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

Step by step

The question

Question: find the inverse of A.

A = [2, 1; 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 = |2, 1; 4, 3|

So D = 2 × 3 − 4 × 1.

D = 2 × 3 − 4 × 1

2 times 3 is 6.

4 times 1 is 4.

D = 6 − 4

6 minus 4 is 2.

So D = 2.

D = 2

2 Swap and change signs

The top left and bottom right swap places: 2 and 3.

The other two stay where they are, but change their signs: 1 becomes −1, 4 becomes −4.

Then we divide by D.

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

A⁻¹ = 12 × [3, −1; −4, 2]

Careful!

Only the other two change sign.

2 and 3 just swap places.

Careful: 2 and 3 swap, keep their signs

3 Divide by the determinant

Every entry over 2.

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

Then we make each one simpler.

The answer: rows 3/2, −1/2 and −2, 1.

A⁻¹ = [3 over 2, −1 over 2; −2, 1]

4 Check

Let's check: A times the swapped matrix.

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

= [2, 1; 4, 3] × [3, −1; −4, 2]

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

We work out every times.

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

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