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

Step by step

The question

Question: find the inverse of A.

A = [3, −2; 4, −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 = |3, −2; 4, −2|

So D = 3 × (−2) − 4 × (−2).

D = 3 × (−2) − 4 × (−2)

3 times (−2) is −6.

4 times (−2) is −8.

D = −6 − (−8)

Taking away a negative is adding: −6 + 8.

−6 − (−8) = −6 + 8

−6 minus (−8) is 2.

So D = 2.

D = 2

2 Swap and change signs

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

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

Then we divide by D.

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

A⁻¹ = 12 × [−2, 2; −4, 3]

Careful!

Only the other two change sign.

3 and −2 just swap places.

Careful: 3 and −2 swap, keep their signs

3 Divide by the determinant

Every entry over 2.

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

Then we make each one simpler.

The answer: rows −1, 1 and −2, 3/2.

A⁻¹ = [−1, 1; −2, 3 over 2]

4 Check

Let's check: A times the swapped matrix.

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

= [3, −2; 4, −2] × [−2, 2; −4, 3]

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

We work out every times.

= [−6 + 8, 6 + (−6); −8 + 8, 8 + (−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