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