2.

A⁻¹ = 12 × [2, −2; −3, 4]

Careful!

Only the other two change sign.

4 and 2 just swap places.

Careful: 4 and 2 swap, keep their signs

3 Divide by the determinant

Every entry over 2.

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

Then we make each one simpler.

The answer: rows 1, −1 and −3/2, 2.

A⁻¹ = [1, −1; −3 over 2, 2]

4 Check

Let's check: A times the swapped matrix.

It must give 2s down the diagonal, 0s elsewhere.

= [4, 2; 3, 2] × [2, −2; −3, 4]

= [4 × 2 + 2 × (−3), 4 × (−2) + 2 × 4; 3 × 2 + 2 × (−3), 3 × (−2) + 2 × 4]

We work out every times.

= [8 + (−6), −8 + 8; 6 + (−6), −6 + 8]

Then we add: 2 down the diagonal, 0 everywhere else.

= [2, 0; 0, 2]

Divided by 2, 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