Inverse of [6, 2; 4, 3] Step by Step | Inverse of a 2 × 2 matrix

Step by step

The question

Question: find the inverse of A.

A = [6, 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 = |6, 2; 4, 3|

So D = 6 × 3 − 4 × 2.

D = 6 × 3 − 4 × 2

6 times 3 is 18.

4 times 2 is 8.

D = 18 − 8

18 minus 8 is 10.

So D = 10.

D = 10

2 Swap and change signs

The top left and bottom right swap places: 6 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 10 is the same as times 1/10.

A⁻¹ = 110 × [3, −2; −4, 6]

Careful!

Only the other two change sign.

6 and 3 just swap places.

Careful: 6 and 3 swap, keep their signs

3 Divide by the determinant

Every entry over 10.

A⁻¹ = [3 over 10, −2 over 10; −4 over 10, 6 over 10]

Then we make each one simpler.

The answer: rows 3/10, −1/5 and −2/5, 3/5.

A⁻¹ = [3 over 10, −1 over 5; −2 over 5, 3 over 5]

4 Check

Let's check: A times the swapped matrix.

It must give 10s down the diagonal, 0s elsewhere.

= [6, 2; 4, 3] × [3, −2; −4, 6]

= [6 × 3 + 2 × (−4), 6 × (−2) + 2 × 6; 4 × 3 + 3 × (−4), 4 × (−2) + 3 × 6]

We work out every times.

= [18 + (−8), −12 + 12; 12 + (−12), −8 + 18]

Then we add: 10 down the diagonal, 0 everywhere else.

= [10, 0; 0, 10]

Divided by 10, 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