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

Step by step

The question

Question: find the inverse of A.

A = [3, 2; 4, 3]

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, 2; 4, 3|

So D = 3 × 3 − 4 × 2.

D = 3 × 3 − 4 × 2

3 times 3 is 9.

4 times 2 is 8.

D = 9 − 8

9 minus 8 is 1.

So D = 1.

D = 1

2 Swap and change signs

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

The other two stay where they are, but change their signs: 2 becomes −2, 4 becomes −4.

Then we divide by D.

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

A⁻¹ = 11 × [3, −2; −4, 3]

Careful!

Only the other two change sign.

3 and 3 just swap places.

Careful: 3 and 3 swap, keep their signs

3 Divide by the determinant

Dividing by 1 changes nothing.

So the answer is rows 3, −2 and −4, 3.

A⁻¹ = [3, −2; −4, 3]

4 Check

Let's check: A times the swapped matrix.

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

= [3, 2; 4, 3] × [3, −2; −4, 3]

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

We work out every times.

= [9 + (−8), −6 + 6; 12 + (−12), −8 + 9]

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