Inverse of [1, 2; 3, 4] Step by Step | Inverse of a 2 × 2 matrix

Step by step

The question

Question: find the inverse of A.

A = [1, 2; 3, 4]

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 = |1, 2; 3, 4|

So D = 1 × 4 − 3 × 2.

D = 1 × 4 − 3 × 2

1 times 4 is 4.

3 times 2 is 6.

D = 4 − 6

4 minus 6 is −2.

So D = −2.

D = −2

2 Swap and change signs

The top left and bottom right swap places: 1 and 4.

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 over −2.

A⁻¹ = 1−2 × [4, −2; −3, 1]

Careful!

Only the other two change sign.

1 and 4 just swap places.

Careful: 1 and 4 swap, keep their signs

3 Divide by the determinant

Every entry over −2.

The bottom is negative, so each sign changes.

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

Then we make each one simpler.

The answer: rows −2, 1 and 3/2, −1/2.

A⁻¹ = [−2, 1; 3 over 2, −1 over 2]

4 Check

Let's check: A times the swapped matrix.

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

= [1, 2; 3, 4] × [4, −2; −3, 1]

= [1 × 4 + 2 × (−3), 1 × (−2) + 2 × 1; 3 × 4 + 4 × (−3), 3 × (−2) + 4 × 1]

We work out every times.

= [4 + (−6), −2 + 2; 12 + (−12), −6 + 4]

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